INTRODUCTION
TO
NUMBER THEORY
BY
TRYGVE NAGELL
---
---
Professor of l\larhemmics
l.1niversi'Y of Uppsalo
JOHN ft1'IJ..EY d- {)rlS, IIi c.. :SXW YORK
I _.... ..
. 1":-  . , .
.Ll\IQVIST & WIKSELL, STOCKHOLM

r::- I ..-
--- --
-,
I' "i ( '..
Printed 'in Su.eden
1J:?PHADA'. ' i15i · " : . : : ., :
. .,.a.. .., ..
.,,1 j.' . 'I.. .." .t_. .
A.L1IQ\"lST & WIKSELLS BOKTRYCKERI AB
" .
. "
. .
"0 ·
. "
. .. '" . .

PREFACE
N a.t.ural l1umber is the original matheulatical concept and the
1110St fundamental. Speculations about the nature and prOI)erties
of '\vhole l1umbers doubt.less constitute the oldest form of Inathe-
Dlatical thought.
It is knowl1 t.hat the SUlnerians and Babylonialls as well as the
Ancient Eyptians had a fair kl10wledge of the properties of
natural nUlnbers. But first in connection \vith the Greeks is it
possible to speak of a proper theory of l1ulnbers. Pythagoras
(circa 500 B. C.) and his pupils pursued extel1sive studies in the
field of iutegers. The first systeulatic presentation of results in
nunlber theory \\""ith proof is to be found in Euclid's lh'nl(.l1ta
(circa 300 B. C.). Alllong the later Greek luathenlaticians, Dio-
phantos (circa A. D. 350) was of the g-reatest significance in the
development of nnmber theory; six of the thirteen books of his
lriflHllefir({ have been preserved.
It is also certain that l1uluher theory has a very old tradition
in Il1dia. \vhere it tlourished during the period between A. D.
500 an d J :!OO.
\Vesterl1 Europe becalne ac<[uaintell with Greek Dlathelnat.icB
1l1i1:inly through the agency of the Arabs. But development was
slo,v, and ,ve cannot speak of an independent "7' estern theory
of nUlnbers before the seventeenth ce11tury. The French Inathe-
111atician Fero1at (lfj01-1()uf» nlay rightly' be regarded as the
father of Ulore recent llumber tbeory. Its further develolnnent
before the nineteentll century "Tas associated chiefly with the
llanles of Euler (1707-1783). J.Jag-range (173tJ-l>;13), I.Jegendre
(1752 1833) and Gauss (1777-1855). The first textboo1\: in the
theory of llulnbers ,\Tas published in 17H)3 by Legendl.e under the
title l,",""ai sur la t/uJorie de8 1201JllJres. But the really basic work
is Gauss's book Di,q'Ui8iti(jne. .....4.rithnu.fieae, ,vhich apl)eared in 1801.
'Vith that. work nUDlber theory becanle a s)"steDlatic science.
Gauss hiInself considered that it ,vas the g-reatest of all his works.
\ \  -1

6
PREFACE
Hi opInIon on the importa.nce of nUluber theo!'}' is expressed in
his remark: .'Mat.helnatics is the queen of the sciences, and the
theor..V' of nUlnbers is the queen of mathematics."
The last hundred Jears have been characterized by an intensive
development of number theory in 111anJ different. directions.
It is the aim of this book to give the reader a brief introuuction
to the most important results ill the elemel1tar)'" t.heory of numbers.
rfhe book reproduces, in tIle Inain, lectures ,,,which I ha.ve given
at the Uni,"ersity of Uppsala. It should he possible for those with
only the elemeutary college foundations of arithmetic and algebra
to read the greater part. Sections 27, i and 29 toget.her with
(hapter V anc1 VII re(l11ire a sligh.tly wider kno\vledge of algebra.
In Sections 13.. 16 and 17 and ill Chapt.er VIII SOlne sinlple
results from anal)"sis are used. 1\'lost of the exercises are not of
a rOllt.ille character but. are really intended to supplement the
theory with known and new results which are not otherwise
included in the t.ext.
I should like to express Iny warnlest thanks t.o Professor Dr.
Ernst J acobsthal and to !)r. Sven Gellel"stedt for t.Ileir valu
able help
Uppsala, Decembel- 1950.
TRYGVE NAGELI".

CONTENTS
CIIAPTER I
DIVISIBILIT\-
Scrtion
1. Divisors. . . . . . . 11
 . Remainders . . . . . . . . . . . 1 :!
3. Primes . . . . . . . . . . . . . 13
4. The fundamental theorenl. . . . . . . .. ... 1-l
n. Least common multiple and greatet COlnnlon di,""isor . 16
It il'loduls, ring's and fields . . . . . . . . . . . . 1 fJ
. Euclid's algorithm. . . . . . . . . . . . . . . . . :?l
K Relativel)' prinle nUlnbers. Euler's cp-fuI1ction. . . . . 23
9. .Arithnletical functions . . . . . . . . .. ... o
10. Diophantine equations of the first ueg'ree . . . . :?9
11. Lattice point.s nnll point lattices . . . . . . . . 3:?
1 . Irrational numbers. . . . . . . . .. ... 34
1 i}. Irrationality of the nUlnbers  and:r . .. ... 3S
1',I.eJ'l'i8rR (1-40) . . . . . . . . . . . . . . .. -!o
Pac
. . .
CIIAPTER II
ON THE DISTRIBUTION ()F PRIMES
1.
1 ).
16.
17.
18.
Some lemluata. . . . . . . . . . . . . . .
General relnarks. The sieve of Eratosthenes . . .
The functiol1 :r (,1;) . . . . . . . . . . . . .
SOlne eleluent.ary results on the distribution of priules 57
Other problems ana result,s concerning primes . . . . ')-1
. -
"*.
)l
5.+
CHAPTER III
THEOR)- OF CONGRUENCES
19. Definitions and fundalnental properties. . . . . . .
O. Residue classes anu residue systems . .
w
21. Ferulat's theorem and its generalization by Euler.
fjH
Hn
71

8
CONTENTS
Sectiun Page
2:!. Algehraic congruences and functional congruences .. 73
' _  3. L o '7 () "
Inear congruences . . . 0 0 . . 0 . .. ... I
2-1. .Algebraic congruences to a. prinle IDodulus . . . 79
25. Prime divisors of integral }Jolynonlials . . . 81
6. Algebraic congTuences to a COIllposite nlo<.lulus ... 83
7. .Algebraic congruences to a prinle-power D10dulus 5
8. NUlnerical examples of solution of alg'ebraic congruences tO
2U. Divisibility of integ'ral polynoD1ials ,\-ith l'eg'ard to a
p r i n1 e n10 d ul ns . . . . . . . 0 . . 0 . . . . . 93
30. \Vilson's theoreIl1 and its generalization. . . 99
31. Exponent of all int.eger modulo 11 . . . . ° 102
3, l\Iocluli ha,in prinlitie roots . 0 , . . 0 0 . . . . 107
3i3. r.!'he index calculus . . . . . . . .. ..... 111
34. Po 'vel' resid neSt Binonlial cong'ruences ° . ° 1 ] 5
35, Pol"nonlials representing" integers. .. ..... ° 120
3u. Thue's renlainder theorenl and its generalization by Scholz 1:!
E'J.'{'rc.i,sles (-l1-8}) ° . . . . . 0 . 0 0 . . 0 . . . . 1:!4
-l;). The roots of unity' . . . . , . . . . . .
-!-n. rrhe cyc1oton1ic polYl1onlia.l. . . . . . 0 .
-ri. Irreducibility of the cJ'clotolnic pol.noll1ial
3 ....
- 4,
3t),
39.
-!O.
-l1.
42.
-.13,
..t: -1.
CHAPTER IV
THEORY OF QUADRATIC RESIDlTES
The general ({uadratic COllg'lwuence . 0 . . . . .
Euler's criterion and Leg'endre's s.n1bol. . . . .
011 the solvability of the cong'ruenccas .{,2 == + 2 (n1o<.11 j )
4 . 1
(- u uss s en1111a . . . . . . . . . . . . . . 0 . . .
'rhe q nadratic reci pI'ocit.y la \\. . . . . . . . . . . .
Jacobi's sJ'mbol and the generalization of the reciproc-
i ty In \v . . . . . . . . . . . . . 0 .
The pri Ine diyisors of quadratic polYIlon1ials. 0 . . .
Primes, in special tt'l-ithuletical progressions
CHAPTER V
_t\.RITHMETICAL PROPERTIES OF THE ROOTS
OF UNIT)-
. . . . .
. . . . .
. . . . .
13
133
136
139
1-!-1
J-!-f>
149
153
15f)
158
1 t)()

8('(.tjoll
CONTENTS
9
])age
. The prime divisors of the c)"clot.olllic POlJl101uial. . . 1 (j.t.
49. A theorelD of Bauer on the prime divisors of certain
pol y n 0 In i al s . . . . . . . . . . . . . . . . . . . 1 68
00. On the primes of t.he form Il 11 - 1. . . . . . . . . J 70
51. Soule trigonolnetrical products. . . . . . . 173
5. A polynoDlial identit.y of Gaus . . . . . . 17+
53. The G1:11Ssian SUll1S . . . . . _. ... 177
J"'..v('rci8('," (90-122). . . . . . 180
CHAPTER VI
DIOPHANTINE E(JU:\.TIONS OF THE SECOND DEGREE
5-1. l'he represen tatioll of iutec'ers as SU1118 of integral
t"I
8f!UareS . . . . . . . . . . . . . . . . . 18
-- Baehet's theorelu 191
DO. . . . . . . . . . . .
f)t1. The Diophantine equation ./2 _ D y2 == t . . . . . . 1D5
The Diophantine . }) ') - 1 :?O1
n 4. eq ua tlOIl :r -- .'/ ti4 - - . .
5S. Tllt Diol )ha.n tine elJ uat.ion ,,2 _ I) ,.2 = (' O-t
59. Lattice poin ts 011 conics . e) 1 ')
. . . . . . . - -
HO. Rational points ill the plane and 011 conics . . . :!lH
() 1. ffhe Diophantine equation 2 l.) ') 21
a x  j /I"" - (" .... == () . .
( 'e)
)_.
li3.
tj.
'if).
6li.
H7.
oR.
fi9.
CHAPTER VII
DIOPHANTINE EQUATIONS OF HIGHER DEGREE
SOllIe Diophalltine e(!uations of the fourth degree with
three unknowns. . . . . . . . . . . . . _ . . . .
TI D . I t . t -) 4 4 - _2
le lOp Ian .llle equa 1011 _ ,I - !I ---  . . . . . . .
The qUtdratic filds K l JilL K CJ/=-:1) and K CV ;\)
The Diophantine equation 3 -:- 1}3 -j- ;3  0 anu analogous
equations . . . . . . . . . . . . . . . . . . . .
Diophantine equations of tht. i third degree ,vith an in-
finit. of solutions. . . . . . . . . . . . . . .
The Diophantine equa tiolJ .1.,; + .'1; + .:; :.-:= o. . . .
Fermat's last theorelu. . . . . . . . . . . . .
Rational points on plane algebraic eurve. iordell's
the 0 re nl . . . . . . . . . . . . . . . . . . . . .
7
3
:!36
.t.l
:?-tt3
:.?.t.
';)- 1
_D
25H

10
CONTENTS
Section Page
70. Lattice point.s on plane algebraic curves. Theorems of
Thue and Siegel . . . . . . . . . 60
E:l'erci.e.q (123- 171) . . . .. ... .. :?65
CHAPTER 'YIIJ
THE PRIl\IE NUMBER THEOREl\tI
71. Lemn1at.a on the order of D1agnitucle of SOUle finite
s 111118 . . . . . . . . . . . . . . . . . . . . . . :! 7 5
72. Lenlmata on the Mobius function and some related
functi 0 DS. . . . . . . . . . . . . . . . . . . . . :! 7 
73. FUltller lenlmata. Proof of SelbE'rg'8 formula . . . . 23
74. An elelnentary proof of the prime DUlnber theorem. . :?8fi
Exercises (172-180) . . . . .. .... .. 98
Table 0,/ prhnifit"e roots . . . . . . . . . . 300
l:'l(nda1nntal ..::ol,t£on of equations x 2 - I) .II'.!. = - 1 301
]V al1ze index . . . . . .. .... 30
Subject index. . . . . . . . . . . . . . . .. 30

CHAPTER I
DI\ISIBILIT\-
1. Divisors. -- The eleluental"Y theory of nuulbers deals prima-
l'ilJ' with the properties of tbe lJositire infe9r. or natural n-urnb('r8
1, , 3, 4, 5, . . .
In order to simplify reasoning and the mode of pl'esenta tion, it
is, ho,vever, preferable to operate ,vith the larger s"stem of all
i llteg(}1"s
0, + 1, + 2, + 3, + 4. . . .
'l'he SUDI, the difference and the product of t\\ro integers are
t.henlselves integers; but the quotiellt of two integers is in general
not an integer.
There are ouly a finite l1uluber of natural Ilulnbers which are
less than a given lluDlber. In a set of natural numbers there is
always a least l1unlber. Eit]ler al] natural llun1bers have a cer-
01
tain g'i ven propert:r, or there is a least natural 11 Ul11 her w hic]}
does not have this property.
Theorems in nUlnber theor" are OftPll proved bJ 1naflu J l1zafil'al
ineluctioll. By this is understood that the proof proceeds according'
to the folio ,,"ing' schelua:
1. By trial or in S0l11e other \aJ' ,ve are led to the hypothesis:
Every natural nUlllbel' 12 bas t.be property E.
2. '\Te show that tbp nunlber n =- 1 has the property E.
3. ,\r e aSSUlnp that. the natural nUI11ber 11Z (or, if necessary, eyery
natural nUJUbel. < 1n) lIas t.he property E.
.J" "... e prove by Ineans of this assu1l1ption that the llUl1lber ?Jl + 1
has the pro pert).. E.
U. We are tben ent.itlecl to conclude that all natura] nUDlbers
have tbe property E, and the truth of the hypothesis in st.ep
1 is estaLlislled.

12
CHAPTER I
If a, band c are three integers such that b  0 and
a -= be,
,ve say that. b is a dirisor of a, 01" that a is diz'i",'ible by h. ".. e
say also that b divides a, or t.hat a is a ,nllltijJle of u. To ascer-
t.ain whether the given b is a divisor of the given a, ,ve ha\"'e
clearly only to determine if a is anlong the (infinitely 1l1uny)
numbel's
o. + [J, +  b. + 3 b, :t.t b, . . .
The multiples of the number 2 are the el:eJi nUD1 bers; the other
in teg-ers are called ()d c/.
Every integer a, which is different fronl zero, haR clearl)- only
a finite l1umber of divisors. and these can he deternIined by a
finite nunlber of trials. The nun) berR + 1 and .i. a. the so-called
trh-ial divisors_ al\yays occur among the divisors. A di\"'isor d
of a is a proper divisor when 1 < d < li. All integ'ers are divisors
of the n UUl bel" zero.
If  is a rational number ;F- O. a possible common divisor
> 1 of the integers a and b CaJl be deternlilled (by trial) and
divided out. We can thus al,vays aSSUDle that a g'iven fraction
is irreducible.
2. Remainders. - Let a und 1J be two integers. ,)  0, al1d let
us consider the 111ultiples of b whicb are < a. Let 1, q hp the
greatest of these. The nUluber r = a - h q is t.ben clearly <:: 1') I
bllt > 0. 1 Thus we have
Theorern 1. (a aud hare intrgPf,'-: and b  0, a u.nique ittege1. q
eJ.:'isfs 8fU:h flia t
(1)
()
a = 11 q + 1-,
'rher(
() < ,. < I b I.
The lluluber r uniquely deterlnined by the division of a by 1) is
called the least nou-uegatil:e renulinder or jJl-incilJalrel1Ulinder of a
1 By 11,1 we l1Daertnnd th(a ah()l11te vulne uf 1). In w'ords, if h 0, W(,--
Ita \"e II, I = lJ; if b <: 0, \ve have I b I = - h.

DIYISIBILITY
13
lnodulo b. The remainder r is equal to zero if and only if a is
divisible by h.
If c is a real nUll1ber > 0, we denote by the sJmbol
fcl
the greatest integer < c. Thus ,ve have for example
{ l: i ()] = 3, [ . J = 0, [J ] = 1.
Assun1e that the numbers a and b in Theorem 1 are positive.
'fhen the quotient q in (1) satisfies t.he inequalities
a a
-1< q < .-,
b -b
and is therefore identical with the greatest integer < ; thus we
have
q = [l
In equation (1) we can also consider q so chosen that the
reJnainder r fulfills the condi tion
(3)
- .1. I b I < / . .......'" 1 Il) I
2 = ........  ·
This is done by letting b tJ be the n1ult.iple of b which lies near-
est G. If a lies half-way bet,veen two n1ultiples of ii, ".e choose
the g'1"eater n1ultiple; this case occurs only ,vhen b is an even
nUlnber and a is the product of  b find an odd number. The
renlainder which fulfills couditioll (3) is called the /('a8f ab..ol'llff
,.(JI,ainder q( a 1120dul() b. I t is uniquel). deterll1ined by II and b.
3. Primes. -- If the natural number '2 (>... 1) has only trivial
di,-isors_ it is aic1 to be a prhne uUluber or sin1ply a jH-inzc. All
other natural nUlubers > 1 are called rU1n}JfJsit() nUlubers. It is
clearly possible to decide ,vb ether a given number is a prin1e or
composite by a finite nUInber of trials. Among the first ten
natural nUlnbers, 2, 3, [) and 7 are primes, and -1, 6, 8, 9, 10
COD1 posi te nUlnbers.
Every natural nUluber 1l (> 1) has at least one jn.il12e di'ris()r,
i. e. a divisor which is a prime. For, the least divisor > 1 of 12

14
CJlAPTER I
must clearlJ be a prime q, and the 11umber 12 can now be ,vrittell
in the form n = q1n, where tn is a natural number.
The following theoreln was proved in Euclid's Ele'11Ze12ta (9th
book) :
l'heUl.e11Z 2. There is an ilfinity oj" 1JriJlle,s'.
Proof It is sufficient to show that, for every given prime,
there exists a prime which is greater. Let us arrange the primes
in ordel" of ascending magnitude, and let us nunlber them ac-
cordingl)9, so that we put PI = .. 1 1 2 = 3, ))3 = o. etc. If we l10W
}Jut PIJJ2'."}Ju = P, the nun1ber P + 1 is clearly not divisible by
any of the first 12 prin1es. If q denotes the least prime divisor
of P + 1, then q > pn. The theorenl is thereby proved. rrhe
method of proof, the san1e in principle as that of Euclid, also
provides a possibility fOl" determining increasingly large pl"in1es.
4. The fundamental theorem. - \Ve begin by proving a lemIua:
Let )J be a prime and a a natural number not divisible by 1).
TheIl only the follo,ving positive multiples of a are divisible by p:
(1)
a . p, a. 2 p.. a. 3 p. . . .
Assume" in particular, that a. iU is the least positive multiple
of a which is divisible b)'Y }); then clearly 1 < '}n < p. N o'v let
(l · h be an arbitrary positive ml1ltiple of a divisible by p. Ac-
cording to Theorem ], ,ve Dlay put
J. = h - nl q,
where q and r are integers, 0 < J. < 'Ill. Thus the number
ar=ah-atnq
is divisible by p. But according to the definition of the nun)-
ber 111, we Inust have 1. = 0, and h is therefore a multiple of nl.
Since a 1J is di visible by p, n must be a divisor of the prime p;
and since tn > I, ,ve must have l1Z = p. From this we conclude
t.hat every positive multiple of p divisible by a is included iu
the sequence (1).
Froln the lemma we obtain at once

DI'7ISIBILITY
15
The01'e1n .!:J. l.f the )J1"l11le p dil'ides the produ(.t a b uf fhe naf'Il'J'al
Jltl1nbers a and b, then if Haist dilide at lea8f one o.l the tlro
.lactuts a and b.
'or, if the number a were not divisible by 1), then, according
to the lemlna, the number b must be a multiple of p. Theorem 
i8 a180 to be found in Euclid's Jlcnzenta (7th book). Euclid's
proof is based. however, on the algorithu1 nanled after him. His
algorithm is gi ven in Sectioll 7.
After these preparations we no\v continue with the proof of
the ./;lndauzental theorenl, of nlunber fheor!/:
TheureHZ 4. Efel"Y 'natural 1lUl1zber n (> 1) cau be expressed as the
lJ1"()clilcf of ptinles (prinze factol"s) in the .1'o1"'l1z
(:?)
n = PI P2 . . . pr,
(r >- 1).
l'llfrt is ollly uue such ei:pres8iun as a product (de(OOl1Zposition
info prinze factoril): if the ()J"der of the factor." is }{()f taken int()
t:oJ28idel-atiou.
Proql: The first part of the t.heorem is proved b:r induction
in the following way. It is valid for the nunlber:!. Assunle
that it is valid for all natural numbers < 12. Theu it is valid
for n also. For, as \ve have seen in Section 3, Jl can be written
as a product. n = PI HI' where PI is the least prime di\"'isor of n.
But, according to the hypothesis, the uatural number 111, since
it is < 11. can be written as a product of primes in the form
III =-= })2]Ja . . . }J I ,. a,ssulniug that it is > 1. Thus expression (2) is
valid for u. The number r of prin1e factors is of course finite"
AS8ume UO\V t.hat, besides (:?L we also have a decoDl position
of Jl in to prime factors as follows,
n = '11 q2 . . . qx,
where the factors qi are primes. If ,ve no,v apply Theorem 3 to
the identity
(3)
PIP2 . 0 . pr = '11 q2 . . . q:-"

16
CHA.PTER I
\ve see that the prillle f)1 nlust divide one of the prilues })i; if
we take this prime to be 1)1, t.hen \ve must ha,?e Pt =-= Ql' On
dividing (3) by PI, we obtain the identity
})21)3 . . . })I' = fJ2 q3 . . . 'ls.
By analogous reasolling we see that })2 = Q2' Continuing in this
way, we have final1y that r = sand tJlat the nUDlbers Ql' Q2'
. . .. qr coincide \\"ith the n UI11 bel's PI, P2, . . ., }Jr, disregardillg
the order. The second part of tIle theorem is thereby proved.
Fronl the first part of the proof it is easy t.o deduce how the
prilne factors of a gi,"'en uUlnber call be deternlined and the
number expressed in the fornl (:?).
Let 1)1' })2, . . ., lJ., deuote all distinct prime divisors of n; we
may then express Ii in the forDl
( -1)
l'
Ji = II pi.
i=1
where Uj is a natural nunlber \vhich depends on }); and 12.
TheorelDs in nUluber theory can often be proved by 111eanS of
lJlllltiplicatire induefioll. By this \ve nlean that the proof proceeds
according to the follo,,"illg schenlu:
1. B- trial or in SOUle ot.her way we are led to the hypothesis:
Ever1' natural nunlber n (> 1) has the property E.
:!. 'Ve show that all th primes have tJle ]1ropert.. E.
B. '\Ve assulne that the natural num bel" 1H has the propert.)1' E.
J.. 'Ve prove by l11eans of t.his aSS1tID ption that the nUII1 ber III p
has the same property, if 1J is an arbitrary prime.
D. Then, by Theorenl  lfirst part) all integers > 1 have the
property E, and the truth of the hypot.hesis ill step I is
established.
5. Least common multiple and greatest COIDD10n divisor. -
If the Il integ'ers at, a2, . . ., an are all different fronl zero, they'
have an infinity of comnlon Inultiples; e. g. one of these is the
product al a2 . . . an. Consequently there must. be a l('at".t j}()sitirr
ronZ1non 'I1utltiph J of t.he 12 numbers; it is denoted by the symbol
(1)
{Ol,02,
., Un}.

DIVISIBILiTY
17
If Ql' a2, . . ., all are Jl integers, not all zero, they have but a
finite nUlnber of conln1on divisors; the llu1l1bers + 1 a.l,vaJs occur
al110nO" these. There is a greatest conznzon dil'isor of the n 11um-

bers: it is denoted by the synlbol
(2)
(a 1, a2, . · ., an).
It is a nUluber > 1. \Ve shall also speak of the greatest com-
mon divisor of the numbers in ltll infinite set of integers.
We have the follo\ving theorenls:
TheoJ"elli :;. The lea.'lt l)ositi1.:e ronnnon nzultiple oj' the i1ltegeJ"s ai,
a 2, . . .. an i.'!J' a di 'i8or of all the C(J1nnzou. 'niultiples of tlze8e
12'll1nber..,' .
I.Jroof. The sum or the difference of two conlmon IDultiples is
itself a corou1on n1ultiple. Let nl be the least positive comnlon
multiple. If lJI is an arbitrary COlnmon multiple, by Theorenl 1
,ve can \vrite
1" = 111 - 11l q,
,vhere q and r are integers and 0 < .r < 111. Since 'In q is a com-
1110n multiple, }" is likewise. But r < 'In; hence, from the defini-
tion of In it follows that r = O. Therefore lJI is a multiple of tn.
Tlu,orenl 6". (d = (ai, a2' . . ., an), fJu-r(.1 e:I.'ist Ii intege1".'l Xl' .T2, . . .,
:i'" surh that
(3)
al Xl + {(2,7"2 + ' . + a,"£ XIl = d.
J'l'('r!l eo;;unoJl diri.t?or of tlu) inte{/('I' ai, a2, . . ., an is a di,.i..;;or
uf d -=-= ((11, a ..., all)'
Proo./ Let us consider the (infinite) set M c011sisting of all the
integers of the for01
(-1)
al ,{'II + a2 X2 + . . + lln .:t"n,
w here Xl' X2, . . ., Xn run throug'h all the integral values 0, + 1,
+ 2, etc. The sum or the difference of auy two numbers in M
is itself a, nU111ber in M. In particular, M contains all the num-
bers al, a2, . , ., an. The numbers in M clearly have the greatest
2 - 516670 Trygt'e }....agell

18
CHAPTER I
con1IDon divisor d. For d is a divisor of all thc.1se llUlnbel's, and
no number d l > dill M has this property.. since such a rnlI11 LeI'
d l ,vould be a divisor of all the nUI11 bers at, a2, . . " an. Let do
denote the least natural nUDlber in the set M. 'urtber, let. .....\r
be any number in M. "... e shall show that ,... is a Inultiple of do.
By Theorem 1 we can ,vl'ite
r == --,... - do fJ ,
wllere q and J' are integers and 0 < r < do. The nUll1ber r be-
long's to M, being the difference of two numbers in M. But,
since 'r < do.. this is only possible for .,. = o. rL'hus all nUDlbers
in M are multiples of do. Hence do is t.he greatest COD11DOl1
divisor of the lluDlbers in M, and therefore do = d. Thus the
first part of the theoreUl is proved. If ell is an)'" COUIJ110n divisor
of the nUll1bers tll. ([2' . . ., an, it follo\ys frolI1 e(lua.t.ioll (3) that
ell l11USt be a divisor of d.
Let c be a natural nl1J11ber. From TheorelDs [) and tj ,ve del'i'fe
the rules
( a I, a 2' . . ., a 1') r, = ({( 1 r, a 2 (", . . ., an (0)
and
{ {(I, a 2. . · ., an} (" = {a If, a 2 (.  . . ., an (,"}.
For two integers it is easJ t.o prove
1'heoreJJl 7. 1.1' a anfl U l11YJ natural 1211'Jlber" Iff? haf(.
(a, lJ) · {a  b} = a b.
Proof. By Theorem 5, the nUluber
ab
'Ii = { 1. t
ll, UJ
b
is an integer. Then, a must be a- diyisor of a. (a, b) ' and b a
d . · f 1_ a I a!J · ] t . 1
IVlsor 0 U. (a, lJ) · n consequence, (a, i) IS Q, C0111m0l1 IUU IP e
of a and b. Hence, by Theorem 5, {a, b J :....: fl II is a divisor of
ni
ah
( a, b) ' and therefore (a, b) is a divisor of m. On the other hand,
since the numbers

DIVISIBILITY
19
a ..__ {a, b} and  = {a, h}
'/u b lit a
are integers, the nUll1 bel'
Thus, by Theore111 G. 1n is
ouly ,yhen 11l = (a, b).
E',x'tlJJlple. If a = 12 and b = 10, ,ve have
112 is a C0J11I110n divisor of (( and h.
a tlivjsor of (a, b). But this is possible
Q.E. D.
(a, h) = 3 al1d {a, b} = 60,
and in accordance with Theorelu 7
3 . t10 = 12 . 15 .:. - 180.
6. Moduls, rings and fields. - A set of nUD1Lers is called a
nzudul ,yheu it hUrs t.he follo\ying properties:
1. 1'he set contains nt. least one nUlnber  o.
2. If the n U111 bel's II u,nd b belong' to the set their difference
a - b nlso belong"s to the set.
Eacl1 motlul contains the nUJD her O. If a Jl10dul contains the
nUlnher a, it also contains the nUluber - a, since - a = 0 - a.
If a Jnodul conta.ins the nUDlbers a al1d h. it also contains the
11Ulnbpl' a + b, since a + b = a - (- b).
Exan1ples of moduls are: 1. 'l'he set of all integers. 2. The
set of all even integers. 3. '1'ho set of all rational num bers.
4-. The set of all real n urn bers. 5. The set of all cOluplex
nUl11bers.
But, the natural llulubers obviously do not for111 a n1odul.
Let ai, U2. . . ., aI/I be any nun1 bers  O. The set of all nun1-
bers of the f01"ln
al t'l + a2 ;)"'2;- ,+ am }"m ·
w11ere )"1, .1'2, . . .. m are int.egers, forll1s a I11ouul, ,vhich 'fe de-
note bJ M(al a2,. ... am) or, Dlore briefly, by [al" U2,.. .. am].
The nun1 bel' ysteln (Ll. a2. . . ., am is called a g( J2frai htrl sy..dr-nl
of the I110dul. If a Inodul has the generat.ing systenl PI, (J2, . . ., PI
of r nUl}) bers. but no g'enernting' systen1 of ..:: lluJnbers, for . < 1\
we say that the tnodul has t.he ranla: r. The SYStClll Pl. t12, . . ., pr
then forms a ba,....is of t.he modul.

20
CllAPTER I
'Vf..) shall }n"o\"e
ThfOre})i 8. _-11//1 ,uf.Jdul M of (ratioual) inli)!/(JJ'S cOJlsj.t.... lJj" all
1iU tlfi.p 1(.,....' 0.( the '('(o.d jJo8it;,.e JlundJ(JI" in M.
Proof. Let d be the lea.st naturaJ nun1ber in M. If a IS a
nunlber in M, t.her( exist t\VO iutegers q and r such that
l' = II - d () ,
,,,'bere 0 < r < d (Theol'eUl 1)_ N O\V, it i clear that the nUll} ber
r belongs to M. But by the detinition of d this is }Jossi1Jle only
if r =- O. Thus u is a nlultjp]e of d. It. is clear that. there is no
other nUluher in M having tbe saIne propert)- as d.
The 1110uul [1 J consists of the set of all integers. If t.he nU111-
bel' 1 belong's to the Inodul M contnillill}.! only integ-erR, \\'e oL-
viousl)" ha ve M ==- [ 1].
An ilunlec1iate consecluellce of Theoreuls (j nl1d 8 is
Theore"nt Y. 1./ ill, a2, . . ., a ll arf int(Jl}(TS  0, tfe hllt.(
[ aI' ll2. · . .. a 11 J = [(a 1. a 2" . . " a /l ].
Hence, any [uodul coutaining' onl y illteg'er8 has the rank 1. Thl
.J , .. .. ...
IDOdul fl, V 2) hus the rank ::!; for the equation (a + b V 2) x = 1 ,
\vhere ll, band .i? are integ-ers, is possible only for b = 0, since
)/2 is irrational.
.
A. Inotlul is ca.lleu a. riur} ""hen it has the follo,,"ing pro})erty;
If a and 1J belong to the 1l10tlul, the product 1I b also belougs
to t.he 11lOdul.
When J) is an integer. t.he 1110dul [1, WI is a ring". This is
a.pparent frolu the l"elnt.ion
((l + h ) / D ) ((. + d) 1))  1I e + !J (1) + (a d -t- h () JI n ,
Fronl TheorelU H we obtain
Throrelll 10. 112!1 'Jnodul (:uutainil2!1 onlJI (1 4 ational) iufriy('rs i.s. a tin.q.
This result is not ""alid for Inouuls in g'eneral. Thus t.he
modul [11 2] is not a ring, since the }Jroduct. 11 2 . V2 = 2 does not
belong' to the modul; in fact, the equation 2 = V2 · f is not. pos-
sible for au)? integ'er t.

DIYISIBII..IT)....
21
r\ ring. is caIIcl [1. jl(4rl ,vhen it has the follo\t'ing propert.y:
If a and b belong to the ring', the quotient  also belongs to
the ring, })rovided b  O.
Exaluples of fields are: 1. The set of all complex nUD1bers.
2. The set of all real IHlnlbers or the r(,al.field. 3. The set of
all rat.ional number or the rational .1ip1d.
There exist rings which are not fields. Thus the set of all
integers is a ring but not a fiel<l.
Every field K contains all the rational nUll1bers. For, let II be
an r nUlnber  0 in K. Theu K contaius the nUluber l!- = 1.
. a
Thus, applying' addition and subtraction, ,ve see that all integers
belong to K. Finally, applying diyision, it is clear that all rat.ional
numbers belong to K.
Let. (J, be any nUluber ;# 0, and consider the set of all nun1bers
of the form
ao + at a + 02 a 2 + ... + llm a)lI,
ho -i- hi a + b 2 a 2 + + b,z an
where Ok, hI:- 171 and 11 are iutegers 1 in > 0 and II :> O. TlJis set
is obviously a field; we denote it by K (at Thus K (1) is the
rational field.
If J) is a rational nUluber which is not the square of a
rational nunlb(r.. thl' nuu1ber VD is irrat.ional (for the proof see
Theorenl 1 g in Section 12). K ( VD) is said. to be a quadratir
.1it-ld. EverJ nun1ber in such a field may be written in the forn1
p = a + 7) 111) where a. b, (' and tl are integers, Multiplying
r+ill / n '
nUlnerat.or and denoTuiuator by (' - d Jl n , we have
1 --
P -- 2 _ LJd 2 (u('-bdj) --- ad - bc)YD) = If + r rD,
".here u and ,. are ratiollal nUlubers.
'l'he field K ( V lJ ) is J"(.al ,,hen ]) is positive; l111a[linar,1j when
D is negative.
7. Euclid's algorithm. - Let a and III be natural l1ulubers,
a > fit. If a is not divisible by OJ  ihe principal ren1ainder a2

22
ell APTEn r
of f1 1110dulo al is a. posit-ire number < {(I. Dividing (11 hy a2,
\ve get the principal l"eu1ainder aa of ([1 lllodulo a2. If as r 0,
we Il1ay in the satne 111anner find a ne\v 111incipal relnainder
([4 < as. Repeating t.his procedure a certain nunlber of tiu1es,
we obta.in a seqnence of nccessjvelY' decreasing" integer > 0:
al > a2 > {fa > a4 > .. ,
alld we Inust final].. arrive at tt di\"'ision for \vhich the principal
remainder a,'+1 i:o; equal t.o zero. Hence we ha\e the following-
system of relations:
r a = {II '11 + a2. (J < {(2 < 111 '
III = a2 q2 + a3. O<aa<a2.
( 1 )
I . . . . . . . . . . .
a J - :.- aa -1 f)J-l -I- a., 0< ((,. < {[,-1,
I ((..-I  a.. q... fl J -.-1 =- = 0,
where Ql, fJ2, . . ., q... are positive integers.
It is easy to see that t.he nutnher il J is the g.reatest ('OUlnl011
divisor of the nutnbers a and £l1. ]'01', froln the last relation
in (1) it is clear that a", is a divisor of a.-I: from the preceding
it follows that al' is a divisor of a J '-2. C'ontinuillg in this 'T'a.y,
we finally see that a J . is a di\"'isor of both a1 and a. On the
other hand, if d is a comtnon di"f'isor of (fl and a, it is evident
fronl the first relation in (1) tl1at ,/ is a di\"isor of a2; frolH the
second relation it follows that d is a divisor of aa. Contiuuing
this argU1l1ent, \\.e finall.y see that d is a <liviRor of ll... Hence
a JI =-= (a, a 1 ).
This n1ethod for deterlninil1g thl"\ greatest COIUDIOll divisor of
two integers is called the E-uclid{}(fJ/ algoritlnn. It is g'ivel1 in
EucliJ's ]"leUleJlfli. 7th book.
EXanl)lle. For II = 288 and al =--= 158 we ba ve the following'
alg-orithnl:
:?H8 -= t u 1 + 130.
1 f)8  130. 1 + :!8,
]BO = :!8.J + 18,
:!  ==::- 1 t< . t + J 0,
t 8  : to. I + .
1 0 - -: 8. 1 + .
8  2 . .!.

DIYIIBrLTTY
23
Hence the g-reatest comnlon di,isor of :?8R and 158 is 2.
It is ea.sily seell that, in the algorithul (1), the principal re-
mainders tnay be replaced by the least absolute relnainders. In
this way the algorithnl l1lay' clearly be nluch shortened.
EXaUlj)lr. 'Ve consider once 1110re t.Ile case a = 2S8, 0] = 158.
Using the least absolute retl1ainders tIle algorithnl takes t.be form
288 = 158 . 2 - 28,
15R= 2.6-10,
28=-= 10.3- 2,
10 = 2. 5.
Thus the number of divisions in the algorithm is reduced from
se\"'en to four.
8. Relatively prime numbers. Euler's q. .function. - Let {(I- a2, . . .,
a Til be integers ha Vill g no COllnnon di risor > 1. Th en
(aI, a2, . . ., am) = 1,
and we sa.y tllat the nunlbers are ,ylatil.('l!J prinl('. The 11ulnbers
are said to be r('lafir(:lll lJrhne in pairs if (ai. Gj) = 1 for all z"
and.i, i .i.
JX{nnl)h'. The nUlllbers 8. 9. 10, I are relativelJ" prime; the
nutnbrs 5. 8, 9. ] 1 are relatively prin1e in pairs.
"'rhen (a. b) =-= 1. we say that a is )Jrhlle hJ b and vice ,.ersa.
\Vhen (a. b) = - 1 and (a, r) = 1. we clearlJ ha e (a, be) =- 1. For
(a. b r) is a divisor of both a e and be and therefore of (a l" be)
=-= ( (0, Ii) == c; further. (a.lJc) is a diisor of a and thus of
(a. c) = 1.
'Vhen II is a natural llulllher, W denote by cp (n) the nUD1ber
of natural nUlnbers < n ,vhich are prinle to u. (E'll[,'r's f{':/ilJ1C-
lioll or foti('nl of n.) For the first fie integers '"f'e find
9'(1)==1. cp(2)--=1, 9'(3)=2, p(-t.)=:?, cp(5)=-+.
'VhC1l 1) is a prime, we clearJy have
tp (p) == jJ - I,
since none of the numbers 1, :?, 3,. . p - 1 IS divisible by ]1.
'Vhen l)a is a power of the pritne jl, we find
rp(prr) = pa-1 (1 1 - 1).

24
CHAPTER I
For, between ] and pa there are the following m.ultiples of p:
}) . 1, }J. 2, J)' 3, . .. ., }J .. ])'! -1,
and no other natural number < ])a has a divisor in coronIon
with pa. which is > 1. Hence we have
rp (1)") = p" - ))a-l.
We shall establish the g-eneral result:
Thf.'01'"e11l 11. (N -lb' a natural nU111lJer 1fith till' ditfi'rcut ]Jrin1C
fartol'"S PI' j J 2, . · ., j)r.. then
(2)
r(i\}=.J.'T ( I- )( I-.! ) ." ( 1-  ) .
I J I ))2 l)r
Proof We use multiplicative inuuction (Section 4). Theorem 11
is true when .J.,-r- is the po,ver of a prilue. Suppose that it is
true for a cert.ain natural n u In ber l.'T". Then \ve shall sho,v that it
is also true for the number ..L\r 1 = ...'1). where }) is any prime. If
Q], a2  . .. ., a." denote the natural nnnIbers <}.." ,vhich are prinIe
to J.r, theu , = rp (..i'T). I t is easily seen that the }) (p (T) nnlnbers
(3) a i + 1z i\ , (i = 1, 2, . - -, "; 1z = 0, 1, 2, . . .. 1 J - 1 ),
constitute a.ll natural nUlnbers < ...''''p ,vhich are prilne to 1. For
if r is prime to ...,.,., so is the principal reIuainder of r nIodnlo ").".
If ;.,... is divisible by}), the nnInbers (3) are also prin1e to 1). 111
this case ,ve ha\e by (2)
cp (]{l) = l' rp (.\'") = "'\'1 ( 1-  ) ... ( 1 _1 ) ·
))1 }J r
Hence Theorem 11 is true for J..Tl.
Suppose next that ...V is not divisible hJ )). '}'o deterluine rp l...'''I)
we have to CODIpute how man)"" of the nUlnbers (3) are divisible
by p. The following' natural numbers < 'Tl = .J..\T j ) are DlultipJes
of 1):
( 4 ) 1 .}} , 2"]1. 3. 1 J , . . ., 1'1'.}) .
Here (l.p, ]t) = 1 if and only if (I.., ..,-r-) = 1. Thus exactly cp (l
of the nUlnbers (4) are priIl1e to .,. Hence, the number of

DIVISIBILITY
25
llluitipies of 1J anlong the numbers (3) is also = rp (N). Finally
by (2) we get
CP(N I )=lJffJ(N)-CP(:f)='\'-1 ( 1-! )( t- I ) -- ( l-! ) "
}J PI 1Jr
and Theorenl 1 1 is proved by multiplicati\Te induction.
From Theorem 11 we ilumediatel:r derive
Tizeore111 .12. /' the J2atln.al JllUnbel.S .1JI and Nate relath.cly prhnc,
then
rp (.ll ..'T) = rp (M) cp (.\T).
A direct proof of this proposition will be given in Sectio11 20.
Finally we prove
Tlu!orClll 1/J. If "1\"'" i. a natural l1UJilblW, thrll
(5)
 rp (£I) = )..-,
(I
the 8UJ12 being extended Oller nll lJositit'() dil.iso'r8 d 0.1' N.
Pro(/: Consider the sequence
(6)
1, 2, 3, . . .. J.,T - 1. Y.
If d is any positive divisor of N, t.here are in this sequence
N 1 " 1 f 1 I } b
d mu tIp es 0 (nall1e y t Ie nun1 ers
J.t
1 · rl. 2. d.. . . ... ;r. d.
\Vhich of these nUlubers }JaY'e 1 he g'reatf'st rODIIDOl1 divisor d
wit.h ....Y? Thp greatest Conll110n divjsor of kd and J.'. is d, if and
only if (,{" n = 1. There are 9' ( : ) numbers hi with this pro-
perty. Since all nUI11bers (lj) have u. great.est coronIon divisor
,vith .V, we have
 ( "}l ) _ T
'P II -.

26
CIL\PTER I
But, when d runs through all positive divisors of X, so does . ·
Hence relation (5) is proved, even without appl).ing Theoreu1s 11
a:nd 1:? 'Ve shall show in Section 9 that rrheorelll 13 lUay be
..
used for proving Theorern 11.
Exan1j)lp. ),7 = 60 has the positive divisOl.S
d = 1, 2, 3. 4, 5, 0, 10, 12, 15, 20, 30, 60.
Further
qJ(1)  1, q;(2) =.-= 1, If(3) = 2, cp(-I.):-. 2, </(5) = 4. 9'(6) = 2,
cp(IO)  -l (p(l:!) = -l r(15) = 8 q: (()) = 8, cp(HO)-=-H, (p(f30)=--16,
and ,ve get in accordance with Theoren1 13
1 + 1 + 2 +  + 4 +  + 4 - 4 + 8 + 8 +  + 1 () = (-iO.
9. Arithmetical (unctions. - A function 1(") defined for all
11atural llunlbers J1 is called an aritlzntrfiral jilllCfiol1. Examples
of such functions are: Euler's function q; (n) and 12! = 1 · :? · 3
· . . (n - 1) 11. 'Ve have already 1net arithmetica.1 functions of
several ,aria.bles such as the functions (aI' a2, . . .. an) and
{aI, a2, . . q all}. The principal relnainder of a modulo b is an
arithmetical function of two ,"ariable integers a and u.
'Ve c1enote by T (n) the arithJuetical function which indicates
the nunlber of positive divisors of Jl. If 12 has the distinct l)rime
factors ]Jl, ]J2, . . ., }Jr and if
,.
II f('
J1 = 1) / '
z'c::l
we clearly have
(1 )
r
T (Ii) = II (1 + fLj).
i=1
It is eas to establish the following relation for T (11):
(:!)
" 71 [ ]
T(k)= I ;; ·

DIY(IBILITY
27
For, if f{n) is the SUDI on the right., we have
[  ] - [ u - 1 ] = 1 or 0
h h '
accol-ding as h is a di isor of 1} or :not. Hence
I(n) - .1'(n - 1) = T (11).
From t.his the relation (2) follo,ys at Ollce.
.A.11 illlportant arithuletical function is the 11I(jlJ-ill, function
,a (11), defined as follows:
,It (1) = 1; It (n) == 0, if 11 is divisible by the square of any prime;
,U, (PI }i2 · · . }J r ) =-- (-- 1 y, if })1, P2, . . .  1Jr are distinct pritnes.
Thus \ye have: ,u (2) =-=,Il fa) == - 1  ,u (-l) === U, II (6)  + 1, /l (10)
 + l p(30) = -1, etc.
An il1teg'er is called a sqUQ}'(I-j}'('(' 1lUnJ ber if it is not divisible
b.r any S(ll1are > 1.
\Ve prove
1'/zCO/,(-11l 1,1. F(j]. el'(!1"!1 natural I2zonbel' n > I 1l"e hate
(3)
5n = L ,u (d) = 0,
,f
tlzr SIHn beill[) extend(.rl orer all }Jo...:itire dicit""o"s d oj 11.
!)rooj: '\Ve 1leed onl.f extend the SUID (3) o,er all IJositie
square-free divisors cl of u. 'Ve pro\"'e the theoreUI by multi-
plicati \"e induction. I t is tl'ue when 11 is any prinle 1), since
S}I =-= ,II (1) +. !I (p) --= O. Suppose that it is true for n = UL Then
'ye shall sho,v tl1at it is also trne for 11 = J11}J, when ]J is any
prilne, If H is divisible b" 1J, it is easily seen that mJ) contains
t.he sanle terulS as ."m. Since, bJ hypothesis, 5'm = 0, we also
hae Smp == o. If In is not divisible by p, we clearly have
9m l' =  ell  -t ,Ll P D ) ,

the SUIH being" extended over all positi,"e square-free divisors 
of 112. Since /-1 (p c5) = -,1(. (), it follows that Sill II = O.

28
CHAPTER I
,\1 e shall applJ. Theorenl 14 for proving the Jlli;l,ills int'ers£on
f01.JJlula:
Th(lo,.e,n 1:). If 1{Y(11) 18 an lIritlnnefieal.lllntfion q/' 11 and If
(4)
(i (11) =-  1 fT (t/).
d
thpu CrJ12t(J)"S('!1I
(5)
F(/I) = /tG;) (;(d),
the .un1.c: heiug ext(?lu/(l o.el. all p(J;jfh.e dil.i,,:ors oJ' u.
Proq{. A pplying. the fplatiol1 (-1-), we can \vrit.e the right-hand
side of (5) as a double sn In
(6)
" ( Ii )  } ' .
 II. - . {Ie}
...;..J. d  \.'
i..l I'
the outer SUIJ1 being' extel1deu oyer all positi-re divisors tI of }l
and the inner sun) over all posit.ive di,isors c of d. This double
sunl luay be \Vrittell
 2fI (L1) · If' (L1) ,
.J
,v here .J runs t.hroug'h all positive divisors of 1/ and \vhere zP (L1)
is a function of LI to be deter)lilled presently. In the expression
(11) r taks a certain given 'talne .J if and only if d is of the
f _.1  h 5.. .1. . f 'It H h
ornl LI. U, were U JS a n.v uI \ Isor 0 j. ence we avp
Ifl (.1) = .f P. (' b) ·
the sum being' extended over all positive divisors b of . But
according to (3) this sun1 is equal to zero except whet1 L1 = u.
Consequently since P (u) = 1, t.he double sunl (0) has the value
]1'(12), and Theoren1 15 iN proved,
If W apply Theore1l1 15 to fOrll111la (5) in Theoren1 I;} 've
obtain tllP £ollo\ying expression for Euler's function:

Dl \'1SIDILITY
29
Cf ( Ji) = L,ll ( :n · tI .
d I
or
m ( ut = n - )'.  + )'  - ".
T 'I.,
i Pi 0 Pi ]Jj
the first sun} Leing" extended over all distinct prin1e factors of }I,
the second sunl being extended o,.er all pairs of distinct prill1e
factors of 11, etc This forlnula is clearl)' the same as forI11ula (2)
in Theoreul I J "
10. Diopllantine equations of the first degree. - Froln Theorelu tj
follovrs at once
1"lzeure1u lfJ". l'!te ne("el"sar,11 aud suf.lit'ient conditioll fur the lin(.ar
equation
(1)
al ,1'1 -1- ({2 ,l'2
-r {{ It J.',.  c,
ffith integral l"Ol!l i ,.itJ i ls ai' 1I2,.... lilt and i', to be s()lt'able
in int(!!lf:J'S ,)'1, .1'2' . . ., .i n i. thai tlte greafe.,.t Cul1HJ20n dil'isor
(al,ll2"."' (ill) tli,.ide f.
In the case of t".,o uukno\\"lls \\"e further IJrove
T/{(I01Tl1/ 17. I}" llu) liuf l ll1' cquation
(:?)
{I..y': + b! I = (',
(filII inle!lfal ('oe./iicit"utE; ll, baud (.: ha.,,: tlu: integral sollttiu'1l
.I' =, y = 1}. n'(' oblain all :sulutiun..... in illtege1"s. x £Ind !I /Jy
I Ii C ../Ufl1t l( I tJ, e
f ,) =  + (a u) f.
1 /1=r]- (llU/ '
ll'her£' i rlfn thrOll!lh all inte{ler,,\
J)ruoj: I t is easily seen that ,,{. and .11 gjY'cn by expressjons (3)
satisfy equation (:!) considering' that it E - 1) 1] -=-= (, Now suppose
that ..l:, 11 is au arbitrary intt"'gTal solution of t:!). Then
(3)
a .1.' +- b ,;{ = t = II  -j- 1J 'I} .

30
CHAPTER I
Hence
(.t)
a (,i' -;) -:. -- IJ C'I - 'I}).
If d = (a, b) is the gratest conlInOl1 divisor of a ana b, it follows
b
frOl1l this tha t d is a c1i visor of J.' - .:. therefore
b
J-=df
where t is an integer. Finally we get fronl (.t)
a
.lI - '17 = - d t .
Thus eqnation (2) lIas an infinite number of integral solutions
wIlen it is solvable. A solution 111aJ always he found by trial.
'Ve shall show ho\v to find a sOllltion by Euclid's alg'orithlll.
There is 110 loss of generality in SUPl>osing' C == (a, h). 'V e vrite
the algorithll1 (coD1pare Section 7) ill the forn1
(5)
[
. I l(,-  .- (;.-  (;. -'1 . .
a l -2 -- q.'-1 U J -1 - ((. - - ( .
a - '11 h =: ((2,
b-q2 11 2=aS-
Eliminating 111'-1 from the two last. equatious, 'e obtuin the
relation
Q}'-2 (1 + q,'-1 '1,,-2) - a.-3 (b-1 = C.
Eliminating' a 1 '-2 from this equatioll and the third e(juation fron1
the end in (5), we obtain a relatiol1 of the form
a l -3P + {l.-4 Q = c.
where P and Q are integers. By continuing this procedure and
by successive elimination of Q ..-3, a 1 --I. tc., \e finally obtain a
relation of the fOl'ln
II __1 + b B -= c"

DIYlIBILITY
31
,,,,llere J and 1J are integ'ers detern1ined by q1- 1J2, . , . - fl)-I-
This Ina r easil \' be verified by inductiou, In this ,va y 'e have
...,. .,
found an integral solution x -==- ...1. Y -= B of equation (2).
Ec{(nlple. If (( = 15, b = 11, c -= 1. the u.lgorith1l1 is
15 - 1 . 1 1 =-=..t- ,
11 - 2 '"* -= a,
.t - 1 . 3 -- 1
By elimination ,,'e finu
-1.;}--11.1-=1.
Ju 3-11.4-=1
and thus "'Ie ha\Te the solution {' = 3, y  - 4 of the equation
15..i' + 11 y --= 1.
The probleul of solving" the equation (1) in n ( > 3) unkno,vlls
Iua)" be reduced to the problenl of solving an analog'ous equation
in only II - 1 unknowns. Suppose that we know a. set of solu-
tions
')" - I:
tAl -'!!II.
 -- t  - I: II - 17
wi 2 - .., 2. . . .. oA. Il -  - - ':I" - 2  L -
of the equation
(0)
((1 f1
ll2 ,J' 2 + . i (( Il - 2 X' , - 2 -! d!J --= (',
,v here d -= (an-I. ((Il)' By the Inethod j llst developed, we can
deternline a set of solutions
C -- 1: .. - 
.. 11 - 1 -., Il --1, .1. Il -- ':I n
of the equation
ll'l-l Ll'll-1 + au :C" == d'l.
Then equation (1) clearl)- Las the solutions .1.', = , (i = 1,2, ..., u).
The general solution of (1) is finall)' gi ven Ly the systClll of
fornlulae
x, = i + b l t l + b 2 t 2 +
+ b n -If ll -1, (i -= 1,:?... n)
where b 1 , b 2 , . . ., b 11 - 1 depenu on the nlllllbers £(1- {(2. . . ., a'l only 
and where th( puraluetcrs 11. 1 2  . , . l,,-] run through all integ-crs,

32
CHAPTER J
It is easy to show this bJT i11duction froDl Jl - 1 to n startil1g
fron1 Theorel11 17.
Equations (1) and (2) a.re the sin1 plest eXall1 pIes of so-called
Diophantine or indeterminate equations. Let J" (.L\ /, Z, . . .) be a
polynoDlial in the variables L1., / z, etc.. \vith integral 01" ratiol1al
coefficients. The problem of solving the equation
( 7) J' (:(', ]I, Z', . . .) = 0
in intpgral 01" rational numbers .1", 11.. fe, etc., is called a Dio)Jlzal1tiue
problem; (7) is said to be a Diophantine or indeterl1linat( equation.
Simultalleous systeu1S of such equations n1ay also be investigated.
III his work ..41.itll1netira the Greek mathematician Diophantos
examined and solved a great nUll1Uer of indeterlninate equations
of the first four degrees.
In Chapters VI and VII we shall treat SODle Diophantine
equations of higher deg'rees.
11. Lattice points and point lattices. - III number theory it
is ofteu advantageous to make use of g'eometrical ideas and
interpretations.
The integers may be represented as a set of equidistant points,
unit distance apart, 011 an infinite straight line in the plane, the
so-called real uu?nb(.J}" axis.
Let us consider a plane rectangular or obliqne Cartesian coor-
dil1ate-system with the abscissae ;).' and ordinate y; x and !I need
not be measured bJ' the same unit of measurenlent. A point
(x, y) wllose coordinates x and ]1 are integers is called a lattice
1J0-1 nt. The set of all lattice points is said to be a plane point
latt'ice. Let us draw all straig.ht lines parallel to the coordil1ate
axes through the lattice points. The geon1etrical figure thus
obtained is called a plane latticr'.
Fig. 1 represents a point lattice in ordinary rectang'ular coor-
dinates, and fig. 2 a point lattice in oblique coordinates.
The equation
a"C+by=c
represents a straigllt line. The problem of solving this eql1ation
in integers x and y is equivalent to the problem of finding' the
lattice points situated 011 the straight line in question.

DIVISIBILITY
33
.
.
.
]1
o
.
.
o
.
o
o
.
.
.
.
.
.
.
o
.
.
.
.
.
.
o
x
x
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
...
.
.
o
.
o
.
.
l.'ig. 1.
}t'ig. 2.
E:A:anl}Jlr. The straight line 4 x - 311 = I passes through the
lattice points (1, 1), (4: 5). (- 2, - 3) (- 5. - 7), etc. (see fig. 3).
o 000 0 0
!I
o 0 0 000
000
o
o 0 0 000
000
o 0
o 0 000 0
o 0
o 0 0
o 0 0 0 0 0
o 0 0 0
o 0 0 000
00000
x
00000
o 000 0 0
o 000
000 0 0 0
000 0
o 0 0 000
000
00000 0
o 0
o 0
000 0 0 0
o 0
000
o 0 0 0 0 0
'ig" 3.
Suppose that the straight line has integral coefficients, or,
what is obviously the saIne, rational coefficients. Then, by Theo-
rem 17, the straig"ll t line passes through an infinity of lattice
points if it passes throllgh one lattice point. This does not hold
for a straight line witb irrational coefficients a, b, f, except when
there exists a positive n unl ber co such that the l1um bers a co,
b OJ, C OJ are all rational. TIlus the straight line
3 - [j I UHiO T'"!}fll'l;' .\"tl!/ell

34
CHAPTER I
'!J =-= 11  . L"C
only passes throug'h Ollt lattice point, nalnely the orlg'ln.
If a straig-ht line passes t.hrough two lattice points, the coeffi.
ciellts of its equatioIl are determined by the coordinates of the
lattice points as rational l1ulnbers, and therefore the line passes
tllrough infinitely lua.ny lat.tice points.
This can be g'eneralizec..l to three dimellsiolls. Thus we can
speak of lattice points anu point lattices ill s})ace. The equa.tion
a:c + 1),11 + cz==d
represents a plane in a Cnrtesiall coordinate-sJsteul with the
coordinates :1;, I and .?:. The problenl of soling' this fquation in
integers ."" 11, z is equivalent to the proLlenl of deterlnining' all
lattice point.s situated on the plane. 1\lore reuerally one n13)W
study the distribution of lattice poiuts on a given curye in thp
plane or on a g-iyen surface in space; this leads to the probleul
of solving Diophantine equatiolls in t,vo or three unknowns.
1\linkowski has creat.ed u: theory treating' questions of the
follo,ving t)ype: Find approximations to the number of lattice
points included inside a given closeJ curve in t.he plane or in-
side a given closeu surface in pace. The starting-point of this
theory is the follo\\-ing" theorelll: If a parallelog-ralu ..:-1 'v hose
Iniddle point is the orig'ill has the area ..... then besides the orig.in:
there is at least one lattice point inside .4 or on its boundary.
This theory will, however, not he developed in the present
voluDle.
12. Irrational numbers. - The g'eueral theorY' of irrational nUD1-
bers belongs to analysis. But special tYVes of irrationaillulll bers
are of great iluportance in elenleutary nUlllber theory.
The irrationality of the number J/i was doubtless discovered
by some disciple of P.ythag'oras. LQ,ter Greek nlutheulaticians
probably proved the irrationality of other quaura.tic surds. But
only the proof of the irrationality of 1'-' :! has been preseryed: it
is to be found in the 10th book of Euclid's E"'le111('ufa. 1'he
Euclidean proof is a special case of t he proof of Theorenl 18
given belo,,,,.

DIVISIBILITY
35
In general it is v'ery difficult to decide whether a certain given
real nUDlber is rational or irrational; the l1UDlber D1a. for exaDlple
be givell as the SUD1 of all infinite series. According' to school
arithluetic\ the necessary and sufficient condition for a real
11Ul11 bel" to be rational is that its decinlal expallsiou is finite or
periodic. But, the real difficulty consists in deciding whether
the decimal fraction is periodic or not.
By the follo,ving' theoren1 ''Ve can construct special irrational
nUD1hers.
Theor()}n 18. If' f{.J.') = x/l +- aI.1'n-l + .,. "7" a'l i,,: a 1)olllU01nial in
.C leitll int(Jgral cu(j}ieif]2f. aI- a2, , . .. au, and ij'  is II rout
(l' the ('fJuation f (..I.:) .::= O. thell  is pifhf!J' an ;nleY(:f or an
il.ratiuuul nUll/bel'.
].Jroof. 'Ve Ul8J suppose an  O. Putting'  == I , where}' and ""
.....'
(r6 0) are integ-ers having' no conlD10n divisor > 1  we have
]'n -:- al }'1l-1 8 + "
...;- a .II:: -- 0
. II '0 .
If s were di visible by the prin1e 1), I" ,vould also be divisible by }J.
But, since this is contrary to hypothesis, we nlust ha.ye 8 :...- 1,
a11d the theoreUl is proved.
In particular, we have
Theo}"e}J 1 1.'). If" 11 and Jv art> natural J111?>dJcr.,', and if T is not
t Ii e n tit J J(JU' (, f 0. r a 12 a t It ra I II III n her, the 1/
n
l /J\"
i' an i,.,.atio}lul nIl/Jibe)".
The irrational lluulbers constructed by nleal1 of TheorenJs ] r;
aud 1 H belong' to the exteusive class of nUlubers "hich are called
alg'ebraic l1UD1bel"s. ""'hen j-r(,} is a polJDolllial ill x with rational
coefficients any root of the e<tuatioll j(x) == 0 is called au a7ge-
In'at'c nunlbcl'. In particular, the rational nUluhers are alg'ebrnic.
Any 1l1unber wllich is not al'ebl'aic is said to ue frans(.('ndentai.
The existence of transceudental numbers ''''as pl'oved ill 1851 L}
Liouville.

36
CHAPTER 1
Let a be any real nUlnber. \Ve divide the iuterval 0-1 in t
equal parts  in each 811 hil1t.erval the left entl poiut only is in-
cluded. 111 fig". 4- t = 8
I
c.
.
- .
.--1.
I
1
Fig. 4.
No\v let !J be au iuteger > 0, and let x be the least integer
> a '!I. Then
O < x--ay<l.
Putting y = 0, 1,2, . . ., f, we get f + 1 nunlbers x - ay, which
all belong to the intervnl 0-1. Since tIle nU111ber of subintervals
is t, t.here is at least one Sllbinterval containing at least two of'
the nU111bers r - ay. Thus there exist, a natural number h < f
and two pairs of integ-ers x', !I' and .r", :1/', such that
h-l , , h
::; {(: - a 1 / < -,
t - .1 I
!t - -..! < j/'
t =.0
" h
- a 11 < -.
· t
Puttill oo
b
'I' ",
: = .J" _u:l. .'1 = !I .'I .
've fint!
(1)
1
I .t" - aI/I < , .
If \ve sUI-pose .Il' > '!I', then .'1 is oue of the nUll1bers 1, :!, . . ., f.
Hence ,ve have l)l'oecl: There corresponds to every real nunlber
a and to ever. 11atural number t at least one pair of integers
J: and 11 such that 1 < Y < t and such that. inequality (1) is
satisfied. Frout this \ve also 0 btaill the inequality
(2)
x 1
- a < -..
Y !l 2
III (1) as well as in (2) ,ve can obviously suppose that ."t and y
are relatively prime. If a is ra.tional and =, where (a, b) = 1
and b > 0, tllen inequality (2) has only n finite num ber of solu-

DIVISIBILITY
37
x
tions in relatively prhne nUlllbers L'r. and !I. For, if a  .. and
!I
I > 0, we get
.
x
--a
y
x
n = Ihx-a,l/ l>!.
b b '1 = h 11
. ,
y
Hence, if (2) is satisfied, ?J < b.
On the other hand we have
Theoly-nz .()o /" a i,r.: {t real irrational 12U1Ul)(Jl., inrqllalify (2) has
an i}tillif!1 (( .'?olllfions in relafirelH IJri'lne infegel"s :r: and ?/.
Proof Let '1 be a natural nunlbero .ApplJ'ing the result just
obt.ained, \ve then determine a pair of relatively priJne integers
Xl and .111 such that
.r] 1
1]1 =. .- - (!. <-
.'/1 !It t 1
\\' here 1 < !11 $ f 1 0 Since (( is irra.tional, '171  O. Then we choose
1
a natural nunlber to):"> -. and deterJuine the relativel .. y P rime in-
.. 1]1
tegers ;T2 and 1/2 sue h that
iC2
1]2 ; if 
1 1
a < - < < YJl
1/2 / 2 - '2
with 1 < Y2 < 1 2 . Repeating' this procedure we obtain an infinite
sequence of successively decreasing positive numbers
1]1 > 172 > '1]3 > · 0 > 1]l >
where the nUUl bel"
.r;
1). = - - n
,IIi
satisfies the inequality
'YJi < -;; 0
r I 1 1 -'
01 Z
rfhis proves Theorem O The procedure just developed readily
gives an infinite sequence of successivelJ better approximations

38
CHAPTER I
to t.he real irrational nUDlber a by means of t.he rational num-
x.
bers _I.
II -
,1 1
111 the proof of inequalit..v (1) we applied the so-called Dirichlet
lJ(J.1; lJri12 (. ip If..' : If l1lore thall t objects are distributed in f boxes,
at least one of the boxes 111USt cont.ain t\VO or more objects.
This ext.remel)- sill1plc prillciple has ne'\"ertheless been verJ effective
ill many mathematical proofs.
13. Irrationality of the numbers e and :T. - Let e = 2.;18R . . .
be t.Ile base of the systenl of nntural log'arithnlR, and let.
 -= 3,1415 . .. be the length of the cil"cunlference of a cil.cle
with the radius .
'Ve then prove the following theorems.
TlzeoreHl 21. Till.' JiUfn'ber e is irrational.
Proq{. In the introduction to analysis it is proved that. when
(1)
111
=--- 1+ : +..,+
.... 71 - , T )' I " '
1. oJ. '12.
t.he inequalities
(2)
O ..- 1
<:' -- ...:.. -
- f' S n .:::.=. I
}i-Ii.
are satisfied for all natura] nUlnbers n. Suppose that e is rational
a
and = b' ,vhere a and b are natural l1unlbers; we can clearly
suppose b > 1. Now, choosing t.he number n in (1) so large that
b divides J(!, we obtain fl"ODl (2)
() n ! , 1 1
< a - 12. .n < < .
b Ji
}I'
But. t.his is ilnpossible.. since both a I)' and J2! Sn are integ"el"s.
Thus the nunlber e is irrational. This theorem ,vas first established
by Euler.

DIVISIBILITY
39
Theorenl 22. Tlzr 12111uber 'J"l is irrational.
a
Proo.f. Suppose that :t = i.J' where a alld b are natural nunlbers.
For an)''' natural nUlnber 12 we define t.he two po]ynoDlials
f(:.r.) = x n (a. -- h x)n
71!
and
l(.):') = f(x)
, . "j ( ) . , " .4' l )
- - . yo + · .J: .
. ,
(--- 1 )nj'2n) (x),
where j'(m) (x) is the Hlth deri,ate of ,((..1:).
The polYl101uiai Ji! ,/(:1') has integral coefficients and 110 t.erms
in x of degree less thn.n the nth. It then folloW's tJlat f(.1') ana
all its deri,ates take integral values for x: = 0; and, since
f(.r) = f ( - x) ----= fi:r .- ,1'),
this also applies for ,l' = :t.
By computat.ion \ve 011d tha t
_ 1 ft [F' (:) sin x -- II'():) COS ,c] -= [F" (x) + 1-'(,1.')] sin,r =f(x) sin .L',
t .r
Hence
 
(3) f f(:c) sin :L d x  [ 1 1 " {or) sin x - I.'(:t) cos x] = F{n-) + ]' (0).
o 0
Here F('J"l) + }(O) is an integ'er, since j"!TIl) (n) and .ffnz) (0) are
integers. But. for 0 < .() <:t.. we obviously have
:r1l an
o < f(x) sin x < " , .
Ii.
Si nee the in fini te power series
. ;(' , :r 2 , .1'3 0
(.r =- 1 -r - -r. -j- T
I! i! a!
IS convergent for all :J::, we can choose the number 11 so large tbat
x)} (:t a)u 1
-=-< .
Ji ! U ! :t

40
CHAPTER I
Hence, according to a mean value theorell1 in elementary integral
calculus, the integral (3) Dlust have a positive value < ]. On the
other hand, we llave just shown that the l"ight-hand side in (3)
is an integer. Thus the assumption :It =  leads to a contradic-
tion, and Theorem 2 is proved.
This result was first established in 1761 by Lambert. The
proof reproduced above is due to Niven and was published in
19-47.
III reality, the numbers e and :r are transcendental numbers.
This was proved for e by Hermite (1873) and for n by F. T...inde-
mann (1882).
Exercises
1. [f n is an integer, show that the product
JI (n -- ]) (2 1/ - 1)
is divisible b)? 6.
2. If u 2 is tIle square of an integer which is divisible neither
bv 2 nor hy 3.. sho\v tbat the l1umber le 2 + 23 is divisible
.  
by 4.
3. Show that the 11umber 3 l,2 -] can never be a square, ,vben
n is an integer.
4. If jJ n denotes the nth priule in the sequence of natural num-
bers, prove the inequalit."
_ ,)71
1J Ii <;... - ·
Suggestion: Apply the proof of TheorelD :2.
5. Show that the least absolute remainder of a nlodulo 1J IS
a - b[ 2, ] + b[il
where a and bare na.tural numbers.
6. If 9 is a natural number > 1, show that anJ natural number
.i.\T ( > g) can be represented uniquely in the fornl
,- .)
.4' :-:: t'o + C-l 9 + C02 !( +
. + C (. 7 71£ ..
'm... .

DI'VISIBILITY
41
where the integral coefficients ri satisfy tbe conditions
o < Ci  (I - t,
o < em < 9 - ].
(i = 0, 1, 2, . . " 111 - 1)
7. The nUDlber
g' + gS + ,92 + .q + ]
is never a squa.re, when the natural number 9 is rf 3.
The number
g' +. 2 9 s t- 2 g2 + :?.'I + 5
is a square only ,,,,hen tbe natural number 9 is equal to .
Prove this by means of the result in tbe precedillg exercise.
8. If n is a natural number and a > 0, prove the formula
,,-1 [  J
 1I+  = [na].
k=O n
9. If 12 is a natural number, and if T denotes the number of
positive di\Tisors of 12, show that t.he product of all these
di visors is equal to
,
Ii': T .
10. Let 1'1,1)2' . . ., }Jr be the distinct prilne factors of the natural
nunlber 12 and suppose
r
11 =- II pi.
t=1
Show that the sunl of all positive divisors of u is equal to
n i+l - 1 .
jE:1 lJi - 1
11. Su ppose that })1  12' . . . . llr arE' t.he distinct priule factors of
the product a b, and suppose that
r
a = llpt; (at > 0),
;=1
r
b = II p1 i (Pi > 0).
;=1

42
CHAPTER I
Further, denote by Vi t.he least of the two exponents ai and
Pi and by Pi the largst of the same eXI)onents. Prove the
formulae
.,.
({1, ,)) = IT}};',-,
2"=1
r
{ a, b} = II 1):' i ·
i=1
12. Prove the rela t.ions
(aI' a2'... an) = ('aI, a2-.... ak  ".. (am..... an ),
{aI' a2.... an} = {{aI' a2..." ak}".., {am.... an}}.
13. Prove the relations
(a {h, rn:-7. [(a, b). (a. c)},
{a, (b, r)J = (ia, iJ}. {a, c}).
1-+. Prove the relation
( { a, b}, { a, (.}, {b, c}) = {(a, b) , (a  (- ) , ( b , c)}.
15. Show that the nUluher of irreducible fractions between 0
and 1 whose den0111inators do not exceed the naturalllulnber
12 is
1)
 cp (m).
m=l
16. Let N, 111 and 12 be natural numbers. Find by nleans of
the Euclidean algol.ithnl, t.he greatest COlnnlon divisor of the
numbers
...,Tn _ 1 and ..'-7ft - 1.
17. Find al1 natural numbers 12 such that
tp{u) ==- 2-*.
18. Find all natural numbel.s 'In < 100 such that the equation
If (11) = Hl
has no solution.
19. If n is a natural nunluer > 1  sho\y that
'"' -1 ()
a-2.12<p 12,

DIYISIDII..IT'Y
,t3
the sun1 being extended o,er all natural nUlnbers a wbicb
are prime to Jl and < n.
Find all the natural numbers n satisfying the illequality
2  'f {'11) < n.
Let F(n), G (u) and 1[(n) be three arithn1etical functions
which satisfv' the conditions
.'
(; (n) = 2: H (d),
tl
the sum being- extended over all positive divisors rl of 11, a.nd
'I}
F (n) .=  (; (1.-).
1::==1
Show that
F(12) =  r;] H(f).
What formula do we obtain by putting ill the preceding
exercise 1/ (n) = 1 for all J2?
'Vhat formula do we obtain by putting in Exercise 21
(; (12) = log n and applying' Theorenl 15?
Pl'ove the fornlula
. U (III) [ n J : 1
....- I H1
711 : 1
valid for all natural numbers J2.
Detern1ine the arithmetical function tp (12) defined by the
relations
tp (1) . -=- 1 an d ,u (n) =-  tp ( d),
(1
the sunl being extended oer all positive divisors d of 'Il.
Suggestion: Apply the inversion fornlula of 1\'lobius.
Show that there are infinitelJ many prilnes of the form
..t Jl + 3, i. e. leaving' the reJnainder 3 on division by ..t.
Suggestion: Apply the same method as in the 11roof of
Theorem 2.

44 CHAPTER I
27. Sho'v that there are infinit.ely nlal1J primes of the form () n + 5,
i. e. leaving t.he renlainder f> on division by 6.
Suggestion: Appl). the sanle Inet.hod a.s in the proof of
Theorenl .
28. If the number
2 P -1
is a priJne, then p is a. priIue. (l\lersenne prinles.)
29. If the nUlnber
2 n + 1
is a prime, then Jl is a power of 2. (Ferlnat prinles.)
30. Sho,v that 0\"'ery odd l1Ulllber can be ,vritten as the difference
of two integral squu,res. In ho'v IlHtllY 'VRJ'S is this possible?
31. Solve conI pleteJ.\F tile DiopllRlltine equation
11 !-J.r -- 29 1/ = 8
.
in integers  and !I.
32. Solve completely tile Diophantine systenl
-) "-+- 0 ., 1 '1 -- 1 I ;; - 1
... ... I . , _ -- ,
&l' - 1::?!1 + 7, = 2
in integers .r, y and z.
33. A natural l1umber that is equal to half t.he sum of its
positive divisors is said to be a }Jf.JJ1ect nU1nbpf. The least
perfect number is clearl.r 13.
Prove the following tbeorem of Euclid ('len7ellfa, fJth book):
If 2 m - 1 is a prilne, then
2 711 - 1 (2 m - 1)
is a perfect nU111her.
This iR t.he case for 11? = 2. 3, 0, 7, 13, 17  19, 31, 61, 89,
107 and 1:27. No other perfect nUl11bers are known than
these twelve.
34. Prove the follo,ving- theorenl of Euler:
Every even perfect. nUluber lllust be of the form just indicated
in the theorenl of Euclid.
No odd perfect numbers are knO\Vll.

Dl VISIBILITY
45
...
i,f).
Let Ji be a natural nUluber hu Villg- the distint prirue factors
}ll, Ji2, . . ., pr. Prove' the f()l"lnula
",'
.:...Ja 2 = A ({J (u) u 2 + (- l)r A 'P (n) ])1})2 . . · )1,.,
the SUlll being extended over all natural nUlnbcrs II prime
to J/, al1d < I.
Suo'o-estion: Start froln the formula

1 2 + 2 + ... -- (It - 1)2 =  Ii (Ie - 1) ( 11 - 1)
and apply the inversion theorem of Iobius.
3u. Let 12 be an odd natural nUlnher haviug the distinct prirne
factors ))1, 1)2, · . ", jJ r . Prove t.lle fOl'l11ula
 a =  cP (II) II - (- l)r .  tpi - 1 ) (P2 - 1) . . . (1)r - 1),
the sUln being' extended over all natural nunlbel"s (J prime
to 11 and < - 12"
Suggestion: Start froln the forluula
J + 2 + 3 + .., !  (11 - 1) =  (1,2 - 1)
and appl)p the in version thporenl of fobius"
37. Let 1/ = 2 h + 1 be an odd natural nUDlber baving- exactl)'
Ii. disti1J ct prilue factol-s of the forl11 -l- f + 1 and exactl)" "
distinct prime factors of the forll1 4 f + 3. Further, for I -= 1,
2, 3, 4, let L. r denote the number of iutegers prime to 11 ill
the interval 1: (r - 1) II - ! 1"n. Finall)., for ,. = 0, 1" 2: 3,
let Br denote the lluluber of }Jositive illteg'ers < II which are
prime to Ii- a.nd of tIle forln 4 f + t.
Prove for J.t > 0 that
j - ! -- f - A - 1 m (,, )
.J.  1 - ...::! -- ..."1 3 -  4 - 4 T .
and
Bu:--=Bt "-=B:!. "-=B: -:llf'(U).
Prove for Ii  0 that
.. 1 1 "-;:..4 -I :- 1 'P (,,) + (- l)h:? 1 - 2"
. t - A - 1 m ( 'i ) - I - 1 ) It .) J - 2
...  - 3 -- 4" T \ -,

46
CHAPTER I
an cl
Bo == J cp (n) -f- ( - 1 )/, 2)'-2 Bl = ! <p (11) ;- :?1'-,
B 2 :. 1- 9'(n) - - (- - IV' :!.--, Ba = l<p(n) - . :?1-2.
38. The Farey series of order 11 is the ascending seq llel1Ce of
irreducible fractions  satisfying' the follo,,,,ing conditions:
b
(a, b) = 1 and 0 < a < b  12.
Thus, the E'arey series of order G is given b.
U 1 1 1 1 2 t :t 2 :i I !) 1
i' tj' :.' l' ;; , Z".'  ' :) , :;' ..' [" t.' 1.
Prove the follo\ing theorems
1. If  and  are two consE.>cutivE.> fractions in n. Fare)" series,
a c
b < d' then
bc--ad=l.
a c j'
2. If , - and are three consecuti¥e fractions in n Fnrp)p
b d .tJ
. a (" r
sel'les < - <. -, then
b d !l
( il + r
"=' --....- 15. .
d b,-!!
39. Show that cos 1 and sin 1 are irrational nUlubfl'S.
Suggestion: Like the proof of Theo1"enl 21.
40. Show that the nUlnber e is not a root of a quadratic equLl-
tion with rational coefficients.
uggest.ion: Prove by a slight ulteratioll of the proof of
Theorem 21.

CHAPTE R II
ON rrHE DIS 1 I{ ] B II TION OF P l{J!\I E S
14. SODle lenlmata. - ".,. e first prove
l'heo7 e e1n i2.:J. If x is > 0, 1112d if' a i' a natural Ull1Jlber, then
(1 )
e;]=l [] ]'
Proof Since [[':]J is the greatest integer < [:] . we can write
[x] = [ [x] J + l' ,
a it a
where ,. is an integer > 0 and .< a-I. Pll tting x == [.).] + e.
,vhere 0 < Q < 1, 'e have
.r = [x] + (! = [ ['l] J ;- (! + ,
a tI ,t a (l
Fronl this the theorem follo,t's at once, since
() :::;  + ': < 1 +  - 1 .--: 1.
a. a
Theorern ,J. L(.t Ji be (/ natlll.al nltnZbf,., and lri p be a }Jri1ne.
Then the e.'}Jo12enf o.f flip hig/u!si IJo/fr-r q( jj lrhich dit.:ides
1/ ! 1 . 2 . 3. n i, ffj II a' to
(2)
[ n ] r_1 ] [ n ]
.N= jJ +l.1}2 + p3 +'.
Pronj: The series (2) contil111eS so long as tlJt }Jo,ver of 1) is
< 12. If /z" denotes the nUl11ber of terlllS in the seqnence 1,  3,
. .. II ,vhich a.re uivisiLle Ly 1/"', tIle required eXIJOl1ent _,T is

48
CfiAPTER II
obviouslJ
which are
equal to hI + h 2 -f- "3 -t- The natural nUIl1bers < n
divisible by pI' are
1 .. ,)", [ Ii ] v
· p " 6J · P , . .., ---:- . p .
1)"
Thus we have h = [;1. J. and the theorem i proved.
By Theoreul i3 we have
h,.t! = c: ] ·
Hence, we can determine the nUlnbers h successivel)"" by divisions
by p instead of by powers of p. This gives a quicker 111ethod
for determining N.
Example. If 12 = 3000 and p = 7, \ve get
3000 _ ..) 42. HI 8
,.. - _8 + el' -,:- = U1 -i- e2, -;:- = 8 + es and r- = 1 + e.s,
( 4 . 4
where el, e2, ea and e4 are positive llulnbers < 1. III this case
it follo\vs tbat
...,y = 4:?8 + U 1 j-  + 1 = 98.
By means of forulula (i) \ve p]'ov
Theo1'l";'nt 25. If n 1110 il2' }is, (tl"o, are natural ulfnb(.,.s ,"tuch that
Ji = }II + 112 ! Us +. .
the quotient
(3)
121
Uti U2! us! . . .
is an integel".
Prooj: Let m be a natural number. Fronl the relation
'1Z nl 12 2 ns
-= +-+--
111 lt Il2 1Jl
we obtain the inequality
[ .n ] > [ J2 1 ] + [ 122 ] + [ 11S J -t-
'In n 11J HZ

o THE DISTRIBUTION OF PRIl\IES
49
Let jj be a prill1e factor of n!, and put into this inequality
_ ,2 3
successively 1n - p, P 1 p, etc. By additioll we then obtain the
inequalit.y
(-1)
. [ n. ] > . [ n ] + , [ n2 ] .
..... l - J  i +
i=1 p i=1 P i=1 P
According to Theorem 2-1, the sum on the left-band side is
the exponent of the higllest po\yer of p w llich di vides Il!. und
like\vise, th 'J,th SUIU on the right-lulnd side is the exponent
of the highest po\\"er of p \\"bich divides n.! (v = 1, 2'1 B, etc.).
Thus it follows froIll the inequality (-1) that the highest po'er
of }) \vbicll divides the denoluinator of the llulllber (3) also lIivides
the llUluerator. Hence the l1u1l1beJ' (3) is an integer. Q. E. D.
ThcoreUl 25 Ina)" also be proved b.r combinatorial reasoning.
III fact, the nUJuber (3) is a so-called jJolyno1nial (4oef1il:i(Jnt; nlOl'e
precisely, it is the coefficieut of the po\ver product
Jon! :c""! X 713
1 2 :i
in the developzuellt of
(Xl + X2 + X3 + )11.
By their lllanner of fornlation, the }JolYDoluial coefficients are
necessarilJ' intger8. In particular, \\ hen n = III 1- J72' the quotient
 = Ji (n - 1) . (Ji - 111 + J) = ( ' }j )
n1 ! 112! J .  .  . .. Ul n1
is a biuonlial coejficient.
A set of n objects chosen fron1 a gi,.en set of }l objects,
'\vithout reg'ard to order, is called a conlin nation of JI things
taken 1n at a tin1e. It is eas,'" to sho\v that the nUluber of these
."
combinations is
Jl ( n - --D _ (It - nt__+ = ( }2 ) .
] . :.? . 3 In nz
'tVe have the following corollary to Theorem 25:
The produ.ct ()f 12 conse(.llti'e integers is diritt?ible by n!
This is ob,"iol1s1y true also \vhen the integ'ers are lleg'ati¥e.
4: . 516670 Trg[1lJt )."tlgell

50
CHAPTER 11
The following theorem, ,vhich is due to Sylvester (1883), is
of a very gelleral character.
l 1 helJreu1, 2(1. Let us COli8ider ..,r oLject8 and a l11l'1nlJer oj'llillthenla-
tical jJJ"o]JCrtic8 El. E 2  Ea, etc. IJet '"'ri dl}llote the nlllllber u..(
objects ha'ing the jJ/"ojJerl.'1 E i , let .,r i ,} denlJte the 1Hon/Jer uf
objects having the llro di,\:tinct jJfo]J(lrtits Ei a nd h let J.'\';.j. k
denote the ll11niber l1" objects hacin.lj the three distinct properties
El. EJ and E k , ell'. l'llen the nUI/LtJe/' uJ' Ojl)("tt"; hlll"in.lJ nOJit> of
the propertiel'" El' E 2 , Ea, lJte., i8 equal to
(5)
[t. T _ " ,... -'_ " 7\ T . ,_  '" T. . l - f-
..L'  J,.' l -1  'loJ L.i 'I,), :
the ,llrn.. beil2!! e,£tended Ol"fJr all calues u.l tilt> iudicP8 i, j, k\
(Ife. sati&11Iin{} the conclltions: i = 1, 2 3, l4e., 'i12 the jif,"t Sinn,
i > j > 1 in the ."ec(Jnd 8llJn, i > j > k > 1 'in the third Sl017, etr.
])}'ouj: Let.A be an object having exactly r of the properties
El' E 2 , Ea, etc. Then oJ. contributes 1 to the 11Umbel' )..... If t > 0,
A contributes 1 to r of the numbel"s ...,"',. If r> 1, 4 contributes
1 to G) =  r (I' - 1) uf the numbers Si,j. If ,. > :!. A cont.riu-
utes I to
( r ) = r tr - 1) (1" - 2)
D ] .2.3
of the uunlbers }....i.j, k, and so all. ,V' e finally see that the object
...4. contributes
1- () + () - (;) +
= (1 - I)r = 0
to the sum (5), if I. > o. On the other Land, if l' = 0, it con-
tributes 1. Consequently the number of objects having- nOlle of
the propel-ties is given bJ (5).
Suppose no\v tllat the ...,.. objects in 1'heorenl 26 are the natllral
nUlubers < x, and thus .'9' = [x]. Suppose further that E i denotes
divisibility by the natural nUJ11ber ai. \\7' e then obta.i11 the result:
Theore1Jl 27. £(4 lil, a2  .  q am be natural nllube'.8 such that
(ai, l1 J ) = 1 1f i rf j. Then the nunlber oj' naturaZ o 1l'll1nbl J l'S < x

ON THE DISTRIBtJTIO.:.'i OF PRIl\IE 51
z,.hilo/i are not dil'i:)ilJlf b." lllill one (!( tlu) /iun,her..,> aI, ({2, . . ., alii
is ('flUal to
r _ I [ ] ]
" /' X X
[.r] -. j- - -- i'
.,(..j i. tf(  itj {lj  [{lj{lj ak
the Sltn8 l)f'ilt!l ('J."lended OCf'1" all raluc..... (r the iudil"es i, j, A's
etc.. .':o'ati:-,'.l!lill!J the (.uudiliull.....: i = 1. :!. 3, efl'" ill tlu) .t;rl.d ."U11i,
i > j > 1 in the '-....>(,("olld 8101i i > j > k >. 1 in fhe fhird ......tOll, (.i(,'.
For the uuuILcl' of IHttnru.l l1uluLers < x '\7Ideh are Ji\"isiLle U,
every 011(\ of the uluubers lii. an liT.:. etc., IS obviouslr
r lti {(j  J; . . . j-
If, in Theol'elll 7, t]l( nUI11 hers lII, a2, ({3: etc.. denote tIle
distinct }Jrillle fact.ors of [x], ,ye obtain a ne,," proof of the
formula for Euler's pfunction in Theorenl 11.
15. General remarks. The sieve of Eratosthenes. - It is theoreti-
cally possible to decial \vhether or not a given natural nunl ber
7Z is a prinJe by trying to divid( it by every' sDlaller natural
nUDlber. For, if n is not a })rime, it Inust ha v'e a positive divisor
> 1 and < Il. This Inetllod does not presuppose allY prilne to
be previously kno,,,,n; it is, ho,ve'fer, inapplicable for large values
of n. If t.he prilnes < Y; are already knowll the question can
be decided in a much shorter time by tr.ying vhcther or not JI
is divisible by anyone of these priules.
Provided that 1/ is llot too large, the quest.ion can be solved
by Ineans of a. factor table or a prillle table. The larg'est prinle
table yet published was worked ont hy D. N. Lellnler: it gives
the vriInes up to 10006721.
By inspecting a prinle taule one obsel.ves that the priIne l1UD1-
Lers grad uall y beCOll1e m ore scarce the farth t. 'r one goes 011 ill
the sequence of natural nUluuers. In the ten intervals 1-100,
100-200, ..., 900-10Ul) there are the following' nUl11bers of
prllues:
5, 1. It). 16, 17. 1-l. It), 1-l. 15. 1-l,

52
CHAPTER 11
In the ten intervals each of one hundred nunlbers bet"geen
1 0000000 and 1 0001000 tIle corresponding- 11 lUll bel's are
2, (), G. 6, 5, .t, 7, 10, U, G.
The largest nUlnber kuown art present to be a prinle is
217 -1 = 1701.tlI83.to0-169317316873037158-1105727:
this ""as shown b)? Lucas.
The distribution of the primes in detail is DIOSt irregular. In
an interval of relatiyel)" l11an)" prinles, there 111ay occur long
sequences of consecutive cOlnposite numbers. Thus, there are no
prillles bet\veen 13:!7 and 1361.  gap of this length does not
reoccur until b{t".een the priInes -l-()7 and 8501. \Vhen 12 is
any positive integer, it is easy to construct sequences of 11 con-
secutive COl11posite l1uulbers; for instauce, the numbers
(u !- 1)! + , (n  1)! .t 3. ., . (n + 1)! + n + 1
are all composite. 011 the other hand, pairs of pl.ilnes wIdch
hft ve the difference 2, so-called pri;llr tu:ins, occur relatively often;
we have the following eight pairs of prinle twi11s less than 100:
3, 5; 5 7; 11,13; 17,19; 29,31; -11, -13; 59.61; 71,73.
Among the first hundred priInes after the nUll1ber 10000000l}
there are ten pairs of priIne t,vins. 'rhere are pl'obably an in-
finity of pairs of pritlle twins; but the proof of this conjecture
is at present beyond the resources of JuatheIl1atics.
\Vhell the priIlles < V are kno\vn, t.he priIues  x n1a)" be
found in the fullo\ving ,yay. \Ve write up the sequence of all
integel.s  2 and < c / ... in their natural succession. 'Ve first strike
out all numbers divisible by . then all nUDlbers divisible b\" 3
." . .
further all llu1l1bers divisible by 5, etc., and finally all numbers
divisible L' q, ,,,,here q denotes the g'reatest prilne .< 1 . (;. The
rplnainil1g nUlnbers ouviously consist of all the primes that are
> l" a: and < x. FOl' uch a nunlber cauuot ha\""e any prinle
fact or < ]/ ..v , and it cannot be the product of t".o nunlbers
> J"-.{I. rfhis siInple but effectiYt method is known as Eratosthp-
lles's f:ier(' '}uethorl.

O:'i TIlE DITRInUTI0 OF PRIl\I£S
53
E}'a111})le 1. \Ve consider t.he case x = 26 and a pply the sieY'e
luethod. The priIue nUlnbers < 1 / 26 are 2, 3 and 5, "T" e ,vTit
do\vn the integers frolu 2 to 26  ',?e.-6rst Inark by a bar eyerJ
second number counting fronl 2, tllen e\ery third nUInber counting
fl'OIn 3 and finall. e\"er.v fifth l1tul1ber count.ing from O. Tbn
t.he sequel1ce looks like this:
2. 3. 4, 5. .) , 7, R, , J_O , 11, J 2 , 13, 1'+ , 1 Q , 1 () ,
17, I , lU, ;!O , 1 , 2 , 3, 2d: , 2 , 2.);
The numbers not barred
7, 11, 1 B, 17. 1 nand 23
are the six prinles :.. J/U and <"- ().
J?,r{(1n1)le:t. if we take ,'{  - 300, the prill1es < 1 300 are
2, 3, O. 7, ] 1, 13 and 17.
"
.1\ pplj!l the sitY'e 11lethod we find the following UD
:> 1 300 and < 300:
19 ") 3 ';)0  ] 3 ... -11 ' 3 t  ,.. 3 -«) f 1 .,.. _ 1 - f,\
, _ , _ v, i),  . , -J:. , -:t', D , a.., v , h ,  ,  i),
7 t), 8 a,  f} . !) 7, 10 1 , ] 03, 1 07, 1 OH , 11 3, 1  7, 1  1 ,
137, JHf}, 1'+9, 15], 1n7, I(};}, 167. 17B, 179. I].
Iff1, lta. 1U7, 19), I1. 23. 227, :!:?8. ::?33 239.
..) t l ')- 1 9- - ") t "' 3 ") ( "0 ")""" 1 9 "Ju l ')1-'" ")( )3
_-I- . ....0 , _;). _);, _ )"', _, ,....  , _0 , _()i). _ .
.
prln1es
B). 111eanS of tIle siee Dlethod we ca.n also calculate t.he nU1Jll)(r
of prinlcs which do not exceed a g'ivell linlit ,I". Thi:s 1111111Uer
is usualJy denoted by 7T (.x'). For inst.ance we ha,-e 7C (10) = -+,
7t (V 300 ) = 7, n (300) = H2. After the application of the sieve
met.hod to the sequence of integers > 2 and < ,):, tbere are lft
exactl.r 7l (x) - :r tV x) integers. It is, hO''fever, possible to deduce
another expression for the 1111111ber of integ'ers relnainillg, For
if we replace ai by l1t in Theorenl 7, anu. su ppose tba t P1, 1)2 ,
. , ., 1Jm are all the prinles < 1/ x , we find the following expression
for the nun1ber in question
- 1 + [.:r] -  [ , ] + ) [ --- ] --
 Pi ..... .1JilJ

54
CHAPTER II
'Ve thus obt.ain thf' f01"lnnla
,:r (.r) - :c (1/;') = - 1 +  It (d) [  ] ,
tI _ 0.-
(1)
the SUlll being- extentle<.l ov('r all positi'''l di,.isot"s of the product
PI1 J 2 . · . ])111'
It i, however. possible to impro\"e this resu]t cousiderahly. aR
'vas shown b.y l\leissel. The f01'Jnula f'stn hlished b.," hinl g'ives
t.he best. Jnethod up to no'''' for nUluerica] calculation of n ("C),
rfhe follo\viug' tuble g'i\"(':;) au idea of the \va). in ,vhich the
fun ction ..T (,i:) increases..
11' I :1" .,':
100 0)-
_"J
200 4"
:OO t3
. '0 78
,jOlt !).j
6()(J lOB
700 I:!;)
HOO J :i J
{uau I It-l
loon 10S
ooo :J03
:JOOn 430
:r 1'1
,tt)OO
50( U I
600U
7000
goon
BOl)O
10000
100000
100000.)
1 OOOOOfU.
]OOOOOOOu
1 UOooQOOf).'
5!)O
fj(i9
78:
uno
JOn7
1117
12O
95!)2
78498
un t,j7B
.) 7 (j 1-1-,');')
.')OM4 7 ,I i 8
The ,:tIne of ;1 (10 9 ) ,vas calculated by Bertelsen from the forn1ula
of leissf'l; ;r. (,i't) has not been calculated for values of ,i' larg'er
than 10 9 .
16. The funclion :r: (x). - Legendre and Gauss occupied thelu.
selves \vith the problem of finding siJur.}e functions \vhich gip
good approximations to n (x) for largE"' values of r. Thus, in
his book 1 1 Jz6o)';c dCi.: -uouJbrr,Io.' (17n), Leg'cndre states that the
function
.'
log ;'1: -- 1.0i<3tjlj
g'l \""cs a good a p)Jl'OXinln.tion to ;1 (X).l BJ Ineans of prime tab]cs
1 J lere and in the following, o1-{ dt")}utes thC' nntnral lO3rit}))n.

ON TIlE DISTRIBUTION OF PRIl\IES
55
Gauss discovered that 7t' (x) Inay be ",ery well approxinlated b)1'
each of the functions
x
log .1'
ana
( 1)
:t
Li (x) = r du ·
.. log 'U
2
But he gave no proof of it. The function (I) is t.he so-called
integral logaritlun of .r.
The first delnonstratecl results are.. however, due to Tchebychef,
u.bo (1850), arDong ot.her things, pro,ed that the inequalities
(2)
7 x 9 x
- · < n-(x) < - · --
8 log . J" 8 log .
are valid for all sufficientl" large values of x. He also sbo\ved
tha.t the quotient of the nUDlbers 7(; (x) and 1 x bas the limit
og ..l'
1 for increasing- r provided that the linlit exists. In 189() Hada-
J11arcl and Vallee Poussin, independently of each other, proved
the existence of this linlit and thus the relation
(3)
lim n(x) = 1.
x.. oc :' / log" :c
Their proof of this theorem, the so-ca]lecl ]Jri111C nU1nlJer tlz , eorel1Z
is based on tbe theory of Riernann's zeta function ,(.) defined
bJ the infinite series
(-l)
; (.) = 1
1
...,... -i
. -)Il '
-
1
1
+- -- -+
. R'
U
-- -,-
3 '
for all conlplex values of s = a + it when a > 1. Subsequently,
by ana.lytic eontinuation,  (8) can be defined for aU 8  1.
The connection of the zeta function ,vitb the primes is obvious
froln J'fulcr.8 ide12tzly
(5)
00 J Of.) 1
 (.) -  - - II '
.t -  '}18 - 1 _ } )-B
11=1 P
(a > 1)

56
CHAPTER II
the infinite product being" extended oyer all prin1es }). To pro¥e
this identit. we first ,erify that the infinite product is cOl1verg'ent
and different from zero for (1 > 1. For, since I p-' 1= }}-'J the
series
 .
 }J-
p
is absolutely cont'erg'ent for (J:;: t. Since
I J J-il l < 10)-11 <' 1
="'" 2'
we lls"e
_1 --::::: = 1 -I- p-' -I- p- 2. +
t --}) "
and, therefore, because of the absolute C01)''''rgenc,
II 1 II ( t ' oj  ) , 1
I _ it = ,-f- }1 - ,0.: + p - ...." +, .:;-. 2.J  /
]J.r 1 - P 1)X. JI ti
\vhere }) runs throug'h all priules < IT, and ,,"here n runs through
all positi¥e integers which have no prinle factor > .l". Hence
II - 1_ _  1 ==' t .
1 -- 8  x ," 
p.t - 1) 1&=1 Ji ;r n
Here the absolute value of the right haud side is obviously less
tha.n
co 1
 ',.
n-=x+l li
Hence ,ve have
( t :r 1 )
lill1 II _It -  - = 0,
J'- 00 l'  r 1 -- })' n =- 1 n
which proves the truth of identit. (5). It n'u'" be observed that,
in t.his proof, we n1ake use of t.lle funda.mental theore111 of nun1ber
theol". (TheorelD .t.).
Riemann, perceiviug' the fundall1ental importance of the zeta
function for the study of the distribution of prilues, develolJed
the elen1ents of a thorJ for this function. He also formulated
six hypotheses '''' hich he could not. prot'e. Especially t.he position
of the inlaginury zeros of the function appeared to be of great

O THE DITnInr..IO OF PRI:\JE:-,
,.. ...
t
inlport.ance for the a.pplicatiolls to priIue J)unlber theory. Ac-
cording t.o RieJnann's fanlou hut. till unproved hJPothesi8, a]]
the iInaginary' zeros ha,"(' the renl part a . All the otber
hypotheses of Ri(lnanll haye hpf\u pro¥ed hy ]nter inYestig'ator,
Important. contributions to th( theor,Y haye also l)(en ulade bJ
Mangoldt. Landau, Rohr II nT'tly. Little".oo<1 3nc1 ...\. tIe Selberg'.
These results belong', ho\\""e,er. to the higher allnl.sis and ,,,,ill
not be developcl ill this book. ". e shall on 1.'" l11cntiol1 the fol-
lo\ving result of TitchJnarsh
(Ii)
':1 (.,.') - Li (.r):- 8 k ,I' . (' "J :.r.,
;,
\here (o{.)')- a(log-.}I)I F; it is ynlid for all sufficiently large values
of Ir; e is a positive nU1nber. k and (l are ct"1rtain positi¥P con-
stants, and 0 at"\note a function of :J.: \yhich ¥aries h(,t"'(,Cll the
liIllits - - t and 1. 'his fOl"lUnla. ,vhich "'as pro¥ed in 1 )38.
eX}Jresses the bf'st result U]) to no,,? for the fnuction  (.,t). It is
easily seen froln the forluula that t (.r.a) is approximated by Ii C/')
,vith great accuracy. This is ,.priHed by nunlcrical exanJvles.
For installce, if r == 10000000UO. we have. n part fronl the deci-
nlals,
Li (;c) -  (.t)
1 '"'-'"'
4a... .:
1
this difference is les than _ of the value a (1 OH).
nOOO
I twas 8bo".n b. Littlewood tha t the difference :r (x) - Li (:1')
aSSU1UCS both positie and neg'ative values infinitely often.
rfhere \Y3.S a sensation when rpcently an eleIn(ntary proof of
the prim nuulber theoren1 ,ya8 g'i,.en b. A tIe Splberg: (19..!-8).
The proof is eleJnentar.,. ill tht."' seus(\ that it uses pract.ically no
al1alysis except the sin1plest. properti4?s of the logarithnl. "T e
shall gi\e this J -roof in (1h3 pter VI I I.
17. SOlne elementary results on tIle {listl'ibution of primes. - r.Jt
U put, for ,} > ,
p, = II 1 _ j'
J)X 1 - P
\vhere the product extends oyer nIl prinles p < ,1'. Then we
have

58
CHAPTER II
r [r' + 1 .r
.)" I .
I):r = p I  l. ( l + ) + ! + ... ) >   > / - dll > r clu
.. }J 1)  71 -= 1 11 . U .. ll-
l 1
or
(1)
}).)" > log x.
Hence
(2)
lim II ( I - ! ) =. o.
:1-00 p }J
Thus, corresponding to every d > 0 there IS a natural number r
such that
(3)
IT ( 1 -  ) <  d.
'I' - 1 }))'
wllere PI' PI' . . ., p,. are the first r primes. If B (x) denotes the
number of natural llumbers < x whicll are not divisible bv any
. .
one of the prinles PI' P2' . . ., }Jr, we have by Theorem 27
()
B(.1') =  (d) [J]'
the sum being" extended over all positive divisors of the product
PI 1)<2. . . . pr. Since clearly
'Jr; (x) < 1. - B (:t').
it follows from (-1) that
n' (x) < r +  (1 + P. (d)  ) ·
(1 . (
and, since the nU1nber of divisors of the product ))1 })2 . . . p,. is 21",
we have
'It (.t) < l' + 2 r 1
.r II ( 1 - + ) .
1'=-1 1 )
Then, by (3). we obtain
(5)
:r; (.) < J" + 2 r + ! (} x.

O:N THE DITRIDrTION OF PRIMES
59
Let us choose the number  greater than the rth prime and
sue h tha.t
I . I- -)r < 1. -" l:
...  u .
Then, for all x ;> g it follo\vs that.
1t (.J.:) <  (} ,I' 1  ,1' = (} .1".
'Ve can thus stat.e
T heoreuz 28.
1 . i( (.r) -
11B - = O.
.r _ oc ,r
\Vhen P,,' has the above sig-niticance, "e g'et
 ( 1 1 1 )
loO"P:r= ,7: IO<T 1 + - +  + -- + ." .
  0 }J po. ))3
Since, for any positive ,}\ log' (I + :t') < .1', \\"e obtain
100" ]J.r < '. ( + 1 +- ... ) =   +  1 .
 J' P po. 1) $it 1) 1JP (lJ - 1)
As befort the sU1nnLatiollS <:'xtend over all prilnes p < .c. 1'he
value of t.he last SUIl1 on the right-hand side is ohviously 1('s8
than
OC I DO ( 1 1 )
'1 ; 1 (I/. + 1 ) = t 11 - 11 + 1 = 1.
Hence \ve bave
1
log I';r. <  .) -t- 1.
pxl
Conlbining this result wit.h inequality (1), we can state
Thcorenz 2.'J. Tlte sltnnnation beiuy e..rfel/ded orer all )Jril1ZeS < x,
u:e hare ilzi' inrquolil.'1
, 1
 -  Jo O" 10 0' "I' -- 1
 ,- l"t ,., '. .
]J ",' J)
A s a consequence \ve
primes is divergent.
see tlJat the infinite series )'.  over all
'*-J J J

60
CIIAPTER II
Instead of the function  (x) it is often lllore convenient to
consider the function
{} (.r) =  log i),
1';:;.,;,:
the SUDl extendi11g over all priuJes }) < x. 'Ve shall pro\e the
following theorem of Tchebychef:
TIzeorr1U lJO. Therr r.i:il;t ill.o lJo,,'itit.c c(J}udants c aut! ("J suell that
(6)
C" :r <: 0. (x) c....... C"l,i.:
.for all x > 2.
]Jroqf. Let 12 be an integer "> 2. If pm is the highest power
of the prime )> which divides the binomial coefficient
( 2 n ) = (:? n)! ,
n Ii! 11!
(7)
then, by Theorenl 25, ,ve have
-, ( [ 2 1 ] [ U ] )
111 =-:  - -- -- 2 .
- 1 " j i"
, l' -
,,,here r is the highest integral eX]10nent '\vhich satisfies the in-
equality
(8)
and thus
lJ r < 2 11,
r ==. [ lOg' 2 n J .
log 1 1
The difference
r 2 1,1 ] __ 2 [ . n ]
llJ' })'
has eithel" the value zero or the value 1  and therefore
(B)
nz =::; r.
N ow we ha\e
( :z ) = (I + 1) (fl, 2) . . 2/1 = Ii 2 ":':" h > ",
/,-=1 h

O TIlE DISTRIBtTTION OF PRI]\[R
61
and on the other hand, by (9) anc1 (8),
( ') 12 )
2" < Mil = IT pm < II p",
p<:1» p"'211
the products extending over all primes p < 2 12. Therefore, by
taking the logarithnl,
  [ log 2 11- ]
12 log 2 < .£.J r log})  L.J _ 1 " - log 1).
11" :!n ]1<211 og})
For evel'Y p we have
[ log  12 ] < log' 211 ,
log }) log }J
ana for all p > J/ =l Ii
[ log 211 J
-- _ 1
log 1) ..--.
Thus we obtain for :s = J/  Jl
} 1 I) n
lo ') II ...
n 10 0 ' 2 <  -=- · 10U' 1 ) + )' loo' ) }
  100" 1 )  ..:.J e
p.......11  p,;.lC
or
II log  < J/ n log' 2 12 + -0 (2 12).
lIenee, for all sufficiently large integ'rs 1l,
D (2 11) > 11 log 2 - J / II log 2 12 >  (12 + 1) log' 2.
If 2 12 < .'/.' <  n + 2, we obtain
f) (x) > D (211) >  (12 -1- 1) log:? :...... l,{; log' :?
for all sufficiently lal"ge x. This pro\es the first inequa]itJ in
Theorem 30.
The number (7) is clearly divisible by all prinles }) whicb are
....,. 12 and ..-: 2 1l. Therefore we have
2f1 =  ( 2 l ) > ( 2 l ) > Pii
i =0 l 11- l' I!

62
CIIAPTER 11
and, bT taking the logarithol,
2 12 log'  > {} (u) - O.{n).
If .£ = " (Ii integ'er > 1), it follo\vs that
.0 (,) = (0 :!.'I, - ,(j ."-1 ) + (/7 :!"-1 -- 0 (:?"-:!) + ...
< li 'i + 2"-1 !
" + ..» ) 10 <' !O) / 9" + l Io n' !O) = ,> , { ' Io u' ..)
- h----- - -. _.
Further, if JI-l < J: < ": ,,'e get, for all ,I: ;c '1,
{J (.{') < {} (/i) < 2" + I log. 2 < -! 'log 2,
\\"hich proves the second inequality in f).'heorel11 30.
"r01l1 Theoren1 30 ,ve easily deduce
Th('o,.en ,o £t. 1'/u.'re eyist t'fO }J(jsitil'e l-untant f';! and c 3 su(.h that
( 10)
X ,I'
c2 l - < :n;(') < cS l
og.l' ogx
far all J.' > .
In fact we have
-0 (.c) =-  logp <  (x) log' L/'.
px
and so by (u)
( )  (x) x
jt' > --'>e .
. -log'x logx
011 the other hanu. when \ve put 11 -= V;;,
1jX
{} (x) >  log')) >  10g:L' [(:i') - :t (1/ J:')] .
py
and sillce 1C (VX) < JI,
  (.e) "1/ .i"
n(.c) < 1 + v:t:<e"s l -.
og :J.." og .1.'
Q. E. D
Starting froIll the relation
lo Il! -  J[:]  [ ; ] + ) log p.

ON THE DISTRIBUTION OF PRll\lES
63
,ve shall prove
Thror(Jl1z .:/0 bo The Slun brill!! extpnded OZO(,I' all j.Jri11zes < :J.', zt.°e
hazoe thp ..t'o}"l1lula
(II)
"'" lou' P
/  = 100':1.' + 0
.... } ')  )
1'::" of'
,,'herc 0 is a /tuu.tiull (l ,i' 'lll'h that 101 i8 le. than a })()Bitil°(6
(:O}lstanto
'Ve have
n( r;;2 ] + [ ;3 ] + )logp
 ( 1I '1/
<L.J 2+3+"
p& Il}) }J
Fnrtller, ,ve see that
) log P = 1/  lo P .
pnJJ (1) - 1 )
 100' P  [ n ]  100'}J
12   > L.J  10gjJ > 11  . - 0 (n).
1):a: II jJ p n 1 p:i..1t 1
Consequently, appl.fing 'rheorem 30, 've obtain
1  100" P
log' 12 ! -   = a ,
}l 1)n jJ
(12)
,vhel'e a is a fUllction of Ii such that I a I is less than a positi¥e
constan to
For ever\" inteo'er It > 2 \ve have
. 
log 1z =-= h log It - (Ii - 1) log' (II - 1) - (h - 1) log (1 + il ]J ) ,
where the last term is less thau 1. Hence
Jt
n log n - u - 1 <  log 1z < Ii log Jl 0
It -= :I
Combilling tIlese inetjualities with (12) ,ve obtain the formula (11)
for all :r: > .
Finally we sIlall prove that the priIl1e number theorenl is
equi valent to the theoren1:
(13) lim t? () = 1.
.r- 00 X

64
CHAPTER II
"Te have
o (x) < :r; (c) log' (:
and, for 3 < !I < .f,
 (x) _ II <  (.t') -  (1/) < {j (x t  0. C ) < o.(;l ;l.
 "log !I log !I
Hence
(x) <  (x) 102' x < !llo!:!,.1': ;- log' ,r . {} ('d .
x - ,(' - :1.: log .'1 .1"
Choosing y = .l,d, where 0 ':...-:: 1 - 1 - -, we obtain
log log' .L'
,
;t (.J:) log- x .. 1Ilo'.1" 1 - J
o < -- - - -. - 1 s - + .
- D (It!) -- {) t,l') 6
Since, according to Theoreul au. () (r) > (; ..i.', ,ve baVt1
o < x) !? .r: _ 1 < x J lug-,x + 1
- {} (.I") t ..c log log .1' - 1
Here the right-hand side tends to zero for increasing' '. Hence
I ' :r;(,.C')loQ'," 1
1111 - --.. - =
,I' oo {} (x) ,
which proves the theorelu,
'rhe proofs of Theorenl 30 und of the followil]g' results in tllis
section g'ivp an iJea of the eleluelltal"Y luethods applied bJ' Tche-
bychef. It is apP1trent 110'" the constants , (1' f2 and l.3 Inay be
JeterInined nUlllerically. ""rith our siruplified ulethod it is not
possible to obtain values of the COllstants as good as those fouud
by Tchebychef.
18. Other p.toblems and results concerning prunes. - .A.. pol)"'-
nonLial in .t',
f(x) -= lIo + al'J' -I- 112 :t o2 -j-
+ all .1,11..
,v here the coefficients {.(o, £11-. .. a,l. are illtt:gers, represents in-
tegel.s for all integral values of the ¥ariable ...,
\Ve prove

ON TIlE DISTRIB UTION O}' PRIAIES
65
Theorel1 31. No lJo1!lnOl'nial f(x) f{,'itlz integral coe.l.fieients, ll:hiclt it;;
not a constant, (:an represent only j1rilnC8 for all illtl)gral
X > Xo.
Proof Let Xo be an integer such that f(:r o ) = p is a prilue.
\Va consider the identity
7l
j'(x o + p !I) -IVr o ) = ,!; ai [(:to - p y)i - x]
1=0
={li[(DX-l P.l1 + G)X-2 (py)2 + ...J.
If U is an integer, the last SUIn is an integer ,vhich is divisible
bJ }J. The integer j(..r. o + P 1/) is therefore divisible by 1). From
algebra, we know that the three algebraic equations
j(J..o + 1) Y) = 0 and f(J,'o + P y) = + P
have at most 3 Il solutions '!I if j'(.'A:) is of degoree 12. Thus, for
sufficiently large integ'ral ¥alues of y the !lumber f{x o + p!/) is
an integer ,vhich is different froln zero and different fronl + )).
Since the number J'(ro + P :tJ) is divisible by 1 J , it cannot be a
prime. Thus the theoreln is proved.
Tllere exist, ho\vever, special classes of polynolllials \vhich
produce primes for a long- sequence of consecutive integral values
of the variable. A remarkable exanlple is the pol)'nol11ial
&{' - .J: +  1 ,
which produces prin1es for the following o consecutive values:
oX = 0, + 1, + 2  . . ., + 3 Band .10.
Recently (197) 'V. H. Mills succeeded in proving the existence
of a positive constant ...-1 such that the expression
[.A 3:£]
Jields Ollly primes for all integral ¥alues of:L\ The proof is,
llo,vever, based 011 the 1l10st advanced results in prime number
theory and will not be reproduced in this book.
5 -- 5166'i0 1.'rygve Nagell

66
CHAP1'En II
Let Ii be a natural nUlnber > 3, and let r be one of the
nUll1bers 1, 2, 3, . . ..}1 -1. A necessary condition for the existence
of infinitely mauJ primes in the infinite arithmetical progression
r, r + il, r + 2 12, r -f- 3 n, . . .. r + :J.... 12. . . .
IS obviouslJ that (r u) == 1. According to a fan10tlS theorell1 of
Dirichlet this condition is also sufficient. Dirichlet's proof of
this theorel11 requires nlethoc1s fron1 higher ullal)"sis. A tIe Sel-
berg showed (1 -t9) that the eleulen tar)" 1l1ethod w hicl1 he de¥el-
oped for proving the prime nUlnber theoreIll, could also be applied
to prove Dirichlet's theorem. In the SUllie year another eleJuentary
proof of Dirichlet's theorenl ''"US published by H. Zassenhuus;
it depends 011 the theory of algebraic nUlllbers. None of t.hese
three proofs will be given iu this \'olulue. However, iu Chap-
ters IV and V we shall prove the tbeorelll in SOllIe specia.l cases.
The question ,vhether a polynoD1ial of second elrgree in one
variable will repre8ent an infinity of prilnes for iutegTal values
of the variable has not Jet been solved, not even in any spe-
cial case.
There are l11auy other solved ana unsol,ed proLlems concerning
priInes. We Jnelltioll the follo,ving' conjectures of Goldbach (17'+2);
1. Every even number .1';  tj is the SU111 of tnTo odd prilues;
2. Ever)" odd 11ul11ber J.'. :: 9 is the sunl of three odd priInes.
If the first conjecture is true, the second one is also true.
By' Ineal1S of methods froll1 hig-her analysis Vinog'l"adov proved
(1937) the truth of Goldbach's second conjecture for all sufficientl)""
large values of N. The first conjecture has not yet been proved.
But N. Pipping has verified that it is t.rue for all N <: 100000.
Viggo Brun has generalized the sie,e method in differelt ways
in order to luake it more effective and applicable to other se-
quences than that of all natural numbers. By Iueans of his new
sieve n1ethod he showed (lH19) that every positive even integer
N can be written as the SUIll of two positive odel integers Ql
and Q2,
V = Ql + li2,

O THE DISTRInrTION OF PRIl\IE
67
\vhere (il and t,!2 are the products of at Ill0St nine prinle factors.
I t has since heeOllle possible to replace nine by four il1 this result.
BrUI1 also applied his method to the stud)" of the frequency
of prilne t,vins. He sho,ved ill 1921 that, if T (l') denotes the
nunlber of pri1l1e t,-rins ,vhich are .< x, then
l ' ( ) 1 00 X
.1' < _ (I ) " ·
og ."t' ...
}t'r01l1 tllis it follo',",8 that the sris
g + ) + C \  / ) + ( /7 + 1 ) ) + ( 19 + ;:1) + ...
is COllyergent or perhapR finite. (CoJnpare the reJllark after
rrheOre1l1 9.)
For proving- a theorenl in the theory of 'roups Bertrand con-
jectured (1845) that there is at least one prilne > x and < 2 x- 2,
,v hen x > 4. It is easily seen that the truth of this conject.ure
follows fr0J11 the ine(lUalities () of Tclleb}'chef in Section 16.
From thiR the problem arises of nnding still smaller intervals
,vhieh contain at least one pritne. The best ans,ver to it was
given il1 1 H37 by lng-banl, who proved the result: 'here exists
a positive constal1t k such that there iR :tt least one In-iDle > .t"
and <.1" 1- l. :1.: 5 / 8 .

CHAPTER III
THEOR\- OF CONGRlfENCES
19. Definitions and fundamental properties. - Let n be an in-
teger =rf O. rfhe integers a a.nd b are said to be congruent mo-
dulo 'J, or for the '1n()dllllll Jl, when their difference a - b is divi-
sible by 12. To express this fact we 1\'rite
a == b (lllod 11),
where the synlbol == is to be read his congruent to." This rela-
tion is called a congfllenl"e 1uodulo n. The nUDlber n is the '1l0dlll'U8
of the congruence. 'Vhell the difference a - b is not divisible
L)'r Jl, we say tl1at {( and h are incongruent Jnod illo 11 and write
(t  b (mod u).
Thse concepts and notations are clue to Gauss: lle introduced
thenI in llis Di,"'qui8iliolles aritll1netira('.
From the defil1itions we easily g'et the following" rules for
operations \"fith cong-rnences:
I. J" a == b (nlod n) and b == C (ulod n). fhcll a == c (mod n).
For. since a - h = h II and b - ( = hI n, by addition we baye
a - c = (Ii + ht) u.
II. 1}9 ((. == b (1110d '2) and t = d (mou n), then a + t == b + d (DIOd n).
For, since {( - b = Ii nand r - d = ht 11, we have a + e
-- (b + d) = (II + Ii 1) n.
III. 1.1. a == b (n1od n) and c == d (mod II), then ac == hd (lllOd u).
For, since a -lJ = It Ii and (' - d == hI J2.. we have ar. - bd
= (b hi -1- cl Ii ...L. h hi 12) n.
In particular, bJ putting e = d we get
I v. lJ a == b (lnou Ii). then a (" == b t (nlod n).
B.r repeated use of Rule III with c == a Ilud d ::-: h, ,ve find

TlIEORY OF CONGR\iECES
69
v. I.f a -== b (Inod 11), then (('nl == b 711 (mod 11) JUl. any po,i ti l"C ex.
ponent nl.
By using successively Rules V, IV and II, ,ve get the
following more gel1eral result:
VI. I.f f(x) is a pol.lJlloHzial in x ,rith infp{]ral cop,f.tlcienfs and if
a == b (mod 12), then f(a) == ..((b) (ll1od n).
According to Rule IV a congruence nlay be nlultiplied by an
arbitrary integer. But, in genera], it is not pernlitted to diyide
a cong'ruence bJ' an il1tegcr, e\Ten if the quotients are integers.
This is illustrated by the follo,ving- example. ""'" e have 5 . 9 == 5 . S
(111od 10), but since the difference 9-3 is not divisible by 10,
the common factor [) cannot be cancelled. \Ve have, however,
the following rule
VII. If In a = 111 b (nlod n) and l:r d is the (Jreate,....f eonHnon dil'i8or
of tn and 11, then a = Ii (mOd 5).
Hence, in particular,
VIII. If In a = 1nlJ (mod u), and 7:l HI and n are relafh:ely }Jrinl(',
then a == b (mod n).
The Rules 1- VI are quite analogous to those valid for equa.
tions in ordinary algebra.
Suppose that a and b are congruent nlodulo u. Then it is
evident that t.hey have the same princil)al relnainder D1oclulo n,
and vice versa. Either of the nunlbers II and b is said to be a
re8idue of t.he other 111odulo lI. It is clear that (a. u) -= (b. n).
In the sequel the ll10dulus is always Sllpposecl to be positive.
20. Residue classes and residue systems. - The integers a and 1J
are said to belong' to the saIne residlle cla.,'s 1nodulu n ,vhen they
have the sallle principal rCluainde1" Dl0uulo -n. I t is evident that
there are ill all }l classes DIOduio Jl, corresponding to the u
possible values of the prillcipal reluainder
0, 1, 2, . . ., n - 1.

70
CHA.PTER III
The necessary and sufficient condition for the integ g ers a and b
to belong' to t.he saIne residue class Inoc1ulo n is obviously that
a = b (Juod n).
Any set of 11 illtegc1. 01, a2. ., ,all representing all the residue
classes 1110dulo 11 is called a (.o111J}lef() }"f'8id"e Sll.Yfl'}}l 11lodulo 11.
The shnplest syst eIIl of this kind is 0.. 1, 2,. ., ii - 1.
TheorC1U ':J:d. J' flu' J/atural 1111JI1bers 1n and 1/ are relafirr7!1 }J,.i1nc
and 7. i. U 11 i1l fel}'" the 11 JlllJ,t!J(',',,'
( 1)
r. ill -- r, 2 111 + i',. . . (II
1 )", : - ,.
j;/J' Ii I (1 ('O,i1 ph f(. ,.('s;,J Uf' ,"!I/?! (In; ,uoc/lil U 1/.
])rooj: It is sufficieut to fo\ho\\' that 1he nUlnhers (I) are iu-
congruent JJ1ounlo II, If , suppose
h In + ,. == k))J  l' (lllOtl nj
\vith h  k, the1) (,h - k) Il = 0 (Jnod n) and.. sinc€' (111. 11) = 1, h = It.
(lllOt1 n), ,y hich i cOlltrnry to ll)'pothesis,
Moreover ,ve can prove
1'!l(1(t}"fJU .1.:1. J./ Ihe natural 121l1UbelW In and 1l a1'f J'lativel!J prinzf1,
I)" ,c rUll,v through a ro'tnplet( ,"psiduc 8!/sfrnz .'Jlodulo '12 utld t)'
.'1 J"ll118 fhroll!lh a ("oll1j)leir re.9idue 8Yidnl }}Iorlulo 111, then the
nz n 11 ulnlJcrs
(2)
Ul ;i: + 11!'
fornz a C011zplcfr' residue '''.!I.c:ienl. 1uodulo HZ 11.
].Jroo.t: It. js ufficjeJlt to show tbat the l1u111bers (2) a.re In-
cong'ruent 111odulo 111 n. Suppose that
'lnx + nY=1nxl + 'Ill/! (nl0dnln).
TheIl
Jll.l' = 111:C (mod n) and n 'Y  11"!1I (ulod n,),
anu, since (l1l, n) -:: 1.
:)' === Xl (1110<.1 n) Llld !1 =:.:: !It (uloLl III).

THEORY OF CONGRUENCES
71
,
If a residue class modulo n contains a number prime to n,
all the numbers in the class are pl'inle to n. A residue class
111odulo n with this property is said to be prime to 12. It follo,ys
from Section 8 that there are qJ (12) residue classes prilne to 11.
Any system of q:» (1/) integers representing all tIle residue classes
prime to 11 is called a 'reduced ,-esidue 8!lsfe111 12odlilo '12. We con-
clude by proving
Theore11l B-1. I.f the natural 12lunber.t; 1n and u are 'relatit.el!ll)}'iJ2e,
"I. II runs throllgh a redllC(!d residllf 8!1....tenl 11zodulo 12, and if
/' ruus through a redul'ed rpsidue s!lstr1Jl nodulo In, then the
q:» (?n) rp (12) i ntegel..
(3)
111 'It + 1l'
fornl a l'elLu,crd residue .5'!/Itpuz 'nzodulo In 11.
Proof Since (nl l u) = (li. 12) -= (.. 1n) = I, \ve have
(uz u + n l'. nz) = (n ", 1n) = 1,
(nzu + nz., n)={Hllt. u)= 1,
and thus
(nz II + n l', 1n n) = J.
The llllinber 11l.£ + n!1 is prin1e to nl Ji only when x is prime to
nand y is prilne to Ill. Since. by Theore111 33, t.he numbers (2)
form a complete residue system nlodnlo nll1, the nombers (3)
fornl a reduced residue systenl modulo n'n.
In particular it folloW's that
q:» (nz1i) = qJ (n) qJ (n),
when (?n. n) = 1, giving a new proof of Theorem 12.
21. Fermat's theorem and its generalization by Euler. - FerInat
stated the follo\ving theorem without proof in 1640.
Theorem 3.? If]) ioY a prhnc and a an intege. not rli'isible b.'1 p,
then the ditference
a P - J - 1
is di fitibl() b!1 p.

72
CII.\PTER III
Proo./ It is sufficient to show that the cong'ruenc
(1)
x P == x (nIod ji)
is satisfied for any integer [1'. It follows from Rule ,,\11 that. we
only need consider the values .1." = 0, 1, 2. .1) --- 1. 'Ve use
matheDIatical induction. The congruence (1) holds for {t = O.
Suppose that it holds for ;1' == a. Then it nIay be shown that it
also holds for .t' -= (I + 1. The biuomial coefficient
( p ) =-= 1) (p -- 1). . (p - - I.' + 1)
k k!
is obvious1y divisible by the pritne 1J ,vhell k ;--: 1, 2, 3. . , '. lJ -- 1.
Hence
(a + l)p = {1P + 1 + ' ( 1: ) ak = a P + 1 == a -f- 1 (nlod p).
k=l It.
Thus Theorem 35 holds for anJ natural nUtnbf\l" and thprefore
for anJ integer. The first proof of it was given by Euler in
1736. He established later (1760) tIle DIore g-eneral l'f\sult:
TlzeU1.elll Bri. 1.1' n i.9 a uafural lunube1' and a f.o:: ]Jri,nr to 71, thrn
(2)
a(f':ni == 1 (nIocIu).
Proo.f. Put rp = 9'(12). Let at, a2' . . .. air be a reduced residue
SysteIn modulo n. Tben the nU111bers
'"
al 0, 02 a, · . .. air a
are evidently incongruent modulo 1l and prinle to n. Hence tbey
also fornl a reduced residue system modulo 12. Tbfrefore. taking
the product, we 1Iave
£11 a · a2 0 . air a = al a2
air (mod n),
and, if we divide each side b" the product of the nUDlbers ai,
we conclude that
a'" (n) == 1 (mod n).
Rernark. It often occurs that we already hae
(3)
a.f ==] (mod 12)

THORY OF COGRULNCES
73
for a positive exponent f' ,-' tp (n). This is illustrated by the
example n -= 12. fP(12) = 4 and
] 2 = 52 = 7 2 == ] 1 2 = 1 (mod 12).
"le shall treat the quest.ioll of the minimu111 alue of the expo-
nent .f in (3) in Section 3].
22. Algebl.aic congruences and functional congruences. - A poly-
nOJuial in :('
.((::r.) :- ':' (10 ,1: lIt . n] ;r 771 - 1 !- +- am
is called an inf('gral ))ol/l}2u}nial \VhCl1 the coefflcipn1s no al.. . am
are integers. An integ'ral polynolliial represfll1ts iutrgers for all
integral yalues of tIle arjable.l:. The integ-ral pol.non1ial f{xJ
is said to be ))rhnitire when t.he g'reatest ('0111111011 divisor of tIle
coefficients ao, aI' , am is equal to ].
Let f(,.}"') be an integral pol.nolnial and u a natu)"'al nunlber.
If e is an integer such that j'(r) is divisible bT n, t\"(\ ay that
r is a root or .ro:Ohffi(jil of the alg()l.Jraic eO}I,qrueUf.e
(J)
it,)') = 0 (llIod n),
and also that e is a root 0..( f(;:r) Hzodulo u. "Then f' is a solution
of the congruence (1), all yalues x for whi('h ,)"::E (. (mod n) are
also solutions. All the solutions belol1g'ing' to the san1e residue
class modulo }2 as e are considered as a single solution. There-
fore, to det.ermille all the solutions of the congruence (1), we
need only try the values .r == 0, J, 2, . . ., 12 - 1 (or other repre-
sentatives of a cOlnplete residue system Jnodulo 12).
If b m is the first. llunlber in the sequence of integers
b r , b r - 1, , . ., 7J m , . , .. b l , b o
which is not divisible by n, we say that t.he congruence
1)0 + 1)1 x + b 2 x 2 + ,., + b m xTll + .,. + b r 3:r = '0 (u10d u)
is of degree '}I". 'Ve spea.k of linear, quadratic, cubic and bi-
quadratic congruences according as the degree is J, :2, 3 or 4,

74
CHAPTER III
JTxa1Jlple,4;(. The COl1g-ruence
18 x 3 + :1.. 2 - 3,1:' + :! == 0 (mod (j)
IS quadratic and equivalent to tIle congruence
;t,2 - 3 x -1- 2 == 0 (mod 6).
We find by trial that. the congruence
x 5 + 2 x' + x 3 + 2 x 2 - 2 x -f- 3 == 0 (Dlod 7)
has only the solutions
.{' == 1, 5, 6 (ll1od 7).
The congruence
.v 2 + 2 == 0 (DIOd 5)
has no solution at. all.
The problem of solving the cOllgruence (1) is obviously equi-
valent to the problem of solving the Diophantine equation
.f(t) = 11 !I
in in tegers ."J.," and ll.
Let f(x) and 9 (:!..) denote t.,vo int.egral polYl1oDlials and put
.((&{..) - .'J (x) = Co .r l ' + ('1 Xl -1 + . . + (.)'.
If all the coefficients lo, Cl. . . ., r" are divisible by the natural
nUD1ber u, we say that .f(.r) and 9 (x) are ldentically congJ'"ttellt
112odulo 12, and write
(2)
f(x) == !J (x) (mod n).
This relation is a functional congJ'"ueJlce or an identical congl'ue'l2ce.
\Vhen f(x) and 9 (x) are constants, this new notion coincides
with our earlier notion of congruence. If all the coefficient.s in
((x) are divisible by Jl, f(:c) is ident.ically congruent to zero
modulo 11. 'fhe rules for operations with functional congruences
are analogous to those for ordinar:y cong'ruences. If f(x), 11 (x),
9 (x) and 91 (,2:') are integral polynomials, we have as a direct
consequence of the definition tIle following rules:

THEORY OF CONGRUENCES
75
1. If .t(x) == rl (x) and .fl (x) == 9 (::r.), then
.l(x) == !I (:.r),
I I. If I(.x) == ./i (x) and 9 (.1:) = 01 (.r), then
.r(v) + !/ (x) = Ji (r) + 91 (.1.')
r(.r) 9 (x) = /1 (x) 91 (.1'),
and
The 1110dulus n IS here a natural 1lumber the same in all the
congruences.
The identical congruence (2) llolds for everJ J, and therefore
;1' Inay be chosen as au arbit.rary integer. On the ot.her hand,
if ,ve have
f(.IO) = 9 ({o) (mod JI)
for all integral values of Xu' we cannot conclude that the poly-
nomials .f(x) and !I (.x) are identically congruent modulo 12. This
is evident frOllI the following- example. The congruence
(: + 1) (x + ). ,(x  n) == :2 .X (x - 1) . (x - 1/ + 1) (mod Jl)
is satisfied by all integ'ral values of .r, because each side of
it is divisible by 12! and therefore by n (Oorollary to Theorem 25).
But it. is obvious tllat this congruence is not ident.ical for 12 > 1.
I f the int.egral polnon1ials ,{(.r), 9 tx) and h (x) satisfy the
identical congruence
ftr) = 9 (.{') h (.1') (mod 11).
we say that 1(.1'} is rlil:isilJlfJ by 9 (x) I1l0d'lllo J1, or that g (x) is a
rlil"isOI' oj' f(x) ulor/1l10 u. f(x) is a 11Htlti})lc 0.( !J (.x) rnodlilo JI,
Theorel1l .717. ...4. nettssary and 8l(f.1irienf condition f01. c to be a root
o.f the integral polllnoluial j(:r) Jnod'lllo 12 is that ,f'(x) be di'i8-
ible !J1/ x - r nlodlilo }l.
p,.ooj: It is evident. t.hat. the condition is sufficient. If f(e) is
divisible by 11, we have the identical cong'ruence
f(J.')  j'(') - /'((.) (DI0d n).

76
CHAPTJ.:R III
Suppose that the degree of j(,:t) is ,n and that
J1l
.,
f(x) =  a1'll-J' .:t".
1-=0
Hellce
f (x) - fk) =  01/1- ,. x" =- _r"
,r -- c x -- r
1==1
n1
:.=.=  (lm-,' C'l"'-1 + C X,'-2 + C 2 .r"_3 -t
,-=1
+ l'''-I) --= [J (.r),
where !J (r) is an integral pol,'''llomial in &X of degree 1n - 1. Thus
,ve have the idelltical congruence
f(x) = (x - - (:) g (x) (nl0d 11),
which shows that the cOlldition is also necessary.
Renz.ar". One mar even consider algebraic congruences with
several unkno'vns x, !I, Z, etc. t of the type
,f(X.!I, 2 . . .) = 0 (mod 12),
where f(x, y, 2', , . .) is a polynolnial in :1', 1I. z, etc., with integral
coefficien ts.
In a following section ,ve shall consider certain types of 110n-
algebraic congruences.
23. Linear congiuences. - In a congruence of the first degree
ax + b = O (modn)
the coefficient a is (bJ' definition) not divisible by 12. If a and
11 are relativelJ' prinle, the numbers
b, a + b, 2 a + b, . . ., (n - 1) a + b
form a complete residue systeDl modulo 12 (Theorem 32). Hence
just one of these nunIbers is = 0 (uiod n). and we have proved
TheoreJll .:/8. If (a, 11) == 1, then the liuear congr'llencr
a x + b = 0 (Illod 12)
ha& e.ractly one solution llzod'lllo 1l.

THEORY OF CONGRUENCES
77
By means of Fern1at's theorem (Theorem 35) the solution nlay
be ,vritten out explicitly by the formula
..t = - b a(r('IZ)-l (mod 11).
Theorelfi 38 is COl1tail1ed in the more general
TheoreJu .!J.9. The ("ongl'lfencf
(1)
ax+b = O(modll)
has 'nO .oltltiuns or exal-t1ll d = (a, 11) sulut'ions, aecording at\; b
is nut, or is, a J1Utltiple oj' d.
Pro oj: When the congruence (1) has a solution, it is clenl"
that d = (a, u) is a divisor of b.
]f we assume that b is divisible by d, the congruence
a b ( Il )
d:!' + l = 0 mod it
has exactly one solution ;1.'0 modulo  (Theorem 38). Then the
congruence (1) is satisfied by all the numbers Xo + !I  . where .II
is an arbitra.ry integer, and only bJ' these numbers. Hence the
iucongruent solutions of (1) modulo It Ina.y be represented by
the d number
n n }l
;.£'0' .1'0 + d · ,,'o + :! tI' - . ,}'o + (d - 1) d .
Finally \ve IJrove the follo\ving theorelll on sets of simultaneous
linear congruences.
The01-enl 40. If no !It'u uJ" the natura11l11'nzbers 111' J12. . . ., I1r (r > 2)
has {( runnnon dit'i".'or > 1, fhe cu12gr'llenc£:s
(:?) x == al (nlod 111), lor == a2 (n1od 172)' . . ., x == ((r (mod Ilr)
£I 1 Zfll.l/t'" hare COl1l1Hun solufiul/, ,l'. These sulttfi()12s llre .f/iff'Ji IJ.'I
the 1l1unbel"8 ill a c(-rtuin re",itlllp cla,-.:s 'l1zCJl!ulo J.' _-: lit J12. - Ilr.

78
CIL\. (JTER III
Pr(Jo./: If "e put .1'1 = lit 'I'i (i = ], , . . ., r), then clearly
('1'1' "'2' . . ., "r) = 1. No,v, according to Theorenl 6 (1hapter J)
the equation
'1'11/1 + 'J'2 !12 +
+ Vr!11' = 1
llas integral solutions Yl, 1I2' . . ., .'Ir. !)ntting' 1'i!/l = i (i = 1, 2,
. . ., 1'), \ve have
2'1 + ? 2 + ... -j Z'r = 1,
where 2'j = 0 (ulod n;), if j  i, and 2i == 1 (DI0d lii). Thus the
systeln (2) is satisfied by
(3)
x = 01 2'1 f- {{2 c2;- ,-+ {lr Zr (mod ..L,T).
If further .1'0 is an arbitrary integer satisf.ring the system (2)
and if x is defined by (3), the diffel"el1ce J.' - )'0 is obviousl)"
divisible b.r all the numbers Ui (i = 1, 2, . . ., 1') and thus also by
their product 111 U2 · · . 111' = N. Hence "ro belong's to the same
residue class modulo ,. as :1".
This t.heorem was known b. the Chinese Inatbell1atician Sun
Tse (about A. D. 250).
Exanzples.
1. In the linear congruence
ux = B (DIod 10)
a = U, b = - 3, 12 = 15 and thus ,f = (a, u) = 3. Using' the saIne
method as ill the proof of Theorem 39 we find ,I 0  3 (Illod 5),
and hence
.l' -== 3. . 13 (mod If>).
2. In the simultaneous system
x == 1 (mod 4), ,I' == '2 (mod 3), x == 3 (Inod 5),
al = 1, 02 = 2, as = 3, iii = 4, 112 = 3, Us = 5; thus _,r = t10 and
'1'1 -= 15" '1'2 = 20, Vs = 1 . The equation
15 1/1 + 20 .1/2 + 1  !Is = 1
has the solution 111 = 'Y2 = -I, .'Is = 3, hence ?1 = -15, 2 = - 20,
Zs = 36, and according to fOl'D1Ula (3)
.1.: == 53 (nlod 60).

THEORY OF COGRlTENCES
79
24. Algebraic congruences to a prinle fil0dulus. - In the fol-
lo,ving' 'fheorenls 41--1-3, j'{x) denotes an integ'ral polynomial
such that the alg'ebraic congruence
j'(x} == 0 (mod 1)),
where jJ is a given pl.i1ne J is of degree )l. rrhus, when ao denotes
the coefficients of xf1 in f'(,I:), 00 is not divisible bJ p.
Theor()" 41. JrluJn the algebraic cungrlu:nce of dl J gl'ee 'In
f(x) == 0 (nIod }I),
tchel'e p is a l)rhne, ha.9 S infOIl!li'lient 'alu.tion.,i ('1, t2, . , ., e s
'I11odulu p, the jollalting identical congru('nce holds:
(1) f(.r') == lx - (4 1 ) (J' - C2) '(..c - c,) 9 (c) (mod p).
Here lce lzat'e 9 (x) = Uo x m - s + Ii (x), 'Il"here h (x) is an integral
}Jolynonlial of degrel) < 1H - [: -1. In particular, if 112 = ., u:e
haz'e 9 (,i") -=--= (/0
ProD): vVe use nlathemat.ical inductioll. The theorenl is true
for L = 1 (according to the proof of Theorenl 37). Suppose that
it is true ,vhel1 the nUl11hel' of incongruent solutions is < .' - 1.
\Ve shall establish its truth whe11 this llunlber is =-- t. By hypo-
thesis we have t.he identieal congruence
(2) ..{{.J:) = (,{' - el) (,t' - C2)' (x - Cs-1)j (d') (IIlOU p),
,vhere /1 (a.')-aoa.m-8+1 is an integ"ral polJl1omial of degree < l1Z-S.
Since Cll is a root of f(J..') == 0 (ll1od i)). it is cleal. that
(ell - ("1) (rs - C2) . .. (('8 - (.s-l)ji (e,) = U (ulod 1)).
Hence
11 (C g ) == 0 (mod p).
According to Theorem 37 ,ve must have the iuentical congruence
fi (x) == (x - c::) y (x) (1nod}J),
where !1 (x) - (tox'f7I-' is an integral polJnoluial of degree < 1n-s-l.
Combining this congruence with the congruence () ,ve get the
congruence (1). Thus the proof of the theoren1 is cOIlJplete.

80
CHAPTER III
Tlzeo ren1 4. The algebraic congruenre 0..( de!li'ee 1n
(3)
f(x) ;: 0 (IDOdp),
u:here p is a l)ril1l(, has at 11l()st U1 iUC012(lrlient 801ution.-:
lnodulo p.
Proof. Let us suppose that t.he congrllence (3) has the in-
congruent solutions C.1, (42' .., (4m l11odulo p. Froln TheOl'eUl 41
we conclude thu,t
f(x) = ao C-c -- (.1) (.c -- (':!) .. (,.t' - r,lI) (IUOJ 1))'
If there 'vere anot.her solution r illcong'ruent to c; (i = 1, 2.. ., 111),
\ve \vould have
ao (r - Cl) (i' - C2)
(j' - l41}l) = 0 (mod p).
This is ilnpossible since none of the nUDlbers llo and r - Ci is
divisible by p.
The theorenl is not true when the modulus is a composite
number. Thus the congruence
.);2 - 1 = 0 (IDod 12)
has the four incollg1"uent solutiolls .I.: = + 1. + 5 (DIOd 1 ).
Thl'01"tlU 4/1. I::;ulJPO,""l' that the alflcbraic: l"uu!lrU('U('lj 0/' de!Jree Ol
(-t)
.((x) == 0 (UIOU }J),
u'here J) il!l a jJriuze, has nl inC(J11(}I'U(>ut sulutions '1llodulo p.
'uppuse also that !I (,t') is a d icisor (!f' j(.r) lli()dulo i), and that
tlu' COllgrueuce
(5)
[I (x) == 0 (mod p)
i.'" of degrf'(J It. Then fhe lat congruence has exactl]1 It in-
tUJl/)9bleut solutions J1lo(,lulo p,
Proof: By hypotllesis we have the identical congruence
(6)
j-(x) == 9 (.1") Ii (..t') (IDod p),
,vhere h (x) is an integral }JolynoD1ial. Suppose t.hat the congruence

1.'HEORY OF cO'GnUENCES
81
(7)
11 () === 0 (u10d }))
is of degree 'P. Then \ve g.et froIll (6),1( . J'':= nl. Further. suppose
that the congruence (5) has exactly III illcong.ruent. solut.ions
Illodulo}J and that the congrupnce (7) has exactly "1 incongruent
solutions Inodulo 1). Then it follows frol11 (G) t.hat everJ root (-
of f(x) modulo }J is a root eitber of g (x) or of h (.1,) 1110dulo }J.
For if .f(c) is cliyisible by }J, so is [I (l.) It (c). Hence we ha\Te
/11 + Vl > 11l. I t follows frol11 Theoren1 -1-2 that PI < }L and "1 < P,
and thus /11 + "1 < nl. Hence \ve have ,lit  J'l -=}li and PI = Ill.
"1 = v. This proves Theorenl -1-3.
Finally \ve prove
Theore,n 44. Tilt> In.or/uei ()j. lu'u }Jrhnifirf" pul!/noJnials i. also a
jJl.i1nilit.e j)Ol!!J1UJllial.
f.Jroof Let 9 ("f) and h (x) be t\VO priInit.ive integ'ral polYl101nials.
Suppose that t.he product g (:,c) h (.1:) is not primitive. Then t.here
exists a prin1e }J such that the identical congruence
(8)
g r:) It (..c) = 0 (mod 1J)
holds. \Ve no\v foru1 .'Jl (x) froDl g (x) and /1 1 (a:) from h (x) by
rejecting all terll1S of [/ (..r.) and h (x) whose coefficients are divis-
ible b.y p. The polynoll1ials 91 (x) and hI (x) have the following
properties. They are both different from zero. The coefficients
of the bighest powers of .1." are not. divisible by p. They satisfy
the identical congruences !/l (&:) == 9 (:.t) (n10d }J) and h l (:.Y') = h (x)
(mod }J). Hence the ident.ical congruence
91 (x) h 1 (.1.) = 0 (IDOd }))
holds. This is inlpossible, ho\vever, since the coefficient of the
highest power of .1' in the polynoD1iai 91 (x) h 1 (x) is not divisible
by p. Hence the product (J (,.c) h (,) must be pril11itive.
25. Prime divisors of integral polynomials. - Let }-(;,o) denote
an integral polynou1ial which is not a constant. If c is an integer
SUCll that .l(c) is divisible by the prinle p, we say that p is a
p,-z''1ne dil:i.or 0.( the }JolynoJnial ftx). '\Ve prove
6 - 5166;0 Tryyt'e]..7 flgell

82
CHAPTER III
Tlzeorenz 45. Erery integral polYllo}nial ,((.1') lehiell is not a e012staut
has an i}finity of pril1e difisur'.
Ptoof. "7' e consider the integral PolYl10mial
f(,::) -= Go x m + £11 m-l + ... + am,
where tn > 1. \Vhen am = 0, all the ])rimes are prinle divisors
of j-(J:). Thus we Inay suppose llm  o. N O\V aSSUlue that }-(,I")
has the prinle divis01"s 1)1' 112'1 .. , }J.. and 110 others. If
b . PI })2 . . " p.. am,
we get
j' (b 1/) - -: am g (!I) ,
where
9 (y) = 1 + 4l ,,'I + ...1 2 ,,1/ 2 +
. + . 1 Y III
, .. 711'
is an integral polJl1onlial in 11. The coefficients Ai are nIl divis-
ible by the product. }Jl }J2 . . · p.,. Henee none of the prinles
1)1, )12' . . ", pI' can be a prilue divisor of the polynornial !I (y).
Every prinle divisor of g (!/) is a v rime divisor of ./'(,J..'). "T e nIllst.
consequently have g (1/) == + 1 for every integ-ral value of y. The
algebraic equat.ions 9 (1/) = + 1 have, ho,ve,er, at most :2 Hi roots.
Thus it is possible to find an integral value of !I suell that 9 (y)
is divisible by a priule different. froln the primes })l, 1'2, . . . 'I }J.,.
Hence Theorenl 45 is true.
It nlay also be fOl'Jl1ulated in the following' manner:
JJThen f(x) is an integral 1)oll/}2CJH2ial ll'hirh is not a C011stanl,
the ()ongrueuee
f("1."') = 0 (Dl0d 1))
is 8()1'able ..tJ' an injbiitll of l)/'inlCS p.
It is easily seell that all primes are prillle divisors of the lineal'
polynomial a x + b, except the primes 'v 11ich divide a and not u.
For the lineal' congruellce
ax + u = 0 (nl0d}))
is al,va.vs solvable when the prinle 1) is not a divisor of a  The-
Ol"em 38).

TIJEORY OF COl'GRUENCES
83
In Chapter IV "e shall determine all the prinle divisors of
all.V }JoIJllonlial of the second deg'ree, and in Chapter V those
of sonle other t.ypes of polynonlials of hig'her degree.
.. ,'"
J-- :26. AIgebl"aic congruences to a composite modulus. - Let f(x) be
all integral polYl1onlial in x, and let 12 be u, composite integ'er
SUCll that
12 = pl 1)2 . . . p r . (Ui > 1),
\",here })1, ])2' . . ., PI' a.l"e distinct pI"inles. Then it is e'f'ident that
the problenl of sol ",iug the algebl"aic congruence
(1)
J'(;t) == 0 (IlIOd J/)
is equivalent to the pl"oblelD of solfing the s)"stenl of algebraic
congruences
j'(x) = 0 (mod l)J),
J'(,(') = 0 (lllod lJ),
........ .
1(:1.') = 0 (ll1od pr).
If congruence (1) is sol \"able, e'f'ery cOl1gruellce in this system is
sol'\able and vice versa. Suppose that the first congruence in
the system has the solution e1. tIle seconu congruence the so]u-
tiOl1 ('2- etc., ancl fil1all. the }'th congruence the solution (41'. Ac
cording to Theoreul -to (8ection 23), the ysteln of cong'l'uences
f .C = ('1 (11I0t! }JI),
() , -- (1110d p'l)
,I. -= l.,)
'j' . 
. . . .
,e = C r (Inod })r)
bas exactly one solution E l'(presel1ting all the nUlll her in a
certain residue class n10dulo 11. Thus the llUl1Iber ,; is u solution
of congruence (1), ana "e obtain nIl the incongruent. solutions
1110dulo 1l of this cong'l"uence when in sJstelIl () the l1unlber
()i (i = 1, 2, . . . 'I r) ruus thl"ough all the incong'ruent solutions
IHodulo l)i of tIle congruenc
(3)
.fCe) = 0 (IJlod )j,).

84
CHAPTER III
In particular we state
Tlzeoreni 46'. IJ" eongrlU)llCe (3), u)here i = 1, 2, . . . , r, lUI' Vi 111-
congrucn t solutious I1lod itlo pi, then congruence (1) ha...,'
1 1 1 )'2 . . . I'r
incongrUt")nt 8olutiont"; 11lodlllo n.
From the preceding results it is e,ident that we need only
consider cong)'luences ,,,,hose moduli are powers of primes.
Applying Theorenl 46 we prove
Theore,n .fi. Let n b,- a naturalllllJizbcr such that
}I == ..).'3 j lt l 1 ,.a"} · . . l )a r
- 1 "2 r '
u.,here 1)1, ])2' . . ., ]J r are dil,.tiucf odd lJrillles, and denuti J b:ll \,r
the nu..Jllber CJ..f incongruent sulutiuns rlloclulu 12 of the eongruence
 2 - I == 0 (mod n).
(-1)
Then ..'T = r .fur (3 = 0 a'2d 1, T = 2"+ 1 IUI" fJ == 2, and
..,or ==- 2'+2 ..ful" f3 > 3. This is alsu true for r == 0, i. e. 1()IU}2 u
i.f) a jJou'{Jr 0.( 2.
p1"oqf. As just proved above, we need only cont:;ider the con-
gruences
(5)
L{.2 - 1 == 0 (mod}>fi), (i = 1, 2, . . ., r),
and
(6)
.r 2 - 1 == 0 (ulocl 2.i).
It is evident that the congruence (5) has ouly the two solutions
x == + 1 (mod pr i ). The congruence (6) has for {3 = 0 and 1 only
one solution, which may be represented by the number 1. For
fJ = 2 it has the t,vo solutions + 1. For {3 = 3 it has the four
solutions + 1, + 3. Suppose next fJ > 3. Then ouly oue of the
nUDlbers x-I and x + 1 is divisible by 4. In this case a neces.
sary aud sufficient condition for x to be a solution of the con-
gruence (6) is that x = + 1 (mod &-l); thus we have the four
solutions

THEORY OF COGRUEN"CES
85
.r == + 1, + I + ,-l (nlod 21).
Hence, applying' TheoreIn 4(-), the theorenl follo'v's as stated
above.
27. Algebraic congruences to a prime-power modulus. - Con-
sider a primitive integral polynolnial
1(,1'.) = ao .).,111 ;- al.l,7u-1 + ... + 0nt-! LX + am,
\vhose discriminant D is different from zero. 'Vhen  is a,rbitrary
we have by Taylor's theorelll
(1) /(x) = fW + (x - E)f' (E) ;- (.f: - Ej2 1':) E) + .
.,J .
+ (x - E)m j'(m j (E) .
711 !
In this identity the coefficiellts
} f '{rl ( 1: ) = a ( tn ) 1:r;,-r' ( In - 1 ) l:m-l'-1 -+-
r !'  0 r  I a 1 '"  I
+ (Im-r
are ob,iously integers if E is an integer, Suppose that 1) is a
prilne and .; a root of the cOllg'ruellce
()
,((x) == 0 (ll1od p),
Then Ll() ha the divisor [() -  1110dulo}J (Theorem 37). If 1(:1;)
also has the di,isor (,r - )2 11lOdulo 1) we 8ay that  is a 11zztltiple
roof of tIle conruenee (2). Otherwise E is a sirnple root of this
congruence. If  is a lnultiple root of (2). we obtain by (1) the
identical COl1g'ruence
(3)
,l(:r) = (.)' - ,)2 !I (:/) (1110d p),
whpre 9 (x) is an intel'a1 polynolnial in x of deg're(\ 111. .- 2. Con
versel)". \"ben thiR cong'rueuce 1101ds, it fol10,,,s fl"OD1 (1) that .f' (E)
is divisible by jJ. "Theu ,l() and r' (E) are both divisible by p,
congruence (3) holds. Hence a. necessar.r and sufficient condition
for the root E to he u, D1ultip1e root is that j" (,) be divisible
by}).

86
CIIAPTER III
Fro In a.lg-ebra we kno,v that the discriluinant. of a polynolllial
"f()) is an integral polynonlial in the coefficients of f(x). Hence
it follows from (3) that the discriluinant J) of the pol:'nontinl
.f(It') is congruent 1110clulo 1) to the c1iscri]uinant VI of the polJ-
110Dliai (Il' - .;)2 g (x). ince ])1 -= 0 and J)l = 0 (IlIod })), we also
have D = 0 (nIodlJ) anJ we state
TheOl.eHl 48. 1.1' rOllgJ'llence (2) hllc.' a 111I1lti})I() r(jot, tlu l discrinll"nanf
D is diz.isiblp. lJ.ll the }J1.hne i).
By a similar argull1ent. it is eas' to esta,blish the 1l10r£' general
result
Theorenz 4.9. Let J) 1J(> a }>rhne, and ltt I (l) g (x) and h (,,) b(J
pri1nifirc illtclJral }jolyno112ial. ,,"ati..f!lin(J the identical C011!/J'ucncc
"1'(,/') = (g \') )2 Ii (..I') (Illod p).
If 9 (..r) i,,;: not identically eOJ1grlu.nt to a constant 1nodulo ]},
the dil'-1crllninant ]) o..l f(Lr) is di,.isiIJ1,. bll 1).
If 'e know the solutiollS of the cong'ruellce
(4)
f(Lr) = 0 (n1od pee),
(a > J),
it is possible to deduce the solutions of the cong l rU{\1)ce
(5)
.(l.}) = 0 (Illod 1it+1).
It is clear that e\ery root of (5) is of the form  -I f 1)'':, where
 is a root of (4) aucl t an integ.er. "\Ve seek a. value of t such
t.hat  + tpft will be 8 root. of {f>). By Taylor's tlleorenl we bal'
f( -;- tp") = f(;) + .r (;) f pI: or f'(!;j (I p")2 + ·
or
.f( + t })u) == f() + j' (;) t pel (mod l)a+]).
Hence we hate to solve the cOllgruence
(6)
j W t = - fW (modp).
pa.

THEORY OF CONGRUECES
{',-
Ol
If the lluluber j" (E) is not divisible b)9 p, that is, if E is a sinlple
root of congruence (:2), congruence (6) has exactly one solution
f 1110dulo p (Theorelll 38 in Section 23). If f' (E) is divisible by JJ,
congruence (6) has either 1) solutions lllodulo }J or none at all.
Hence we ha.,e proea
Tizeoreni ;j(). 1.('; is a cin1])1f? 1'OOt of congrtlence (2). (.ongruenc£' (-1-)
has (',cactl!! OJH! 1.oot llzodulo It' rong'ruent to E 'J1l0dlllo p.
Theorc112 51. {f E i,-,: a l1lultiple roof 0.( C012.qru.encc (2). congruence (4)
h((' at nzost 1)'1.-1 roof.,-: 1110dulo pU congrll(>nt to E t1z0dulo p.
('olllbining Theorelns 48 and 50 we llate
Tlzeorp1Jl :')2. SU}JpOl?t> that the discrinzinant D of the lJrinziti'l:e
intt.gral ]Jolynonzial j(Lr) is not dil.i.i/Jlr. by thi! pri'lne 1). Denote
by ...\-r the nu.,nber q( incongl.lleuf S(jllltiol1. oJ' the congruence
()
,l(.r) == 0 (nlod 1)).
Then tll(" lllonbl'1" of iucongl.ueut .01utio12S OJ" tilt> congruence
(4)
.l(x) = 0 (mod 1ft)
is al..'o \....
Suppose next that the prime }J is a divisor of D and that. ]J
is diisible by 1J.II. and not by p.ll-l- 1. If congruence () has no
Illultiple root, Thorell1 52 is still true.
SUPPOS(\ further that E is a root of t.he congruence
.1'(,:r.) == 0 (nlod l)U-i 1)
and a lllultiple root of congruence (2). Then the nUlll bel" .t J (.;) is
at ITlOst diisible by p.LL. For if this number were divisible by
p'l+l, it ,vould follo,v from identity (1) that an identical con-
g-ruence
f{x) == (x - ';)2 g (x) (mod p.IL+1)
must hold, where 9 (.x') would be an integral polJ'Domial in x.
By tbe property of the discrinlinant D just mentioned we would
hate
D == 0 (modl).'L+1),
which is contrary to hypothesis,

88
cnAPTER III
'\Y e now suppose that ..(' tE) i8 divisible by }1' and not by pf1+1 
hence fJ < !-t. Also t.he llUnlbel" .I" (E + t pLl+l), ,,"here t is an ar-
bitrary integer, is clearly cli\"isible by p,i and not by pp+l.
Suppose further that 1 is a root of the congl'l1enCe
(7)
.f(x) == 0 (ulod pa),
".hCl'e a > '2 Ii + 1, and 1 ==  (luodl/ 't !). 'rhcn Utll the 1)' l1u111bel's
(8)
E 1 .+ It pa- ,i (It = 0, 1 'I :!. . . .. 1) - - 1)
are incongruent roots of cong'l'uence (7). For since .f" (1) is divis-
ible by pf J and since 2 a - :? {J > 2 a -  It > a + J. it. follo\vs
that the nUlnber
,((1 -1- Uj)u;-,I) =j'\;l) + 1l1l-IJ.r'(1) t-  (U}Ju- i 3)2j'" (1) -1
IS di ,isible bJ pa.
Hence we see that the incong'ruent roots of cong'ruel1ce (7),
provided that they' are nll1ltiple roots of (2), forD1 a c(rtain nUD1-
bel' of syste1ns of t.he type (8); and eery sJstenl contaills exactlr
}J.j incongruent. roots.
Let us determine the l.oots of the congrllel1ce
(9 )
f{x} == 0 (ll1od l)Ct+l)
correspo11ding to the roots () of congrnence (7). They are of
the forln
E 1 -r- "J/l-:-1 -1 1.'1/" '!
". here II and ,. are integers. "or  ha ,e
j'(E l + U})t-13 + fJlt) = .r{l + 1(1)':-::1) + t"1)Ct. I' (1 + II pl.-i) +
and. since .(' (E 1 + It ]t!- I) is divisible by 1),
."(1 + u lJa-;i + t'J/() == .t'l;l -1 1£ l)a-p) (nlocl}l!+l).
Since 2 a - 2 P  a -I- 1, ,v{' hn ve
.f('l + Ujlt-,:J) ==,1'('1) + ttpa-i-J./,' (1) (IlIod pa+l).
Here the left-hancl side is di,isiblc bJ }l,!-l.l if und on])" if
.J"tl) + II j/r-.J.(' (1) == 0 (mod pat 1 ),

THEORY OF CONGRUENCES
89
where u is deternIined by the congruence
,f" (E 1 ) - I (1) ( d )
u = - - DIO } ').
lJ })"
Since j--' (;1) is not divisible b)9 p{1+1, this congruence has exactly
one solution Ito nI0dllio p (Theorem 38). Hence. corresponding
to the 1)''1 roots (8) of congruence (7) we have derived the following
pi:J roots of cong'l'uence (9)
(10)
'1 + (u o + hp) pa-i + rpa.,
where h = 0, 1. 2, . . ., J),i-l - 1 alld ,. -= 0, 1, 2, . . ., ]J - 1. The
l1unIbers (10) are il1congruent modulo })atl. Every root of (9)
cong'ruent to El I1I0duio }Ja-I must be congruent to some of the
p.'1 nunlbers (10) DIOduio j)a+l. If we put '1 + Uo})a-j'J = '2' we
see that the systenl (10) Inodulo pa+l can be replaced by the
systenl
'2 + f }Ja-.:1+1,
where t = 0, 1  :!, . . " p.3 - 1. This BJstem is analog-ous to the
sJsteuI (8).
Finally we conclude that the nUlllber of roots of congruence (9)
is exactly the sanle as that of congruence (7). This is also true
for the roots \hich are sitDI,le roots of (2) according to Theo-
relll 50. Hence
Theorenl :'),!J. Let J) be a }Jrz')nc and .';"UJ)]J(p:e that the discriuzina11f
D oj' the pl'iHlifire integral polynoJnial 1(."") is tlil.'i.'?tble lJY ]J.ll
and not by p" + 1.: then tlu? C012!Jr'll('nl'e
,t--(.r) == 0 (nIod J)ct),
It'here a > 2 p. + 1, has eJ.--arf11J the sa'U2e nunzber 0.( roofs a.'? the
congruenl'f
(11 )
.(([)J) == 0 (moclJ)2.1(+ 1).
Of course roots nlwa'Js means roots incongruent for the modulus
of the congruence.
If congruence (2) has exactlJ ).. sinIple roots and exactly Al

90
CHAPTER HI
Dlultiple roots it is evideI1t by Theorenls 50. 51, 53 nnd 4:2 thtt
the nuulber of roots of congoruence (11) is at nlost
A. + ).1 p2." < n2 D2.
Hence
TheoreJii 54. L(,t f(J:) /.Jf} a jJrinlifir(1 inlr!]ral }Jul,llnO}llial 0.1' de!l}'pe
1}1 and lfitli a disrrhnill(lnf D d,:tfereut .Ii-(jrn zero; then flip
(:011 (/J'ue /lc:e
.1' (,I') == 0 (1110<.11;'1)
has at. J}108f 1Jl D 2 rools.
This result '",as found in 1 n:? 1 illc1epeuclcn tly by Ore and 1 he
author.
28. Numerical examples of solution of algebraic eODgruences. -
""e shall illustrate tIJe theory de\eloped in the preceding section
b)? SOIll e ex RIll pIes.
1. If lJ is a pritlle: we find by trial tluLt the eongTuence
,L,3 - ",2 - 2 x + 1 = 0 (lnod p)
is solvable for }) =-= 7, 13. D. l, 43. .. 1, 83 and D. and for no
other priule <.: 100_ 1f }i:---" 7. the congl'uence has three roots.
For p = 7 it has t,he triltle root x = - 2 (nlod 7), as we RPe from
,,3 -- x 2 - 2 x i- 1 = (.' -+- 2)3 (tnod 7).
The discriulillant of tbe polJuonljal on the left-haud side is =--: -19.
It is possible to show that the cong'ruence is solrable for }J = 7
and for all prilnes of either of the forn1s 7 t + 1 und 7 t -- 1 and
for 110 other prinles.
:? If 1) is a priule. \ve find by trial that the congruence
a i ;3 - ,' + 5 == 0 (lnod p)
is sol\able for p = 5, 11, 1 a, ] 9, 3. 29, 31. 37, 1, 43 and is
not solvable for JJ = :? 3, 7, 17. For 1) = 5, 9 and 4-3 the con-
gruence has three roots and for }J == 11 t.o roots, one sinl vIe
and one double. For the ot.her values of p there is onl? one
root, which is siInple. TIle discriminant of the polynon)ial on
the left-hand side is -==: - 071 =-: - 11.61.

THEORY OF CONGRUENCES
91
3. The discrinlin3ut of the polynonlial
j'(x) = .3 + 3J + 9
is [) = 0- 3 3 . {) . J 7. Since
.((x) == {.3 (mod 3),
the eongruence
f(x) = 0 (nlod 3)
has on1.r the t.riple root. .r::= 0 (u1od 3). By the Iuethod deYelopd
in Section 27 we find that the cong'ruence
j'(x) = 0 (Juod 3 2 )
has the roots :)J = 0, 3, 6 (Iuod 9), and further, that the congruence
f(x) == 0 (1l10d 33)
has t.hC' roots .1: == U, 15, 2.t (nlod 27). ,\'hich are all congruent to
H llloclulo 9.
Finnlly', ,ve consider the congruence
(1 )
.j'(/:) == 0 (Juod 3't),
whelee a : . 3. If this cong'ruence has t.he root El. clearly it also
has the three roots
l: l: _: ),-1 l: ...! .). !1-l
l. l I.) , l I - tJ ·
Suppose t.hat eongrucnce {I') has only these three roots. Tl1cn
we s]lall show t11at the congrnellcC?
(2)
j.{{') = 0 (ulod 3(8;'1)
also ha s (xactIJ three roots. Eer'y root of (2) lllUst be of the fornl
1 + ll. 3 a - 1 + ,.. 3(!,
W]lere if a.nd t' are integers. The l1Umbel- i' (x) == 3 ('t2 + 1) is
di ,.isible by 3 and Iiot by 9. 'Ve haye
.I'(l  u. 3,:-1 -;- c. 3:')  ,.f(;l t- ll. 3 c !-1) (ulod 3(!+1),
:tnd, since 2 a - 2 > a  1.

92
CHAPTER III
J'(;t) .+ ll. 3 a - I .!'" (';t) = () (n1od 3(%+1)
or
.r 1) . U = _ l:l) (mod 3).
rrhis congornence has exactly one root It. Hpnre, it is ShO'\'ll that
congruence (1) has exactly three roots for all a > 3. For a = 4:
,,"e find the roots .J; = 6. 33, GO (mod 3') and for a = 5 the roots
r: == 6, 87, 168 (mod 3 5 ).
Fro 111 this example ,ve see that it is advantageo.us in special
cases to Inodify the nlethod.
.t. We choose the salue . ((..r) as in the third eXR1npie a.nd
consider the congruence
J(:f) = 0 (n10d 5).
It is obvious frolll
/(.1') = (.1" + J) (.t + )2 (1110d 5)
that the congruence has the double root. 1.' == -:! {u10d 5) and
the simple root x = - 1 (ll1od 5).
Now the congrupnce
.r( - :?) -;- u n.t" ( - ) == 0 (tllod [)2)
or
- f) :- 7f) If == 0 (lDOd ::?5)
has 110 solution II. Renee, it follows t.hat all the solutions of
the congruence
(3) f(x) == () (mod 5 a )
are =: - 1 (Inoo 5) \Then a ;;> 2. This cong'].ueuce has therefore
only one solution.
5. The polyno111ial .ll.p) =- }', - .1'2 + 1 has the discriminant
D = 2 t .3 2 . \\Te find
j'l.l.') = (.1'2 ,- 1)2 (nlod 3).
Th us the congrruellce
.!,(x) = 0 (Juod 3)
has no solution at all.

THEORY' OF C01SCRUENCES
93
G. The pOIJ'l1oulial ..((:x') = .i: 2 + .1' !- 7 has the discriluinan t
D = - 27. \Ye find
j-'(..) = (x -- 1)2 (lJlOtl 3).
'fhus the congruence
Il.£) == 0 (lllOd 3)
has the double root .t; == 1 (1110<1 3). The congruence
.
J'(.r) = 0 (lnoc1 3 ft )
has fot' II =:! the 1.00tS 3.' == t) 4, 7 (DI0d 9), and for (1 -= 3 the
roots x ==..1., 13, 22 (luod 27); for (1 > ..1. it has no solution.
7. If ,ve choose
J'(.e) = .5 + ..('4 - U,3 + 2.r 2 +.l: + 1
and 1) = 7, then we have identicall)"
f(x} == (..r + 1) (,1"2 + t)2 (nlod 7).
lIence the congruence
..((.1.) = 0 (ulod 7)
has the unique root .c == -- 1 (lDOd 7). Thu8 the coug"ruence
j'{x) = 0 (olod 7(')
has exactl)" one root for evel"Y u.
29. Divisibility of integral polyno mia1s with regard to a prime
modulus. - In this section polynomial means integral polynomial
in x. Let p be a given prinIe.
Every polynomial f(x) which is not identically congruent to
zero modulo p may be written in the f01"n1
(1) f(:r) = ao :t,m -+- al.:c m - 1 + ". + am + }J g (x),
,vhere 9 (..l') is a polynomial, where the coefficients Qo. Ql, . . ., am
are integers and where ao is not divisible by 1). Then j-'(x) is
said to be of degree 'In modulo )J. If I(x) is a constant not
divisible bJ' }J, then ..l(.t') is of degree zero modulo 1).

94
CHAPTER HI
The product of two polynomials of deg'rt'es ,M anti v is obviously
of degree It + v 1110dulo )J. Hence the product of t\'fO or 1110rn
polynonlials is ic1en tica,l1y coug'ruen t to zero lnod ulo 1J if and
only if at least olle of the pol"nomials is coug'ruent to zero
nlodu]o jJ. ((;0111pare rrheorelll -+-+ in Section 24.)
T,vo polynoluials \vhich al'e ident.icall}. cong'rueut 1l10dulo jJ
are said to belong t.o tIle saIne ,"e.\:idue clas". lJ2odulo }J. The 11U111-
bel" of polynolnials of degree 1Jl incougruent n1odulo ]J is obviously
()) - I} }Jill. rrhe polynonlial (I) is said to be lJrl1l1arll 11lodulo jJ
\vhen Clo -== 1. The nutnber of prilnary pO]YDoInials of dE?g'l'ee 1n
incong'ruent lnodulo p is IJ m .
In every residue class ulodulo p there is exactly 011e polYl1oIniul
the coefficients of \vhich are > 0 and :5;. }J - 1. 'l'his PolYllolnial
is called the u01 4 Juai )Jolllnol1zial 1HOdulo JJ iu the class. The residue
class ,\""hich consists of all the poly:nolnials ideutically cong-ru( l1t
to zero 1l1odulo p has the 1101'lnal pol)'Dolnial 0 (the COllstant
zero ).
""'"hen the polYl1oDlials ../(.r}, !/ (.r) and h (l') satisfy the identical
congruence ·
(2) f(X') = [I (x) h (,I') (1110U p),
/(.1') is divisible by y (.1:) l1lodulo J), [I (./') is a di'\isor of .l(J:) DJO-
<1ulo .J} anu ..(l.c) is a ulultiple of [J {."J:) 11louulo jJ (definitions in
Section 22). 'V'e sa}. also that ../(x) is tl1e product n1ot1ulo }J of
9 (.r) and h (J); the polynolnials 1I (..t') u ud h (..r) are the factors
1l10dulo )} of the product.
Suppose 110W that Jlr) is Hot ic1l)ut.ically cong'rUeljt to zero
Inodulo }J. Neither {/ (:r) nor h ();) is tl1ell identicall)" c ::.ngl'uent
to zero 111odulo )). If \"e replace the three polynoluials in (2) by
tIle corresponding- nor1nal polynomials \\'e obtain: If .f({!), !J (.J')
and h (.,c) have the deg'rees in, It an <l 'I' Illod ulo }J respectively,
t.hen IIi -= It + J'. Thus, the ueg-ree 1110dulo }J of a divisor !/ (..r) of
.r (.r) cannot be g'ren tel" than t]le degree nlod nlo jJ of J'lJ').
The polyno111ia] .f(..r) has the following' iririal di,.isors Inouulo p:
1. Ever}" il1teg'er c not uiyisihle by jJ. 2. Every polynolniul
H Cl') == cj"(x), where c is an integ'er not ui\""isible bJ p. In fact.
if [l (,i') == (I' (,(.) and (; c' = 1 (IlIOU }J), ,ve havf\ .f(x) =: (/ [J (.r) (lnod p)
UIH] }'()...) == c c' f(x) (nlod 1").

THEORY OF CONGRUENCES
95
A polynoll1ial of degree > 1 is called n jJ}'inze Ji(}ll'iun u1()dulo })
'v hell all its divisors 1110d ulo j) are t.ri vial. It is also said to be
iJ'rpducible lJlodlllo i). E,-ery linear polynoll1ial \" hich is not identi-
cally coug'ruent to zero 1l10clulo J} is a. prinle function u10dulo l}.
Let j'(x) and 9 (.'0) be two polynoluials \hich are not identically
congruent to zero 11l0dulo 11. \,\T e cOllsider the set 1\1 of all the
polJ'noluials G (L/') W bich satisfy the congTuence
(3)
G (:-c)- := J{x) 11 (x) + g.(x.)-h1. (x) (ulod J).\
I _ . .
f
where Ii t,'{") and hI (c) run throug'h all polYllonliuls in :1.:. rrhe SUU1,
the uiffertnce and the prot1uct of t,yO po]ynoulials in M are also
PolJllolniuls in IVI. By putting' "(.1..) :.= 1 una hI (.r) -= 0 in (3) ,ve
see that .((Ll') is an elenlent of IVI; :-5i1uilarly it follo,rs that !I (.r)
belong's to 1\1. There exist ill IVl ccrtaiu polyuolnials ,vhich are
not idellticallJ cong'ruent to zero l1loul1lu p and ,vLich have t.he
least possible ueg.ree. Let d (;(') Jellote Clue of these polynoD1ials;
\ve can suppose that.d (c) is n 1.JriIuary 1101'lnnl polynoll1ial 1110-
U ulo jJ. Then d (x) is uni(IUely deternlined. [n fUCt1 let us sup-
pose that d 1 (..r) is another priUlar)'" norlnal PolYllolnial of the saIne
degree ll1odulo 1) as d (J"). The1) the aifrtJrellc d (x) - tl l (x) i a
polynoluial ill M; its degree 1110<1ulo p is less than that of d (x),
and it i8 not identically COl1gTlH'nt to zero IJlouulo jJ. But this
contradicts t.he definition of d (x),
Hence there are two polynoluials if' (£) and ?PI (x) ,\'hich satisfy
the identical cong-rue11cc
()
d (.£) == j" (.,) "P (x) t [J (,{) '1'1 (:) (lllOd jJ).
Algebraic Jivisioll of an arbitrary l)olJDollliul G (x) i.n M bJ dl-:r)
leads to an algebraic ic1entity of the forl11
G (x) = d (x) !/l (.e) + r (.:{),
,v here {II (.1') and r (.r) are po]ynoluials and whert\ tIle deg'ree of
}' (.i.:) is less than that of d (J..o). Henee
--- -- ---
(,' (x) :::: d (.I")!lt (.)  I' (..) (IllOd p).

96
CIIAPTER 111
The polynoll1ial r (or) belong's to M, and in cOl1sequel1ce of the
definition of d (x) it IDUSt be identically congruel1t to zero mo-
dulo 1). Hence
G (!I) == d (.v) 1./1 (x) (mod p).
Thus all polynomials in Mare lllultiples of cl (.r) lllodulo p. III
particular, d (x) is a conlmOll divisor nlodulo }J of J'(.1') al1d 9 (x).
On the other hand, it follo,vs fronl () that every common divisor
modulo p of f(x) and 9 (x) is also a divisor modulo }J of d (x).
All multiples modulo p of d (x) belong to M.
The polynomial tl (x), \vhicb is uniquely determined, is called
the greatest COln'11l0n diri,lol' '1nodulo }) of f(.v) and 9 (:ll). If d (x) = 1 'I
we say that the polynomials f(J.:) anu g (x) are rlatz.l'ely l)'ri?ue
lnodlilo 1).
Exauzpl(1s. By trial it is easy to sho\v that the polynomials
.3 _ X -f- 3 and x 5 - ;1,4 - ,1. 3 + x 2 - 2 x-I
have the greatest conlll1on divisor ;,{. - 1 ll10dulo 3.
The polynomials x + 1 and .1: 2 + 1 are relatively prime modulo o.
Finally we prove the following two theorems:
Tlzeorp1n 55. Ltt g (x) be a pril1le fuuction lHOdulo i). Then, lj g (J.')
';s a dil,isor l1zod'lll() 1J of thl) prudu(.f ,l(.c) Ii (x) but 'J20t of fhe
polynoll1ial f(x), g (x) ?nllsf be a clZ.L"is01' 'illodulo lJ of the lJo1y-
'12ornial h. (x).
Proof: Since &f(J:) and g (x) are relatively prime modulo J), ":Ve
have by (4)
1 = &l(x) "1' (,r) + g (x) "Pi (x) (lnod 1)),
and, 1l1ultiplying w by h (x), we get
h (x) == /(.1.') "p (.2') h (x) + g (,}') If'l (x) h (x) (mod p).
Since 1(.1') h (x) is divisible by g (x) 1l1odulo p, it follows that
jW(..t) h (x) = g (x) 1j'2 (x) (mod }I),
where "P2 (x) is a polynomial. Hence

THEORY OF CONGRLTE4'1CES
97
It (.1') == 9 (,1' 1 [VJ (:{.) V'2 (..1') + V'i (,1:) h (,).)] (1110d}J),
\yhicb pro\es the thcol'enl.
Theoren1 /;(l. 1. jj"'L'er!1 jJolNnO1)£al .J'(:(') oJ" degrpe In > 1 l1lodulo 1)
ca II b,.. lcri ffe II i J2 tlte fiJJ'JiI
(5)
j" (.r) = r f.Jl (;,.) 1]2 (.i: ' ) , . . fJ.ij (.1') (111 od }J)
llb) a })}")fllu.t ,( prinze jl.lllctif.Ju,." fJJ (.r) IJtodulu p and a natural
lilllnber c.
2. I..r lji (£) is (j. degree 'Pi JllfJtlulo p (i == 1, , . . ., e), thr>}l
'1 -1- 1'2 + . T }'ll = Jj/.
H. If' 0 < t < jJ aud 'f flif prinu, jltJletioJls fJi (x) a}'(! t!i08l'll lis
1.11 4 lJHllr./1 norJlutl pul.1lJlo1Jlials. tllPU tlu' t/eco11lpositiull (5) is 'llJljrzlle
ll)Ulrf j}'Olll the urder o.t the prilJle jZlJ1etio1l8.
l)"{j(/: AnJong' the divisors lllodnio }J of }'(.1.") there are certain
polynoJuials ,,,hich are of the least positive degree. Suell a di,isor
III (.1') 111Ust Le u. prllUP fUllction 1110dulo J), For, if a divisor t (t')
of ljt (.}.) luouulo jJ \\"erp non-trivial, jt ,,,"QuId be of lo\ver degl-ee
than f}l (x) luotlnlo J}. Binee t (l') is also a divisor of j'{.'L'), this is
only possible \\' heu t (.1') is tl'i \7ial. V\1'" e prove the first part of
thp theorelll by inductioll. rfhe assertion is true for lIi ==- 1. Sup-
pose that a decolllpositioll of the t)"pe (5) exists fur nIl polyno-
Juia.ls of deg'ree < 11i lJ1ouulo p. Then "re can prove that it also
exists for a polynondal of ueg'ree 111, > 1. )...s was sho\vl1 above,
,(lx) has at least one ]H'iIuc tljyisor q1 (:) Inod ulo 11. Hence
tfi) j'(J:') = q1 (x) 9 (.r) hnod p),
,vhre [J (.) is a polynoll1ial of deg-ree 'Ill - 1 1110dulo 1) at lllost,
provided that !tLC) is not a prime function. Therefore, by hypo-
thesis, !J (x) can be written ill the forul
9 (J:) == (' fJ2 (x) qs (x) . q (x) llJlod p)
a a proL1uct uf priule functions and a constant. (10nlbining' this
congruenCl\ ,\'ith cOllg'rtH-nce (u), \V get con'ruellce (f)).
The second part of the theorenl folIo Y8 directly froIll a remark
in t.he beg'inning' of this section.
7 - r,ltit.;ill Trygl)(!, .\"agrll

98
CHAI)T E R I.U
It remaiu8 to prove the third IJart of the t heorenl. 'Ve Inay
suppose that the l)rime functions qi (,i') in (5) are prinlary and
normal polynonlials DI0dulo jJ, and that 0 <.._ (" <, p. Now suppose
that t.here exists a second decolu}Jositioll /(3') of the saIne kiud
as (5) I
(7) .l'(x) :::: CI rl (x) 1'2 (.r) .. 1',) (x) (IlIod p),
W}lere 1'[ (.x") are pritnar)" aud normal prill1e functions modulo j)
and 0 < Cl < p. Then clearly ,ve have Cl == c and
(8)
(, ('
11 qt (x) "= IT r;(x) (mod p).
i=1 i = 1
Using Theorem 55 ,vc see then, that tIle prime function 'II (:,)
must be a divisor nlodulo }J of at least one of the prime func-
tions r( (x), say of rl (,.). Since lJI (.1') and /"1 (.1:) are both priIl1ary
and normal, it is evident that ql (::t) = )"1 (x). In the identical
congruence (8) ,ye can divide both the 111enlbers by the })oIJl1oIl1ial
ql (x) = rl (x), since it is not itlenticall.y congruent to zero modulo})
Thus ,ve get the congruence
p (J
IT qi (x)  II J"i (x) (mod p).
i=2 i=2
By the same arguIuent ,ve prove tIlat q2 (,}.) is equal to sonle
ri (x), say to r2 (x). E,iuently this process Inn)" be continued, and
we find finally that qi (.r) = ri L.r) for all i and that e = G.
TIIUS the proof of Theorenl UU is cOlllplete.
Exa1JllJle. Let us take jJ ==- 7 and
f(x) -== x 8 + .1.7 + 2 .{.6 -  iJ5 + .t. ....'-..1 --- x 2 -- 3:{' + 3.
By trial we find the following result
f(x) == (t" + 1) (x 2 + 1)2 (x 3 + 3) (Illod 7),
,vhere x + 1, .l,2 + 1 and x 3 1- 3 are prinlal"J and normal v rime
fUllctions modulo 7.
Renlark. It is not possible to develop an analogous theory in
the case when the nlodulu8 is a cOlnposite number. Theorenls [>0
and 56 are llOt valill in general.

THEORY OF CONGRUENCES
99
30. Wilson's theorem and its generalization. - 'Vhen 1J is a
prinlc it follo\vs frol11 Theorenl 30 that the congruence
x p - 1 - 1  0 (mod p)
has the roots x == 1, 2, 3 . . .  1) - 1. By Theorem -l1 ,ve have
then identically
x p - 1 - t ::: (x - 1) (x - 2) .. (,r - }J + 1) (ulod 1)).
Put (' = 1) ill this relation, and it follows that
(1)
(}) - 1)! == - 1 (mod p).
This result is called JJTilsun's Theorc1n after the discoverer. The
first proof of it was, however, given by Lagrang'e in 1770. The
theoreln may be extended to yield a criterion for primes:
l 1 heore111 ;j7. ....1 necessary llud 8uj}icient ("ondition that an intege1'
nl> 1) is {( priuie is that (n-l)1 + 1 be dieisible b!lll.
p}Oooj: It reulains only to show that. the number (n - I)! + 1
is not divisible by '12, when 12 is a CODlposite number. If 11 is
composite and q is a prime divisor of 12, then q < u, and there-
fore (11 - I)! is divisible by fj. Hence the llunlber (1/ - I)! + 1
is not di visi hle by 12.
.'
It is, howe,er, obvious that the test furnished by ""ilson's
Theorelll is useless for lal'ge nunlbe1'8 12, since (n - I)! increases
too quickl)" with 12.
N o'v suppose that the prime }J is oad and put  (}J - 1) :-::- '1.
()n the left-hand side of (1) replace every factor 11 + q (for It -= 1,
, . . .. q) b)" the congruent It - f} -- 1, and multiply both Rides
by \ - l)tJ; then we 0 btail1
()
(q !)2 == -- ( - l)q (mod 1 J ).
If the prime 1) is of the form .t. 12 + 1, the rightMband side of (2)
is - 1. Hence \'e obtain the first part of
Theorenl 58. If' p is a pri'lne :::= 1 (mod .1), the con!1"lle'ce
(3)
 - f .
.) ......=-= -';'" j · \ mQQ l d . .
. . I

100
CHAPTER III
has the fit'O 5'o/ufious
+ ( JJ :) I ) ! ( 1 )
x =::. 11I0C jl.
1'/te tOU!IJ'uen,."/J t3) Ii a,,,,, 120 ,....Ullltio}l U'/U!}l}J it.: a })l'iI111. = 3
(mod -1).
\Ve proye the second part of the theorel11 inc1irect.l'. If (3) '(re
solvable for }) = -I- n + 3. we would HIld bJ raising' both sides ill
(3) to the qth power
x p - 1 == (- 1)9 = - 1 tJDod p).
But t,his IS impossible. since by Ferlnat's 'l'heoren1 tTheorenl 35)
:1. 1 )- 1 .= 1 t l110d }J).
If the prilue p is of the fOl'ln -I- n + 3. the rig t ]Jt-hul1d side
of t2) is .t 1. Hence oue of the 11nl11 bel's
(J  - 1 () r lJ! 1 1
is divisihle hY)J lIenee \"e have
(-l) (11  l)! = + I (moll J}).
For p -= 3 and. 3 the rig'ht-hauJ side is -: 1; for J. == 7, 11 and
19 the right-hand side is -1. In Chaptel" IV ,ve shall g'ive u
rule for JeterJllining the rig.ht-halld side of (-l-) for any prin1e.
"'Tilson's Theorem is a special case of the following more
general result due to Uauss.
Tizeorein t;!). IJ£'f 1l lJ(} a natural Jilli1iber >:!, (tad let i.,-r delluf(1
the nurnVl'f 0.1" incol1g7.uenf 8ollltioll, of fhr congrlU:lit:r:
(5)
,1,;2 ::=: 1 (lUOU Ii).
1'hen, i.r lll, a2, . . ., a'l1 are '.ep7.(8ent{lt-ives of a red'uced ,.esid ae
.y.telu '1120d'ulu n, ire hal'r
.,.,. . l ,.
,; ' t . It '''  a '-=: '- 1 l:S.'
 .;. 2 cp -- \ J
(1 1 }(Hl Ii).
\\s-

THEORY OF CONGRUENCES
101
]Jroof cp nleans qJ (u). To eYer? n prilne to J2 there corresponds
a unique a', also priu}e to 11, such t.hat
( t ») a a' == 1 ( n1 0 d n).
Hence the numbers ai- 112, . . .. a(f can be divided into pa.irs a, a'
\vhose product is congruent to 1 ulodulo i). "\'Te haye a == a'
(luod u) only when (( is a root of tIle cOl1gruence (5). Denote by
(l) the product of all the .\T incongruent roots of (5), If a is a
root of (5), so is - a; since n > 2, the roots a and - a are in-
cong'ruent D1odulo 12. \Ve ha ve
(7)
al- a) == - a 2  - 1 (nlod n').
Hence
(8)
( :=: (- 1)  s ( 111 od II).
N ow let QI denote the prouuct of all incollgruent numbers
llj i i = 1  2, . . .. 91) 1l10duio n which are not roots of (5), if there
are au)" nuu)bers of t.his kind; other,yise put lJl = 1. B- the
congruence (6) it follo,vs that
(11 == 1 (luod Ii),
and by (8)
p == Q Ql == (- 1)!-'" (nlod 0).
Q. E. D,
Applying the results of Theoret}l -1-7 to tlJe nUluber 1'''', we find
that the product P is cOl1gruent to - 1 111odulo '1/ in the fol-
lowing' cases: \tVhen Jl =.t..; ,\" 11en Ii is a po\ver of an odd pril11e;
when II is t\vice the power of an odd prill1e. In all other cases
J) is cOl1gruell t to + 1 nlod ulo n.
Example. For n = HO ,vc ha.e (p (n) = 16 and T = 8. The
cong"ruence
x 2 == 1 (ulod tiO)
has the roots + 1, + 11. + 19. t 29. Ho,v the residue classes
Inodulo II which are prinlc to 11 In3)' he di\"i<lea into pairs IS
apparent. fro III the following congl'uellces
7 . (- 17) == 13 . (- 3) == 17 - (- 7) = :!R . (-- 1;3) -= 1 (nlod 60)
and
1 - (- 1) == 11 . t - 11) == 19. ( - 19) == \ . (--. :!) = - 1 (mod 60).

102
CHAPTER III
31. Exponent of an integer modulo B. - Let 'n be a natural
llumber > 1 and a an integer prime to n. In the infinit.e sequence
(1)
23.
{t, a , a , a . . . .
there are numbers == 1 (nlod 12), since by Tlleorem 36
(2)
a 'f (n\  1 (mod n).
Suppose that ad is tIle first number in the sequence (1) which is
== 1 (mod 11). Then a is said to belong to the ei.JJoJ2e12f d ,nodu/o 12.
lJ is the ordeJ. of a 1nod1l70 n.
.
C'ongruent numbers modulo n have the saIne order modulo 11.
TheorC1n 1)'0. L(t n be a uatural I1lUnb('}" :,. 1 and a au integer
}Jrhne to n.
1_ I}' a beloJl.fp:! to the eX1JOnent <5 Hzodulo n, thC12 the nUrJzbe.,-s
2
a, (f . . .
. a rJ
are iJZcollgrtlr.nt Jllodulo 12.
2. Further, 'l:r
am == 1 (mod 12),
then 111 'is dit.isible b!1 t5. In 1>Q1-ticular,  is a clirisor qf lp (n).
Proof. Suppose  :> k :> h > 1. If we had
a k == a h (ulod n),
then
(lk-II  1 (mod '12).
But, since 0 ..: k -- h <: <5. this is contrary to the definition of .
To establish the second part of t,he theorenl put '))1 =  q + r..
,vhere q and r are integers, 0 < ,. < c5. Then we have
a r == a Of1 + r == am == 1 (mod n),
and thus, recalling the definition of d, ,}" = O. Hence <5 is a di'\""isor
of '112. By (2) it follows that lp (13) is divisible bJ' <5.
ExaH1jJle. The following t,able for n = 55 gi\es the order b
modulo 55 of all the positive integers prime to 55 and < 55.

THEORY OF CONGRUENCES
103
b
XUJ11bcl'S of order <5 ]nodul0 5;'
1
1 ;
21, 34, ;;4 ;
I:!, 23, 32. 43;
10, 26, 31. :Jfi;
no J1ulnb. U":-:
4, O. 9, J t. 1 O. 4, 29, :In, t I, ,., 49, 51:
:!, :-1, 7, 8, 1 ;;. 17, 1 H. ::!;, :!, :J7, :., .:?, .7, 4R, n2 53;
.)
-
-1
;)
8
IU
:!()
-to lle1 lJlunbPI':O:.
'Ve next prove
l"hcorern (}'l. Let II be ((. natural U'lt11Zber ::> 1 and a an 'lllt(,1ger
]Jrhne to '11. I..f a beluugs tu tlte f:l..poncut <5 nzodulo nand (f t1l
i.5: a natural 12Ul1tber sZlch that (nz, c5) = IU, then am bela12g, to
<5
the eX1Jonent 11Zodulo 12.
It
Proof: Suppose tIlat, the nunlber am belongs to the exponent l'
modulo u. Then J' is the least positie exponent such that
{3) (a m )\' === 1 (mod n).
Applying the secolld part of Theorem 60 we conclude from this
congruence that 1JZ J' is divisible by o. Thus, since (111, d) == ,ll,
,ve Inust have

,,= h
,
p.
,vhere h IS a positive integer. On the other hand we have
tf m
(a.m)U == (ad)." == 1 (mod 17).
since a belongs to <5. Hence we conclude that h = 1
<5
alld '),::::. .
Jt
Q.E.D.
Exanz}Jle. As in the abo'f'e eX1tmpie we take u -=- 55. From the
table we see that tIle number :2 is of order 20. Then, by Theo-
reIn 61, the nUlnbers
'> -! -) 8 2 ] 2 2 16
 , - , ,
are all of order 5. Tllis is verifierl by the to ble since they are
== 16, 36, 26, 31 (mod 55).

104
CHAPTER lIt
Let <5 be It positive divisor of p (n). If II is of order <5 1110dulo 11,
've saJ that a is a lJrh)lifil'() }'(Joi o./" the l'Ongl'ueUl.p
(4)
x rJ == 1 (ulod u).
Theore11l 6"2. IJpf n h(J (t natural nll-'1nl)('r > 1, let tJ be (t lJo",itirf>
di,..iso)' oJ' q; (n) and It-, a 1)(' a root oJ' ,'10' (.OJlf/FU()IUlf (....). 1'heu
'toe hare
1, 1 uece.f;j8arll and Sllfitienf condition &fo}' a fouf II 0..1' (..1) to br'
a jJrinlitirp root of {-t.) i,., fhat the uZl'J}11)f'J's
(5)
,) ,f
(I, a.. . . ... a
be inCUJlJ/fzU nt nloltulo }I.
2. Let a be a j)rhnitit,'e root (!f' (4). .,1 'I1pccb'Sa1'y and :1ufficz"('ni
eo'ndiiiun ..Ivr {[m to lH' (l pr;nzitire root q( (....) i8 that (n" 6) :-=. 1.
H. I.r tluJ CUJ1yrllfUC(J
Xff (11) == 1 (IlIOd It)
has a p)'jJniiil'(1 root if ha. (p (cp n.) pril1zitill roots incougruent
1nvdulo n.
Proof. ] r a is a root of l-t.). a11 th numbers (5) are roots of (4).
The truth of the first part of the tlleorenl is tben all in1111ediate
COl1Seqllence of the definition of priu1itive root and of Theorell1 60
( first part).
To proe the secolld and third parts we oul. hu\re to apply
Theoreul 61 \vith /' = 1 and obselw\""e that the numbers
(I, 0 2 . _. ({P (11)
fornl a reduce<l 1.esidu<.' Systenl modulo Jl ,vhen a is of order q; (12).
'\V' e Ilext consider the special case \vhen the modulus is a
power of 2.
Theurem 6.g. 1. Eve].!J odd intr'ger J) :....atiso.f;,:>s fluJ congriU'}l("('
(fj)
{-2
x 2 == 1 (mod 2i)
1l"hfl2 {J :::> 3.
i!. Tlzp nU1ilber 5 I' a pri,uitive root vf (6) 'lohen p > 3.

THEORY OF CONGRUENCES
105
/J. Tit e n II J n 1)(: 1.,C:
-t - + - ') Q,i-2
'\ ' ) .. _ + __ 0 ....,
.. v, _ t . . .,
.Ior}Jl a reduced rl'sid'uc ,Y8te'n ulod ulu 21 j uollen {J > 3.
PJ"O().f. The first part of the tbeorenl is true for f3 = 3, since
1 2 == 3 2 ::::: 52 == 7 2 = 1 (DIod 8).
If (6) holds, then
.i-2
;}.2 J = 1 + ji t,
where t is an integer. By squaring' t\'e get
-]
3.:. 2 ' = 1 + 2,:+ 1 t + 21:J t 2
and
Q -1 (  ' I )
:1,... .= 1 mod :?;:IT .
\Ve thus concluue by induction that the cong'ruence (6) is true
for all fJ > 3.
To prove the second part of the theoreln we suppose that the
number 5 is of order <5 1110dulo 2. i . According- to (6) alld to
TheorelD 60 the exponent <5 is a di,isor of 2i 3 - 2 . If <5 < 2.-2,
then  ,vould be a divisor of 2,:1-3 al1d
, I)
0 2 - u s 1 (DI0d 21)0
'Ve can, however, sho\v tJlat for {J >... 3
(7)
 "
,-
;-.2' =-= 1 + 2- 1 + 21 T,
\\yhere l' is an integer. This is true for f3 = 30 If (7) is true for
a gi¥en value of the exponent f3, it is also true when (J is re-
placed by p + 1. For. by squaring both sides of (7), we have
i- .)
;-,2' - = 1 + 2,1 + 213-+ 1 (T + 2f.1- 3 + 21-;-] l ' + 2i 1 - 1 T!).
Thus \ve conclude by induction that the relation (7) is valid for
every {J > 3. Hence we cannot have  < 2;3-8, and it is clear
that tJ = 2i-2.

106
CHAPTER III
The truth of the third part. of the tl1e01'enl follo\vs fro III tIle
second part and from the fact that the congruence
OIL == - 5 k (mod 2i)
is not satisfied for any p > 2.
No\v we introduce a ne,v arithmetical fUl1ction 1p(n) defined in
the follo,ving way:
1. '1'(11) == tp(n)
for n = 1, . ..t and u = 1)a., when }J is an odd pril11e;
2. 1p(J1)=- qJ(n)
for n = 21, ,y hell fJ ::--::- a.
3.
1J' (n) :..= (lj' (pl). 'P (lJ2)  ' . ':.
for any n 11aving at least two different prime factors. Here
n -:. jJ 1 )C<J . , , ,
,vhere })1, })2' etc., are the different prinle fa.ctors of u.
{a, b . . .} denotes as in Section 5 the least CODlmon mult.iple
of a, 11, etc.
From this definition follows
T1zr.Ol'fUl fJ4. If 12 if an intt'[Jf!t > 1 and ,:f a is }Jrilne to n. then
a 1p (n) === 1 (ll1od n).
For! b)? Theorem 3() this congruence is satisfied for f2 = 1, 2 -t.
and n = pa, where }) is an odd prime, and by (6) also for n =-= :2i.
Hence, using the definition, we see that the congruence is satis.
tied for any integer Jl.
Theoreln ()-! has the corollary:
Excel,t fa,. the eases n === 1, 2, -!, }/l. an rI 2})r"l, /f.Il{: Fe p is all odd
pri nu', If'e Iza l'(!
a (( in) ==] (ulod 11).
In fact, the llumber V' ('2) is a. divisor of  (f (1), apart from the
exceptions mentioned.

THEORY OF CONGRl'"ECES
107
32. Moduli ha\Ting primitive roots. - If 12 is a natural nuulber
:> I, and if a belong"s to the exponent <p (n) modulo 11, a is said
to be a jJrhnitil:c root 'J110dulo n 01" of the llumber 'll. 'Ve will
no,v determine all moduli which have prin1itive roots. TIle nUJll-
bel" 1 is a priulitive root modulo . The number 3 is a prin1itiye
root modulo 4:. Fron1 the corollary to Theorem 64- it follo'vs that
every inteer n which has a primiti,.e root and is different fro111
 and ..t is either the power of an odd priole or twice such a
power. ,,: e shall prove
Thcorcnz a/i. 1. The natural 71zonlJ('r 11 ::> 1 has lJrinlith.:e routs if
11 has one 0.( the ralltes
n .::.:= 2, 4-. pa a 12£1 2 pC!.
ll"1z('l'(J }J is au odd ]Jl"hne. aurl .j n IUJ other ca,,?(J.
2. The nUl1zb(,,- oj. i 12(OOll[/rllrl1t prhnitirp roots nUJflulo 11 'i. t/zel1
q; (q; 11 ,,) .
.!J. 1..( f1 i, a ])ritllitire root u.f the odd ]),.il1ze p, and If the 1ill1nber
!/}J-l - 1 it:> not dirisibltj uy ))2, t/zen {/ is a prill1itil,,"(: root a..f
))C!, ./ur an!/ positire ('XjJoJu'nt a.
Proqj: The theOre1l1 is true for 11 == 2 and u = 4:. W P now
have to distinguish three cases.
[4'irst case: 11 = the odd prin1e jJ.
Let <5 be a positi¥e divisor of }J - 1, alld denote by Z () the
nUlnhpr of incong'ruellt integers 1110dulo 1J which belong to the
exponent  Illodulo 1). Then clearlJ
(1 )
 X () === )J - 1,
,;
'vhere the SUlllluation extends over all positive divisors  of jJ - 1.
\Ve have
x/J-l - 1 = (xc1 - 1) /z (x),
where h (oi') IS an integ'ral 110lynomial. By Theorem 35 the con-
gruence
x p - 1 - 1 == 0 (nl0d }))

108
CHAPTER III
haR the p - 1 illCOllg'l'Ufaut roots 1. 2, . . ,p - 1 Inodulo }J. A p-
plJ'ing' 1'heoreul 4:B with j'(x) = Xl -1 - 1 a.lH.l .fIl.t.) = ,}],)' - 1_ we
see that the congTuence
(2)
x ,J -- 1 == () (IlIod jJ)
has exactly  illcongru<.:ant roots 111otlulo }J. If this (.OIlg'1'UenCe
has a prhni ti '\e root (I : the n HI)) bE:rs a _ ({2 _ as. , . . _ a r ) are the
con1plete set of incongruent roots 1110c1ulo p of (:?) (1'heoreru ():!.
first part). Alllong these roots exactly <p l) are lH"iulitiye roots
of (2) (Theoreul ()2, second part). Threfore it follo,,'s that COll-
gruence (2) has either <p l) incongruent pl'itniti\"t"\ roots 11l0dulo })
or llone at all. Hence \\-e ha\e either X () = 0 or X (b) = <r (L
It follows froIll 'heorenl ] 3 that
(3)
2: cr l (») = jJ --- 1,
r)
where the sun1uHltiou extends oyer all positive djyisors d of p - 1,
COll1pal'ing (3) and (1), \\'e see that \ve never can huye l (b) = 0;
thus X (b) = cp (.) for all" d. Hence it is proved that eyer)" oud
})riU1e has prilnitive roots.
Second ra.....'.: n = }J'\ jJ odd prhne, a > 2.
Let U be a. lJl.julitive root of J}. If the nUluber yP-] - 1 is di,is-
ible by p2. tllere exists another prilnitive root ,fII = !I + J) of jJ
such that
lIf-l = l.q + p)p- 1 == 0 1 >-1 + (jJ - 1) lI P - 2 1) == 1 - }) .rJl-2 (1110d p2).
I t is clear that the n nll1ber gf-l - 1 is not divisible by JJ2. Thus
\ve call choose the prhnitive root (I Dlodulo }J such that .qP-1 - 1
is not diyisible by 1)2. This condition satisfied, g is also a pritui-
tiye root of pa for Ull)" u. '1'0 show it we beQ"il1 with the proof
of the following lemnHl:
The number
it
gpQ - - J 1 1
is divisible by 1Ja.-1 a11(1 not l,y pal
By hypothesi t his is true for a =:!. L1ppose that
(.t)
y,J a - 2 I p -l} = 1 + r )J'I-I_

THEORY OF CONGRUENCES
109
\vhere the int.oger c is not di risible b9 p. 'Ve raise hoth sides
in (....) to the pth po\vpr and expan(l the l'ig'ht-hund side by the
binoluial theorell1 to ohta III
{jpu-l(P-l!=(l -r ("J/'-))l= 1 + (.Jl'-+ (..2 jJ(})-l)})(ct-l) -1- hjli(t.-l',
where IJ is an iuteger. Since ':l (l - 1 and D'l -- 3 ale > a + 1.
it follo',",8 that
,f/,p-l(1 1 - 1 = 1 + (.} pC!.,
where the illteg'er ("1 is not divisible by p. Hence the len11ua is
proted by induction.
Suppose next that !I IJelongs to the exponent b nlodulo )It.. It
follo,vs frol11 rrheorenl no that  is a Jj\'isol' of q.-' (}/-) = }/:-1 (jJ - 1).
Since !/ is a prinliti\re l'OOt of 11.. the lll1uILer }) -- 1 iR H di'9isor
of  by Theoreln no. Thus  = jij"J (}) -- 1), \\'hE're 0 < fJ < (J. - 1.
[f <5 F l)lt-l (1) - 1 L then 1 Ja - 2 (j) - 1) ,,'oultI be di,"isible by . and
''fC would l1a \"e
fjTll-2(p-l) == 1 (Inod 1/ l ).
,vhich is contrury to (4). Hence c5 = Jl-l (1) - 1). a1ld !/ iR a In.itni.
ti ve root of }l"
l'hiJ'd Ca8(): II .:=: 21J{(" 1) odel prin1e.
Al11011g the }Jl'iInitive roots of jl' there are also odtI nUlIlbers.
F'or, if (I i:"; even, then .'1 - Jl' is odd. Every odd prirrJiti'"f1 root
.f/ of ll is a, prilllitive root of 21)a.. For, if .fI Leloug"s to the ex-
ponent ,) luodulo 2}l', then  is a divisor of (p (2 })c-) = rp (l)a):
flllther, since .'J belongs to the exponellt cp (pa) 1l10dulo }l\  > <p(l)a).
Hence  = cp (pa) = tp (21)a). Thus the proof of the tirt and the
third }Jart of the theorem is complete. rrhe second part is a
direct cOllsequel1CP of Theorem 62 (third part).
The prin1itive roots of a gi-vell nlouulus luay he deterluined by
trial. A t the enu of the book 've g-i\E' a table of thp least priuli-
tive root of the first 150 primes.
J!.:xa lil}J [t's. The prinle 7 has the cp (en -==  ilHongl'Uent prill1i-
t.i ve roots
a and f).
Since the l1Ulllher 3 6 - 1 = t-: . 7 . 13 is not ditisible by 7 2 , the
nUlnber 3 is a pritniti\7e root of an,' po,ver of 7.

110
cnA PTER III
l'he prin1e 13 ha the p (12) --;- -t. incong'ruent pritl1itive roots
, 6, 7 und 11.
TheoJ'eJ}i f/{}. I}" the nataral nunlbfT U l,> 1) has priJnitire ro()f,,
and ill d is a 1)08£ti1.:(' diri'o" o..f cp (n), tlu J Il thp COJl!/rupucp.
(5)
x([ == 1 (DlOd n)
has f)x{lltiy d inl'ou{}i'urnt foots ?nudulo u.
];)rooj: By Theorem 62 this is true wIlen d = cp (n). K O\V let II
be a prinlitive root of n. TIlen the llulnber
cp (nJ
 ([
=={J
belongs to the exponent d 1l1odulo u, a.nd the nUlubers
(6)
 2 rl
",.. .. 
are incongruent Inodulo n (Theorems 00 and 01): hence, tlJese
numbers are roots of (u). N o,v let YJ be an arbitrary root of (5).
Then it follows from Theorem 60 that
1] =::: gh (llIod n)
where h is a multiple of qJ l) . Tberefore we have
1} == 1: (ilIad n),
\vhere k is an integ'er > 0, i. e. the number I] is congruent to
one of the Dunlbers (G) 1110uulo n. Hence, there are no other
solutions of (5) than those given by (tj). This proves TheoreuJ G6.
.A. supplement to Theoren1 GiJ is
Thporfln 6'7. Let 'Jl be a natural nltnhe,. :.> 1.
1. There aitrays (xist int('geJ's u'hieh beluufJ tu the e,I'P0nen! 1p (u)
'1nOdliio 'Il.
2. Erer!/ integer prioze tu n b(.7oll!Js nUJdltlo n 10 au cX1Jf)nent t,.hielt
is a diL'i80r 0..( 1p(u).
H. .A.t least q:' (<5) iutP!Jers inronyruent nioclulo u belong to {[ yic('1i
l)fJ,ilire dil.i'or  oj' V' (u) l1ludulo n.

THEORY OF CONGnrECS
]]1
Proqt: Suppose that n is diyisible b¥ p(i ana by no Itig'her
po,ver of the prituf' }Ij. TJt"t !Ii denote a. pl'inlitivl' root of JJ:i,
,vhen 1);, is odd; for jJi =  and a;,  2, .fit denotes the lllllnber u,
and for Pi =-::  und Ui ::.= 2, {Ii denotes the nUlubcr 3. N 0'\1', ap-
plying' Theoreul --1-0, we determine the COlnnlon solution of the
sirou] taneOllS cong'ruences
.r; =:= 91 (mod pJ), x::::: 92 (luod 1J22), . . ., x  gr (ulod J)r).
,vhere Pl.}J2' , . .. 1)r are all the differell t 1.Jrinle divisors of n.
Let the COllllnOll solution be
.1' ==  (mod }2).
If  belol1g"s to the exponent  nloc1ulo '11, then  is a. divisor of
1p (Il). (Thcorenls G-I: and GO.) On the other hand,  belongs t.o
the exponent (1' (l)i) == '" (l)ci) nlod ulo J/:i, "9 h1l 1)i i odd, and
when lJi =  and ai == 1 or 2; if lit =  and aj > , theIl, by
Theoreul ()3,  belongs to the eXpOl1el1 t  cp (i ai) =-= 11-' (I!i) 1110dulo 2 Ui.
Hence <5 is a C0l11UIOU 1l1ultiple of all the nU111bers 1f' (jJ,'i) for
i := 1, . . . -, r {Theoreul 6U). AccorJing' to the definition of "I' (n)
we havp then (j -= 1f' (11). Thus the first part of the tl1eorenl is
proved.
The second }lart of the theorenl is a di rect consequence of
TheoreIIls 6.t and GO.
Suppose that  belong's to the eXpOl1l'ut 1p (u) 1l10dulo u. Let 
be a positive divisor of '1' (1/) all U l'u t 'P 1) = fj . By Theorem fj 1
it. is then clear that the nUI11ber 1] - = f} belol1g's to the exponent
 Inodulo Ii: aud any nUlllber 11'l, ,,"here Ii is priJlle to (j. belongs
to the saIne exponent () 110dulo Jl. rrhis cOHlpletes the proof of
Theo1'(lID li7.
In the eXallllJle given ill Section 31 the 1l1oc1ulus is II =-= 55,
thus 1p (u) = 20. Fronl the table ,ye see that sixteell numbers
belong to the exponent 20 nlodulo f):), twelve numbers to the
exponent 10 1l10dulo 55, etc. Since the num bers 8 and .to are
not di,iso1'"s of J (55) = 20. no llulubers belong. to the exponents
8 and -1:0.
33. The index calculus. - Let n be a natural l1unlber havil1g'
pritniti,e roots. If!1 is a primitive root of n, the numbers

112
CHAPTER III
(1)
), .fI, .fI2, . . ., .rlr(7t 1
forlH a rec1uced resi<1ue systelu 1110dulo II (rrheorelll nO). I n the
set (1) there are <p (cp (n)) prilnitive roots, and these are the llUll1-
bers !Ic, ,vhere e is priJue to cp (u). If a is all arbitrary integer
prilne to II there exists nnlong' the l1ulubers 0, 1. :? . . .  q; (nj -- 1
exactly one nUlnber Il such that
. .
(( == .rI.'l (UIO<.1 12).
The nU1l1her It is called the h,de,2' oj" the 'J1Ul1lbf'r a lfitlt }"(Jspect
to tlzf lJ{l.se f) }}lodulo J2. ana ,ve ,,"rite
or. shorter,
11 =:. il1d g ({
.
II -:-intla,
,
,vlle1l no Inisunc1erstantlil1g' is possible.
Exan1ple. The nUluber 7 is tbe least positive priluiti,e root
of n = 41. Since ID 3 .,3 (1llod 41), the llUJuber 1 f) has the index 3
\\,ith respect to the base 7 Jlloc1ulo -+ 1,
\Ve readily verify the fol1o\ving' rules for the index ctllcuJU8.
J. inc1 (llll ilH.l a -f- ind I) (luaU. rp(u)).
II. inu (((q)  fJ ' ind II (ll1od q; ('11), \vhen '1 is a natural JIlUuher.
III. iUfll = O independeutly of the choicp of the priIaitive root.
IV. ind y = J. \\'hen .'1 is the prin1itive root choseu for base.
V. illd! - 1) -  c{' tu), if II > :?
The correctness of the last rule follo',,"8 ill1luediatel v' frODl the
eI
congruence
u(f' (n) - 1 = ("i fp (n) - 1) Cq!'1 (11: + 1) = 0 (1110d II).
For since !/ is a primitive root, ,ve lllUst ha ye
!/ fJ (11; == - 1 (ulod n).
The first four rules valid for the index calculus show an ob-
vious analogy to the rules valid for log"arithlns.
l\'lallJ' types of congruence problelns UU1.). be so] ¥ed more ea.sily
b? lueal1S of the index calculus. The condition for this is, of
course, that index tables have been conlputed for all possible

THEORY OF CONGRl.:ENCES
113
moduli up to a certain limit. Gauss at the end of his Disquisi-
tiQ12eS gives tables of indices for I110duli up to 100. The (")anon
aritluneticus of Jacobi contains tables of indices for all prinIe
power moduli < 1000.
In the following exanIple of an index table, the I1I0dulus is
n = 19 ,vitIl the primitive root 2.
6\7 I.:. 10 llll 113 I
X lun h(1' , I 2 3 4 5 14- 15 10 17 IS'
I-;- 3\  12 \ 1.1 \--;- ---
Index . . , . . . 0 I 13 .) 16 17 7 II 4- Ht 0
..,
Since "re have
ind (n - a) = ind (- a) = ! ({J (u) + iuti tl (DIod q: n)),
the latter half of the table IDav be olnitted.
..
If the number II has pritnitive roots, the linear congruence
II x == b (Il1od u),
where (a, J1) = (IJ, n) = 1, can be solved b)9 use of index theory.
In fact, this congruence is equivalen t to
ind a + ind x = ind b (1110U (I' :ii oJ),
and tllerefore x is uniquel)" deterDlined bJ the congruence
ind x == ind b - ind a (mod cp 12').
1,l:an1ple. Let us consider the congruence
9 x == 7 (uIod 13).
The prime 13 has the primitive root 2, and we obtain the fol-
lowing index table.
Xl1mbpr 1 21:.114 5 Ii ; 81 9 10 II 12
Index -; 171-; H 5 71-; 10 ... ()

Then we get
ind x == ind 7 - ind 9 = 11 - 8 = 3 (nIod 12)
an c.l
x == 8 (DIod 13).
8 - 5] 66';0 Tr:J'JI.'e )."Q[Jeli

114
CHAPTER III
rrhe general binoluial congruence
a J;1n == b (ll1od 12)
may be treated in the saUle manner, as will be shown in the
next section.
By using index theory it is also possible to solve the exponential
cOl1g'l'Uence
aJ: == u (mod n),
where (1I, n) = (b, 12) = 1. In fact, if n has a, primitive root, this
congruence implies
.),: · ind a == ind b (mod 97 \)2 ).
Thus it is evident that the number (97(n\ ind a) n1ust be a divisor
of ind u. Hence, in this case, there are just (97 ::n:, ind a) incon-
gruent solutions modulo 97 (12).
Exa'1nple. Find the solutions of
7 Z == 5 (mod 17).
The prime 17 has the primiti\"(:\ root 3, and we obtain the fol-
lowing. index table.
X un1b(1I1' . . . . 1 .) 3 4 5 6 ,., N 9 10 II 12 ]3 14 15 16
... I
- - - - - - - - - - - - - - -
Index ..... 0 Il I 12 1) 15 11 IU 2 3 7 13 4 9 6 8
Then ,ve get
and
11 x = {) (mod 16),
x = 15 (mod 16).
Finall)' we shall show how it is possible, by use of index theory,
to determine t.he order f modulo 12 of a given integer a. The
llumber j" is, by definition, the least posit.ive exponent t.hat satis-
fies the congruence
a.f == 1 (nlod n).
If the modulus n has primitive roots, we have
,.. ind a == 0 (nlod 97 (12)).

THEORY OF CONGRUEXCES
115
Hence, putting
Ii --=- (g 1/. illd ll),
,ve clearl)"' have
 cp{u)
I = - --.
· d
34. Power residues. BinoDlial congruences. Let II be an 111-
teg'er  0, and let a be an integ'er prime to 11. 1£ I) is a natural
lluDlber > :? such that tIle cong'ruence
.1.'1 == II (lllod Ii)
is solvable, \ve sa)" that the nunlber II is a qth lJoily:r :r('sidlle
11l0dlllu 12. In particular: the llumber a is a quadraiit, cubic or
biquadratic residue nIotlulo 12 according as q =- 2, 3 or 4.
ThecJ1"e1JI f)'H. Let p br. a,l odd }Jrint fJ , lt4 a he au int(gel' not di,.is-
ihle ill! i), a lid ltt u = prt. {(ud (j = (f), cP In)). 1'!ieu the con-
grueuce
(1 )
x rl = II (mod n)
IUI," exactl:1J  incYJnlJ1'uf'ni solutiuns J12odul() u. (f iud a is tli.
cisibl() 611 (j. OtheriCiiU) it has no 801ulioll.
Prouj: If we c}loose a priuIiti'f'e root 1110dulo Ii. it follo'\vs from
(1) t.hat
q · illd x = iuu II (IliOd r:p (1/)).
This is a linear cong'ruence in the unknown ind x. Hence, ap"
plying TheoreJn 39, Theorenl ()8 follo',,"8.
EJ'a,njJles.
1. Let us cOl1si<.ler the cOl1gruence
x 8 == 3 (Inod 13).
Here  =: (8, q; .13j) :-:::.t. 'Ve may take {J = 2. Then ind 3 = -1-,
H · il1c1 X == 4 (IDOd 12). t.hus iud I.t ==  (mod 3), and
ind .1' = 2. 5, 8'1 11 (nI0d 12),
and finally
x == 4, 6, 7, t) (ll1od 13).

116
CHAPTER III
2. Let us consider the congruence
Xl2 == 13 (mod 17).
Here 8 = (12, lp(17)) = 4. 'Ve may take .9 = 3. Then ind 13 = 4,
12 · ind:v == 4- (mod 16), and
ind ,"C == 3, 7, lIt 15 (nlod 16),
an d finally
x = 6, 7,10,11 (mod 17).
3. Let us consider the congruence
J..7 = 4 (mod 29).
Here  =-= (7, p 29,) == 7. 'Ve may take g . Thel1 ind 4 =2.
But tIle congruence
7 · ind x = 2 (mod 28)
has no solution. Hence the number 4 is not a 7th power residue
modulo 29.
..1.. Let us consider the congruence
x 3 == a (mod 1)),
where p is a prime > 0, ana where a is not divisible by 1).
If P = 6 '11L - It then  = (3, 6 'In - 2) = 1. In this case the
congruence has exactly one solution. If p == 6 HZ + 1, then
 = (3, 6 'Ill) = 3. In this case tllere are either 110 or three in-
cOIlgruent solutions. An exanlple of the first category is thE
congrnence
.1:3 =  (mod 7),
which has no solution. .An example of the second category is
the congruence
x 3 == 6 (mod 7),
which has the solutions x == 3, 5, 6 (mod 7).
According to Theorm 68 the congruence (1) is solvable if and
only if
a == ,q'trJ (mod n),

THEORY OF CONGRUENCES
117
where h is an integer > O. Hence
tp ()
a -r = 1 (mod .n).
Conversely, if this congruence is satisfied, and if
a = g7 (mod n),
where i' is an integer  0, we have
rr (n)
ro-
g d = 1 (mod n).
Since 9 is a. primitive root, the exponent". ffJ ') is a. multiple of
tp(n), and therefore  is a divisor of i'. Hence we have proved
T1zeo,.e1Jl. 6'9. Let p be an odd }Jrinze, l()t a be au 'inieget" not tlilois-
ible by}), and let n = 1)'" and IJ = (q, 9' (1l'). Thr uecessaJ":1J aUll
sufficient condition for the congruence (1) to be solt'able is that
the COllgrliellCP
qJ (tJ'
a& == 1 (mod n)
hold.
A supplement to this result is
Theoroetn 70. Let }> be an odd prinze, and let a be au integer not
diz,'isible by p. Fll1.ther, Slll)P09() that q is a natural lunnbrr
> 2 not dirisible by p. If the Cotzgl'ue12CC
x q == a (DiOd pet)
is olt'able .fol" a  1, it Z"S also soll'able .for all (intp.q)'ol) (1.1:'-
}Jo)1euts a > 1.
Proof. If \ve put n  pft and 6 =- (q, 9'Cu.), tilen 6 =-- (q, p -- J).
If the congruellce
x q == a. (mod lJlt)
is solvable, we have by Theorem 69
(f (7I )
II d == 1 (mod p").

118
CHAPTER III
Hence
((I ( n \
a-r=.:: 1 + 1/- 1 .
where t is an integer. If \e raise bot.h sides of this equation
to the pth ])o\ver, it follows that
1) (I' (11) (I' (pn)
(I --;r-  /./-----rJ = \1 + pI! f)P
= 1 + f;) }l' ; + (n p2r: [2 -i
= 1 + Ji;- 1 11'
where '1 is an il1teg-er since  a > a + 1. Therefore ,ve have
cp (p n )
a-r = J (DIOd Jl+l).
Hence, fron1 Theoren1 G9 it follows that the coug'ruence
x'l = (/ (n10d p,t:+l)
is solvable. ana TheoreIl1 70 is proved by induction.
Further, we can prove
Thpore,n i 1. I.l p is an udd prinu', and l:I"U = pft afzd  = (q, tp (n)),
there aI"e qJ ') qtJt power l"csidue.'1 illcollgl'uellt modulo /I.
Proof By Theorem 69 the number requiren is equal to the
number of incongruent solutions of the congruence
cp('7I)
Lcd == 1 (mod 1/).
By Theorem 66 this congruence has exactly qJ ') illcongmellt solu-
tions. Hence the theoren1.
Examp7p. If II  17. there are four biqua.dratic residues in the
interval 0 - u, namel). 1, .!, 13 and 1 fJ.
"\Ve next consider the cong'ruence
(2)
x q == (I (mod :?U).
where II is odd, and prove

THEORY OF CONGRUENCES
119
Theore111 72. 1. fJ' q and a are odd 'nltnzbel'. the congrtu:nce (2)
has exactly one 801utioJ2.
2. Let a be an odd nzt.nzbel 4 and q == 2112, u'here '111 is odele Let
the eXjJonent a b( > 3. Then the congruence (2) has four -in-
congruent solutions if a = 1 (lnod 8); otlzerlt:ise it has no solution,
.. Let a be an odd II'll III be1 4 and q = 2 'In, u'here In is odd. Then
the congruence x q = a (lnod 4) has tu:o 'l:ncongrllent solutions
(f' a = 1 (mod); otherlt'ise it has 110 solutioll.
Proof If a > 3, we have by Theorem 63
(3)
(4)
a == (- l)h · Ok (mod 2"),
x = (- 1 ),lI · oy (mod 2 Cl ),
where lz, k, p and 1I are integers > O.
N o,v suppose that q is odd. By introducing (3) and (-1) in (2)
,ve <)let
e.
(- 1 }/l , 5t) y == (- l)h · Ok (nlod 2 a ).
Hence f-l == Iz (ulod 2) and by Theorem 63
q y == k (mod 2 a - 2 ).
This linear congruence has exactly one solution 11. Therefore,
the congruence (2) has exactly one solution x. In the proof we
have supposed a > 3. but the result is clearlJ valid also for
a=t and a=2.
Suppose next that q = 2 'In, 'In odd and a :> 3. By introducing
(3) and () in (2) we get
5 2m !1 == (-l)h · 5 k (nlod 2 a ).
Hence the number 11 is even, and thus a == 1 (mod 4). Therefore
2 '111, Y == k (mod :?a-2).
This inlplies A. == 0 (1uod 2) and a = 1 (111od 8). When this condi-
tion is fulfilled, there are two incongruent solutions .lI modulo :?a-2,
and consequently four incongruent solutions :c modulo 2 1t .
Finally J it is evident that the congruence
3.,.fl.m = a (mod 4)

120
CHAPTER III
is solvable if and ouly if a = 1 (mod 4). "Then a = 1 (mod 4)
it has the two solutions x = +_ 1 (n10d 4). Hence the proof of
Theorem 72 is complete.
The theory develo1)ed in this section D1ay also be used for
solving the general binolnial congruence
II x m == b (mod 11).
According to the results in Sect.ion 26, the probleul can be
reduced to the case where the modulus is a prime-po\ver.
ExaHlple. 'Ve consider the congruence
(5)
11 ;r3 = 17 (mod 56).
The number 3 is a prinlitive root of the prinle 7. From the
congruence
11 x 3 == 1 7 (mod 7)
we conclude
and
4 1- 3 · ind x = 1 (mod 6)
ind ..c = 1, , 5 (mod 6).
Hence
(6)
x == 3, 5, (j (n1od 7).
From the congruence
11 :L3 = 1 7 (mod 8)
we conclude
3 :J: = 1 (mod 8)
and
(7)
x == 3 (mod 8).
Combining (9) and (7) we finally get the following solut.ioIls of (5):
r: == 3, 19, 27 (mod 56).
35. Polynomials representing integers. - An integral polynomial
f(x) represents integers for all integral alues of It,'. Tllere exist,
however, other polynomials with the saUle property. An exan1ple
is the polJ'llolnia,] of deg'ree 11

THEORY OF COGnt.iENCES
121
( :C )  :  (x - 1). ( x - 11, - 1)
11 on!
which, by Corollary to Theorem 25, takes integral values for all
integral values of x.
'Vhell a polynomial represents integers for all integral values
of the \"'ariables, we shall call it, for the sake of brevitJ', an
i. r. polYllonzial (i. r.  integer representing). For such polJnoDlials
in one variable ,ve prove
Throte1n i'3. El.'er..ll i. t. }Jolynol1zial J(x) of degtee 11 in the l'ariablc
,:c mall be ll'ri lten i)1 the jor1n
(1) f(x) = Ao + A 1 () + As () +. + AN (:),
ll'he,.e the co,:f.ticiel1fs .Ao. .4 1 , . . ., .Ll n are iute,gc1.8.
p}.oqf. Ever" polynomial .f(x) of degree n nlay be written in
the form
( x ) ( ''(" ) ( x )
(:?) f(.}") = Co + f1 1 + f2 2 +,., + C"1 n '
\v here the nunlbers Co' fl, . . ., (" are uniquely deternlined. This
assunlption is true for pol)9nomials of degree zero. Suppose that
it is true for all polynol1lials of degree < )l -- 1. Then it is
also true for the polynomial f(x) of degree )1. For, if t.he coef-
ficient of x N is 00. the polYllomial g(x)=.f(.r) - GOIl! () is at most
of degree n - 1. Hence the aSSUID1)tion is t.rue for g (x), and by
induction for all f(.r).
N ow suppose that f(J') is all i. r. polynoDlia.l expressed in the
form (2). Since ,1'(0)  fo. the coefficient fO is an integer. Suppose
that the coefficients ro, C1, . . ., Cr-l a.re all integers. Then the
coefficient rr is also an integer. For b putting x =-. r in (2),
,ve have
.f(I") = ('0 + ('1 () + ('s () +. + ('r-l C.  1) + ('r.
Since ,{(r) is an integer, we see that (r is also an inteer.

122
CHAPTER III
Hence, by induction, TheoreDl 73 is proved.
In particular it follows: If f(,}') is an i. r. lJolynoJnial of de-
gree 11, the polynolnia.l 1J! f(.l:) is a,n integ-ral POIJlloDlial.
If an i. r. polynonlial for all integral values of the variables
represents integers, ,vhich are all divisible by the sa,me integer d,
we say that the polynon1ial has the same .fixed dil"i80r d. For
such polYl101nials in oue variable \ve proye
Theore111 7 4. J"l"e,.y i. r. ]JOl!ln()Jl1ial..f(.i.") of deUJ.er.: J1 in the ca1'iable x,
ll'lzz"ch ha.,- the fixed di l"ilOl' d, 1na!l be lfritten in the .lorn1
(3) f(x) = Ao + At () + -12 e) + · . . + A" e) ,
lfltere the coe..(1icienfl .Ao, 41' . . ., ...4. n are integers dirisible b!J d.
Proof. The integer Ao is divisible by d, since ..((0) = ...4. 0 Sup-
pose that the coefficients Ao, 41' . . ., 4.r-l are all divisible by d.
Then the coefficient. ..:1 r is also di¥isible by d. For.. by putt.ing
x = r in (3), we have
J(r) = .10 + ..1 1 C) .- .1 2 () +
-. ,1"-1 (
,.
r )
_ 1 + _4 r ·
Hence, Theorem 7.t is proyed by induction.
In particular it follows: If a prinlitive integral polynoDlial
9 (..c) of degree 12 has the fixed divisor d, then d is a divisor of Jl!.
For i. r. pol)"nomials in several yariables there are leesults
analogous to Theorenls 73 and 7-1.
36. Thue's remainder theorem alld its generalization hy Scholz. -
The follo,ving result due t.o Axel Thue is erJ useful for many
questions in number theory.
The01'enl 7:j. Let J1 he a nat'ttJ"lll J1lnnfJ('1. > ]. and let c denote the
least inie!1f'r > )/n. Then f01. any illte(lP1' a pri'lnc to n, there
exist f1,.0 uatllral uU'lnbtrs x alia y not eXfJf!f!di1lg e - 1 suclz that
(1)
({.II == + ,1'; (mod }l).
Proof. We consider all nnmbers of t,he fOrl11 ny + x, ".here
x ancl yare numbers in thp set 0, 1, 2, . . ., e - ], Since there

THEOR1 OF CONGRUENCES
123
are in all e 2 > 11 snell l1uJnbers, at least t.wo of them must have
the sanle principal reluainder 1l1odn]o J1. (Dirichlet's box: principle,
see Section 12.) If ,\ye suppose
a !Il + .J'1 == 1l?/2 - '/'2 (nlod n).
,ve call write
()
a (.'/1 - .'12) - 1'2 - ,1'1 lluod n).
IIere
o < 1."1 - ,'21 < e - I, () < 1 1."1 - ""21 < e - 1.
For, if one of the differences J....J - "2 and Y1 -- Y2 were = 0, the
other one would also be = o. By putting Y1 - .'12 -=!I and
.V2 - .2"1 = I ,{' in (2), ,ve obtain a congruence of the type (1),
and the theorem is proved.
Ho,vever, it is easy to see that bJ a slight alteration of this
proof ,ve can obtain the follo\",ing- luore general result (Arnold
Scholz) :
Tlu.'o}"(Jl11 'i'(l. Lr4 n IIr) (I J1afUl-111 llll1)2!Jer > 1, and [pf (' and ..f
del10tt tu:o natl( ral 'n'lt1nber,,' .'filch that
ef> u. e > 1,
> / '
J1 = .
,
Thr'}1 for an.'1 i nt(,.'}(),. (t prl JJ1(' to J1. there (J..cist tlCO oatil ral
11u,nbers x and 91 811eh that
..
a 11 = + .r (1110d 11)
aud
o < 1' < e, 0 < .IJ < f.
If n is a prin1e, we ea.l1 clearly suppose (x, y) = 1 in these
theorenls. Then it is convenient to write congruence (1) in the
form
}'
a == .:t. (n10d 11).
!I
Ixa1nl){{ 1. Putting- 11 =. 7, in Theorenl 75, ,ve have (' = 3 and
obtain the following six representations of (I in the fornl (3):
(3)
1 - 1 ') -:! 3 - 1 t - 1 ;; - :? l' - 1
=J. -=1, =-:!. "'t-= "=-J. O=-J'
In all these congruences the Dlodulus is 7.

124
cnAPTER III
Exanlpl( 2. Putting 12 = 11 in Theorenl 75, we hae r = -1 and
obtain the following fourteen representations of a itl the form (3):
1 -1 ')-2 3 - H -  4_1_ :i n- t
=j, -=1' =t=-:J, 0:1:="3=-2' V=-2,
6 - t 'j - 1 -:1 8 - :i -:! 9 - ! 1 0 - t
=2' .=-S=:!, =-t=H' =-1' =-1'
In all these congruences the modulus is 11.
In Exanlple 1 there is only one l.epresentation of every a; in
Example 2 there are two representations of the numbel's 3, 4"
7 and 8. (Compare Exercise 89.)
We shall use Theorenl 75 fOl. proving theorenls on the represen..
tation of natural llunlbers as sums of squares (see Chapter VI).
Exercises
-tl. Find the highest power of J 2 contained in ] 20!.
42. \Vith how many zeros does the number 20001 end?
-t3. Let p be a prhne. Show that there exist exponents 'In with
the follo,viug pl.opert..: There is no natural nunlber 11 such
tllat }}m is the highest po,ver of ]1 contained in n!. Develop
a method for determining the numbers 'In.
-t4. Find the greatest COmI110n divisor of the binomial coefficients
(';). (), · · ., C  1) .
45. Show that the fract.ion
(m + 12 - 1 )1
1nl 111
is an integer, when the natural nUlllbers n and n are rela-
ti vely prin1e.
4(). Let 1n and fa be natural l1UDlbers. Show that the l1unlber
(,n ]i)! is divisible by the nUlnber
12! (1111)".
47. Sho,v that the fraction
(2111)! (2 n) !
1n! n! (m + 1/)!
is all integer when 111 and 11 are natural nUn1bers. (Catalan.)

THEORY OF CONGRUENCES
125
48. Let n be a natural number and put
n
1.'n = 2 2 + 1.
Show that the numbers }'n al1d F,, are relatively prime
when 12  m.
49. Sllow that the cOl1grnence
.:c
2 2 + 1 = 0 (mod}))
has solutions x for an infinite number of primes p.
50. Find all natul"al numbers x such tha.t the product
x (.r + 1) (x + 2)
has no othel" prime divisors thall 2, 3 and o.
51. Let ,n and 1 be two naturalnuD1bers and 11 odele Show that
the sum
1 1 1 1
-. - + + -- + . . . + ---
'IU m+n 1n+1l 1n+xn
is an integer only for ol' = 0 al1d 1/2 = 1.
This result also holds for all even 11; the proof is, however,
more difficult ill this case, but may be accolllplished by
applying Bertrand's theorenl: The1"e is at least one prime
between x and 2:r - 2 when x > 2.
5. Find all positive rational solutions x and 11 of the equation
XU = '!Ix
and in particular the integral solutions.
53. Let 1(.1') be an integral polynomial in .1:, and let 1l and 'In
be natural numbe1's. Show that there exists an integer Xu
such that eacl1 of the numbers
j'(xo), ./'(.1'0 + 1), . . ., .f(xo + 12 - 1)
has at least ,n different prime divisors.
5-t. Let "n be = 0 01. = 1 acco1"ding as 12 is divisible by t.he
square of a prime or not. Show that the infinite decimal
fraction
E = o. 1'1 '2 "3 1-'ot ' . .
is an irrational 1l111nber.

126 CHAPTER III
55. Let f'{x) be an integ'ral polynomial in x which takes positive
values for all x > 1. Show that the infinite decilual fract.ion
1] = U. j' (1) j' () f (3) f (4) . · ·
is an ilTational nUDlber. If, for exalnple, ..f (.1:) = .l"2, the deci-
ulal fraction is 'I} = O.I,HH ()5:J(j.HH; HH IOU. . .
56. Prove Hermite's fonnula
 [  ] =2! [  J - [J ;.)2.
k=l! k=l A
As before [zJ denotes the greatest integer < z.
57. Let T (1") denote the nUlnber of positive divisors of the na-
tural number k. By means of the fornlula ill Exercise 56
prove the follo,ving relation:
[.l:J
 T (1-) = x log x + (:? 'Y - 1)..t, + c) JI x _
k=l
\vhere 'Y is Euler's constant = lim ( 1 + l + ... + ! -log 12 )
,,-.. j;" n
= 0.5772 . . ., and ,vbere  is a function of ..I" such that I  I
is less than a positive constant for all x,
58. Show tha.t the nUlnber of lattice points inside and on a
circle having centre at tIle origin aud the radius J 'x is
.4. (x) :=::: :t ..t' + t5 V x ,
where  is a fUllction of .l: such that I  I is less than a posi-
tive constant for all x.
Sug"gestion: Sho,v first. that
.1 (x) =  (1 -1- :J [J ..c - u 2 J) ,
It
the SUID being extended over all integers II such that I Ii I < JI  .
The sum can be evaluated approxiulately by considering
the integral
v;-
I J / .r - 112 d II .
o

THEOR1" OF CONGRUENCES
127
. We define the functions D (J:) and tp (x) for any x by the
fornlula
o (x) = L log 1).
l)Sa%
the SUUl being extended over all pl-imes 1) < .l', and
tp (../..) = 0 (.!') + 17 (.l.'l) + D (.t,i) + ...,
\vhere the series break8 off when :i,Ji m < 2.
Prove the following relations:
 11' ( x ) =  ([ :L ] + [ x 2 ] -,- . , ) log )J == log [xj!
k=l  p.r 1) 1)
and
[z] ( :r )
 (- 1)1:-111' k = log fx] ! - 2 log [!.r]!.
k=l
As usual, [z] denotes the greatest illteg'er < z.
By meallS of Stirlingl's fornlula, known fron1 analJ'sis, prove
the following inequalities:
log [x] ! < ;c log J.' - x +  log x -t- log V2n ;- 11:1
and
log [,1']! > :c log x - x --  log x + log V2n ,
and thus, for :1' > ..{.,
log lxJ! - 2 log [1J. I 1! < x log 2 +  log :J. -log l"8n + i 1 :?"
Use tbe latter inequality to l,rove the relatioll
'I' (x) < 2 .r ,
\Vhel.e tp (x) is the function defined in Exercise 59.
By llleans of the results in Exercises 59 and 60 prove the
inequalities
 100" l }
- 2 <   - log .J: < 2,
p:i P
the sum being extended to all prinles p < x. (1vIertens.)

128 CHAPTER III
62. Let j"(.x') == a:r? + bx + e be a priulitive irreducible integral
polynomial, alld let p be a prime dividing D = ,)2 - 4- a c.
Show that the number of incongruent solutions of the con-
gruence
f(:c} = 0 (mod lJ m )
is < 21 JIJj I for every exponent J1l.
63. Sho,v that the number
n! + 1
lias at least two different prime factors for infinitelJ manJ
positive integers }l.
Prove the san1e proposition for the nunlber
}II-I.
Suggestion : Use \Vilson' s theorenl.
6-1.. Prove tIle following theorem of 'V olstenholme: If 1J is a
prime > 3, then the numerator of the fraction
65.
66.
i!"
v 4.
68.
1 1
1+ 2 + 3 +
1
.+
1) - 1
is divisible by p2.
Let JJ be a natural number > 1. If a belongs to the ex-
ponent e and b to the expol1ent..llnodulo n, and if (e,J1= 1,
show that a b belongs to the exponent ef modulo 11.
Prove the follo,ving theorem: If a belol1gs to the exponent
4-  ( integer) for the prime 1l10dulus 1), then - a belongs
to the same exponent modulo 1).
If a is an integer, and if aX - 1 is divisible by p for x= 1)-1
but for no positive value x < p - 1, then p is a prime. (Lucas.)
Let t and 'Il be l'elativelJ prime l1atural numbers and let
n = 2 a · 5i' · 111,
where (10 Ill) = 1. If i' denotes the greater of the integers
a and {1, prove the following' theorenl: In the decimal ex-
pansion of !, the period begins after 'Y terms; if the number
'1l

(}9.
70.
71.
72.
_ 3
 .
7 -1-.
75.
7tj.
 ,..
4 4 .
THEOR'''' OF CONGRUENCE:;
129
10 belongs to the exponent I modulo nl, the nunlber of
terms in a period is J:
Find the J1unlbers whicl1 belong to the exponent 2 1 modulo (t
for 1 < t < a - 2. In particular, if a > -1., show that the
numbers belong.ing to the exponent 2 , !-2 are the llunlbers
E + 3 (n1od 8).
Show that a prilnit.ive root of p", where 1J is an odd prime,
is also a primitive root of p.
Show that if .'1 is a prinlitive root of the odd prinle }J and
if gP-l - 1 is divisible bJ p2, then II is not a prinlitive root
of lit for a > 2.
Show that if 9 is a prinlitive root of the prime p = -l n -1- 1,
then -- fJ is also a primitive root of p.
Sho\v that if .ll is a. prinlitive root of the prime p = 4'Jl + 3,
then - II belongs to th exponent . (p - 1) modulo J}.
Show that if a belongs to the exponent l (p - 1) for the
prime modulus p = -1- n + 3, then - a is a primitive root of i).
Show tbat if p = 8 n + 3 aud q =  (1) - 1) = 4- 7l -t- 1 are both
primes, then ::1 is a prilnitive root of 1).
Show that if]J = 811 - 1 anc1 q =  (p - 1) = -l Ii - 1 are both
prinles, then -  is a priluitive root of p.
Show that if J) = 412 + 3 and q  2 n + 1 (n:> 1) are bot.h
Plimes, theu - 3 is a primitiv root of }J.
Sho\v tllat 3 is a pril11itive root of allY prilne of the form
2 n + 1 (" :> 1).
Show that if S denotes the sun1 of all positive priluitive
roots < p of the prinle p, then
/-, s=: It (1 J - 1) (Dl0d 1)),
where fl is the r<.)bius function.
o. Sho,v that if a belong's to an oud exponent for the prime
Inodulus p, then the congruence
.H.
_ 9
. .
aX + 1 == 0 (mod}))
has no solutions :.t...
9 - 516670 Trygve .:.Vagell

130
CHAPTER III
81. DeteruIillc t.he sol utiOI1S " of the congruence
5.( == 3 x -t- 2 (ll1od 11).
8. For what values of {(, is the congruence
10'" = II (IlI0d 41)
solva ble ?
83. Show that the COllg-l'UenCe
((.];2 -t- IJ .l12 == r (nI0d p)
is al\vays solvable ill integers x and 11 when the number
a lJ r is not di visi ble by the prime p.
8-l. Show that the congruence
{( _3 + b .1/3 = C (DIOd i))
is always solvable in il1tegers x and y when the nunlber 7 a b ('
is not divisible by the priule 1).
85. In Section 35 the notion of .fixed d i ril10r was also defin ed
for polJuolnials in several val"iables.
Filld the possible fixed diyisOl"S of the honlogeneous integral
pol Yllolllial
a .1,-1 -I /J .I:J'I f- (. .t ,/ + d .1..' l/ + (J ll-l
. .. . . ,
which is supposed 11rinlitive. \Vhat conditiol1S nlust the
coefficients fulfill, if the polynonlial shall have the fixed
di visor 6?
8G. Prove the identity
 /l(n) _ 
 nl'l - ,r,. ( 8 ) ,
7l=1 
valid for all 8 = a + it when a> 1.
7. Let Q (x) denote the nUllIber of positive square-free integers
 .1.". Prove the relatioll
f'
(I (:1') =  .{' + )I;,
:t0i0l
where  is a function of x such that I d I is less than a posi-
tive constant for all x.

THEORY OF CONGRUENCES
131
H8. Put
71
f/J (n) =  cp (k)
k=1
aud jJ1'o\-e th( relatioll
(j) l.d = 3 2 :.1.,2 + lJ .I.' log .c ,
j'(,
where d is a function of "t. such that I b I is lpss than a posi-
tive constant for all :('.
89. According. to Theorenl 7;) (Th ue 's theorelu) there exists for
eacll integer a not divisible by the prilJle J' at least one
representation of the following type:
,I'
II = + -- (luod }J),
.1/
\vhere .£ and 11 are natural nUluberR < lip.
8ho'v that there are, for given nUDlbers a and jJ, at nlost
two such representations with (., 1I) = 1. If t.here are tro
representations, the l"ight..hand side hl the congrnence is
,
&!: in t.he oue and -- .v, in the otber.
Y 1I
For p .. 3. 7, 23 ttUU 47 there is for enclJ iuteg'er II ouly u
single represent.utioJl. Sho\v by applyiug' the result in the
precediug" exercise that 110 other prilues l' ha f'e this proprty.

CHAPTER IV
rrHEORY OF QUi\DRATIC RESIDlTES
37. The general quadratic congruence. .-- Let n be an integer
 0 and let ]) be an in teg'er prilue to n. .According to the de-
finition in Section 34 the number D i a quadraft'co re..idlle DIO-
dulo n or of the nUl1lber u, if the congruence
(1)
x 2 :=::= D (mod 12)
is solvable. If this congruence has no solution, ]J is said to be a
q-uadrafif: n012-re8idue modulo 11 or of the number n.
In this chapter we shall treat tIle two following l11ain problenls:
Find the quadratic residues of a given nlodulus n. Finel all
moduli of which a given integer is a quadratic residue.
It is easJ to show that the solution of the general quadratic
congruence
()
ax 2 + bx + (. == 0 (mod n),
(n > 1),
where a, band t are integers and a not divisible bJ" n, can be
reduced to the solution of a binomial congruence of the type (1).
For. if ]) == 1.J2 - 4: a r, the cong'1"uence () is equivalent to
-1.a 2 .:t 2 + 4:abx + -1.ae === (2(1.J.." + iJ)2 - I) = 0 (mod .tan);
b)9 setting 2 a x + b = ,iI, this gives
1/ 2 == D (mod -1. a 11)
with the condition !I = b (mod 2 a).
Consider next the case in which the numbers D and Ji in
congruence (1) are not relati-rely prime. We shall prove
Theol.e1n 7i. E;tI,ppo.t.te (D. 1l) = d, d = e 2 j". D = d III and .u = d nl.
Here r:e f, Gl and Ul are i nfe,lJer.. {(12d j" is square-free. Then
the necet,.sary and .l{l.licient rondifi()12 for the congruencp (1) to ')e

THEORY OF QUADRATIC RESIDUES
133
t"ol'able i' fhat (f, 12 1) = 1, and that f Ql be a quadratic reid'lte
of nl'
Pl.00j: Both D and n are divisible by e 2 . Therefore, if the
congruence (1) is satisfied, x must be divisible by ef. T:':1us, put-
ting x = e.fy, we must have fy2  al (mod 111). Hence (j: 111) must
be a divisor of al. This implies that (f, n1) = 1, and we conclude
from (fy)2 == J.al (tuod 111) that f al is a quadratic residue of }l1.
On tIle other hand, suppose that Cf'! 121)::=' 1, a.nd that .tal is a
quadratic residue of 111. Then there exists an integer .1/ such
that 1/ 2 ::= fal (mod 1l1). If we determine an integer x such that
f x  !I (nlod 111), it follows that .(:(.2 == al (mod 111)' IIence, after
D1ultiplication bJ e 2 f  d,
e 2 .:f2 x 2 = d al (mod d 121).
Putting (' f x = " ve get
Z.2  D (mod n),
and Theorem 77 is provd.
Thus it is sufficient to examine the binomial congruence
x 2 == D (mod n),
where the integer D is prime to the n10dulus u. According t.o
,vhat ,vas shown in the begillning of Section 26, it is sufficient
to consider the case in which n is a prime power. ]f we put
f) = 2 and (( = D in Theorems 70 and 72, we obtain the follo\ving
result: A necessary and sufficient condition for the integer D
to be a quadratic residue modulo Ii is that D be a quadratic
residue of all prime factors of n, furthermore that D == 1 (mod 4),
when II is divisible by 4, and that D == 1 (mod 8) when 12 is
di visible by 8.
38. Euler's criterion and Legendre's symbol. -- In Theorem 69
put q = 2, {( = D alld n = jJ = :2 It + 1 = all odd prime. Then
" = (2, jJ - 1) =:? and cp ) =!t. It follows from the congruence
DP-l- 1 == (D'I - l)(Dh + 1) = 0 (n10dp)
that D II is either congruent to 4- I or to - 1 Inodulo 1). Thus
,,.e can state

134
CHAPTER IV
T/u!orern 78. I.f 1) =  It I ] i' (('1l odd }J1"iou\ the iuf('!!")" D i", a
quadratic rr",iduc or a fju(ldfafi(. UOU-f('sitill" of jJ, aCl"oJ'diug
(I."; Dh i...; ('nng1"Jtent to + 1 or to -] nlodulo 1J.
This theorenl, kuo\vn as E'liler.s c}'iterion, Jl1ay also be proved
directly ill the following- ,va" ,vithout the llse of the theory of
pritnitive roots. ErJ integral Rqua.I"e whicb is not divisible by])
is clearly cOllg-ruent to SOUle of the h =  (1) .-- 1) squares
1 2 -.)2 3 2
... .. . , ,
J 2
, .  .
It is easil. seen that these l1UInUers are incongruent 111oclulo 1J.
Hence, any quadratic residue of p is congruent 111odulo )) to
exactly one of tbese square. But it follows fronl D  ;t 2
(mod i)) that
]J" 2:: ,/.211  .x1j-l == 1 (nlod l J ).
Suppose next that 1) is a quadratic nOll-residue of 1'. and con-
sider the cong'ruence
(1)
./.' Y == D (lnod })).
To every iJltegral value of 11.' in the interval 0 - p there cor
responds exactly one value of 11 in the same interval; and we
never can have ;J: == y (mod p). Thus \ve can write down} (p -1) ==h
congruences (1) in which .1.' runs through one half of the nunlbers
1. 2, 3. .. .}) - 1. while 11 runs through the other llalf. :Wlnlti-
plying together all t,hese COllg'rueuces we ha.ye
1.2.3
{p
I}! == Dit (mod }J).
and by \Vilson's theoren1
D't .:.-.-. - 1 (In od }J).
FroIll this proof we also obtain
Theoreuz 7.9. If' J) i,.... au oilt! }J}'inze, thete fire just as 111aull quadratic-
rflf.."tdI(!6' as JluJ/-r(J.,.iduC8 H10dulo }J, i. e.. thfrf' ore J ()7 - 1) q{
each ki1ld.
This result iR, ho\ve\"er, a corollary to rheoreln U8. as is easily
seen by putting q =, (( = D, n ==}J and () === 2.

THEORY OF QUADRATIC RESIDUES
135
For an odd prin1e p, Legendre introduced a symbol defined in
the following nlanller:
(  ) -= + 1, when lJ is a quadratic l'esidue of p;
(  ) = -- 1, when D is a quadratic non-residue of p.
The following relations are valid for this symbol:
(  ) = Di.p-lj (mod p),
(f) = ( '). when D = D' (mod))),
aud further
( Dl') = (  ) ( ' ).
From the last relation we conclude: The product of t\VO quadratic
residues or of two quadratic non-residues is a quadratic residue.
The product of a quadratic residue and a quadratic non-residue
is a quadratic non-residue.
If 9 is a primitive root of p, the incongruent quadrat.ic residues
are represented by the even powers
1. lJ2. (J4, . . ., gP - 3
and the incongruent quadratic non-residues by the odd powers
g, 9 3 , as, . . ., g, J - 2 .
By putting D = - 1 in Theoren1 78, \ve obtain
Theo}.e111 80. The )lll1nbe}. - 1 is a quadratic residu(. 0.1' alljJrilluw
of the fO}.n1 4- n + 1, and a quadratic Jlon..reidlle oj" alljJri'llle.";
0.1' the ..fo J . n ? 4- n + 3. Using Legendre.s "'l/lnbol ire have
(2)
( p 1 ) = (-l)i(p-l!.
It tollo\vs frolll Theorem 80 t.ha.t the odd prin1e divisors of
the polynomial x 2 + 1 are the primes of the form  I? + 1. This
result is already kllO\Vn from Theorem 58 in Section 30.
.,

136
CHAPTER IV
According to l'elation (4) in Section 30 we ha,e
(3) ( p -; 1 ) ! == + 1 (mod p),
when 1) is a prinle of the forln 4 12 + 3. Let 111 denote the nunlber
of quadratic non-residues of }J in t.he interval 0 - ! p. As a
consequence of our statelnel1ts just made about the product of
residues and non-residues, we can conclude: On the right.hand
side of the congruence (3) + ] or - 1 is to be taken according"
as the nUIl1ber tn is even or odd. For t.he number -- 1 is a non-
residue n10uulo i).
If }) =-= 7, there is one single positive non-residue <  lJ,
namely 3. If P = 19, there a.re three positive non-residues < llJ,
nanlelJ" 2, 3 and 8. If p = 3, there are four positiv.e non-resi-
dues < }), nan1ely 5, 7, 10 and 11.
Suppose now that }J is a prill1e of the forln -1 n + 1. Then
the number of quadratic residues in t.he interval 0 -- ! p is
e(lual to ! (p - ]). For if a is a residue in this inter,\ral, then
p -- a is a residue in the in terva.l i]J - p . Thus, t here are the
saJne nUIIJber of residues in both intervals. Also th nun1ber of
non-residues in the interval 0 - l}J is equal to l (p -- 1).
39. On the solvability of the congruences x 2 a + 2 (mod p).
We shall prove the following theoren1:
ThpfJreHl 81. Thr' 12ul1zber 2 is (( quad,-afic re,,'id lie qf' all })ri IIlP.'$ of
(,ifhf1. of the .fut1118 8}2 + 1 and 8 u + 7; if i.,) {{ quadJ'atir: '1l0)1-
re.,;idue of all }JriJne, o.l (-i Ihe,. (f the .f'ol'n1, H Jl + 3 {I url 8 12 + D.
p,.oof SUPl)OSe first that the prill1e 1 1 is of the form 8n + 1;
then we have
x p - 1 - 1 = (x 8 - 1) {J (x) -= (x' + 1) {Ii (or),
where 9 (.r) and Hi (x) are integral "polynomials in x. Therefore,
according t.o TheoreJll 43, the congruence
,}"' + ] === 0 (lnod 1))
has exactly four incongruent roots:1'. DeterJllining the integer !I
such t}lat

TIIEORY OF QUADRATIC RESIDUES
137
","'" == ;e2 + 1 (nlod 1)),
it follo\vs that
.1}2 == 2 (mod p).
Hence, the number 2 is a quadratic residue of any prime ]J == 1
(mod 8).
Next we shall show that the number 2 is a quadratic non-residue
of all prinles }) == + 3 (lnod 8). This is correct for }) == 3. Suppose
there exist primes = + 3 (nlod R) of which the nunlber  is a
quadratic residue, and suppose f} is the least of these prill1es.
rrhell the congruence
;r 2 == 2 (DI0d q)
,vonld be solvablc. \\Te can suppose that the solution ,}' is posi-
tive, odd and < q, since -:J..' is also a solution. l'hen ,ve should
have x 2 - :! = q.f. where the integer ,/" satisfies the following
inequalities:
(1)
q2 _ .)
o <.« -_:. < q.
'1
Thus, the number 2 would be a quadratic residue of any priu1e
factor of the odd integer "t: Hence. by (1) and by the hypothesis
on q, every })rilne factor of f ,vould be either of the fOrll1
8nt + 1 or of the forDl 8 1n - 1. Therefore we should have
f =- + 1 (mod 8). and x 2 - :! == f}.( = ,1 3 (DI0d 8). But this con-
gruence is itnpossible, since .1: 2 .. - 2 == - 1 (nIod 8). Consequentl"
we conclude that there is no prime q, and that
e) = - 1
for all prilues }) == + a (lnod ).
Finally we show that the number 2 is a quadratic residue of
all primes of the form 8 J2 - 1. By Theorenl O it is sufficient
to sho\v that the number - 2 is a quadl'atic non-residue of these
primes. Suppose there exist primes == -- 1 (mod 8) of l\.11ich the
nUlnber - 2 is a quadratic residue, and suppose q is the least
of these pl'imes. Then the conglwuence
x 2 == - 2 (nlod q)

138
CHAPTER IV
would be solvable. "r e can suppose that the solution x is posi-
tive. odcl and < q, since - :J is also a solution. Then we should
have x 2 + 2 = qj where the integer f satisfies t.he following
inequalities:
(:?)
( "» 2 + ,,)
0<1< q-- - <q.
fJ
Thus, the number -  would be a quadratic residue of any prillle
factor n of the odd integer f. Hence, by (2) a.nd by the hypo-
thesis on q the prilne n could not be of the form 8,n - 1. It
could not be of the form 8 'In - 3 either; for, if :Tl = 8 )11  3,
we should have, according to the result just proved,
( ') )
 11l --- 3 = - 1
and
( - '» ( -- 1 ) ( 4t) )
8»1-=3 = m-3 8111 :'-3" =-1.
Thus, n would be either == + 1 or = + 3 (mod ) and the saDIe
would hold for f. Therefore ,ve sl10uld have :1.. 2 + 2 = g.l" -= - 1
or = - 3 (mod 8). But this is impossible, since :r 2 + 2 == 3 (mod 8).
Hence, we conclude that there is no prime q, and that.
( ')'
p"" ) =-1
or
()=( pl )( :? )=(_1).(_1)= + 1
for all pri111eS 1) of the form 8 12 - 1. Thus the proof of the
entire theorem is complete. Using Legendre's symbol the result
may be writ.ten in the form
(3)
() = (- 1)11])"-1'.
Combining Theol'elns 80 and t\ I,ve get

THEORY OF QUADRATIC RESIDUES
139
Tllt'Off'ln Nid. Tilt, lIunlbr;' - -  i.,' a quadratic- residue 0,( alllJrilurs
({l "ith,'r q( the .!'urn2 f ... 8 H + 1 alul 8 n + 3; it is n qllad?afi(-
nou-'''fl,(,'idue of all pl'i111e.. (( pit!zf>1" o.l tile jorl1l8 8 II + 5 and
8n + 7.
*
'Ve shall also establish SODle results 011 biquadratic residues.
BJT means of the identitr
x' + 4 == ((x + 1 ',2 + 1) ( .r.' - 1 2 + 1)
and Theorem 80, ,ve deduce immediately
TlzefJl't ' }11 8H. The nUJ}11Jl'J'" - .... i.,-: (I in'quadratic 1't>sidU(I of all }Jrhne....'
of' tlu' .Iornl 4 Jl + 1 and o.{ no othel" jJ1"i111e8.
Further we havE:'
T/z('orr.ll1 8.1. The J1ll'lnheJ' - 1 i8 a hifJuad7'ufic 1"el:lidue of all p,"irne8
of the fin"ln 8 n -I- 1 a 12d 0.( no otluT jJ1"i'l12e > 2.
Proof. From Theorem 80 we know that the congruence
(4)
,r' = - 1 (Iliod p)
is solvable only if lJ == 1 (IllOd 4). Frou} the proof of Theorenl 81
(first part) we see that congrnence (....) al,vaJ.s has solutions if
1) = 1 (DIOd 8).
N o'v suppose that )} == f) (Inod 8). If (4) were solvable, we
should have from this congruence
p-l p-l
x p - 1 = (.J. 4 f 4 -= (- 1) 4 = - 1 (nIod })).
But this is contradictor)" to FernIat's theorem.
40. Gauss's lemma.
lenI11la due to Gauss:
\Ve shall establish the following useful
Theo1"l'1Jl 8;). Lt.t p be an odd prhne and D all illfcger nof d1'l'i.iblp.
bll p. If /l (lelote.\: the nU1nlJer qf intf.gers in the .....pquenre
(I)
1 . I), 2. 1), 3.]), . . ., . (p - 1). [J,

140
CHAPTER I'V
,ohose princz})al '''e'naitlde''8 I'IlOrllllo pare >  1J, then 1ft"' ha'l:p
() = (- 1)1'.
/Jrooj: The numbers (1) are cJearly incongruent modulo 1J. For,
the congruence h. n == k. n (nlod p) only holds for h =:= k (mod 1)).
Let al, a2  . . . \ a.1J. be those of the principal remainders modulo }J
of the nunlbers (1) whicll are > 'lJ, and PI, P2, . . . \ Pi.. those
which are < }). Then A + P. =  (p - 1). The nunlbers 1) - aI,
1 J - a2, . . ., p - au are aU in the interval 0 - lp. None of .
these nUlnbers is congruent to a.ny one of the nUDlbers pj mo-
dulo p. For, p - at £: pj (ulod p) and at = rD, pj = sD (mod J})
inlplies r + s  0 (nlod p); but this is impossible, since '1" and .,,'
belong to the seqnence 1, 2, . . ., 1 (p - 1) and have a positive
sum < p. Thus, the I (p - 1) numbers
(2) PI ' P2, . . .. p;., )) - al , P - a2, , . .. }) - a,ll
are all the natura] numbers < ! (p - 1). Hence, formil1g the
product of all the numbers (2), we get
PI P2 ' . . p). · (p - aI) (1) - aa) (p - au)
= ( P ; I )! = (_ l).u . ( p;- I )! Bl(P-I) (mod p).
Since
Di(p-Ij = (  ) (mod p),
it follows that
() = (- l}u,
and Gauss's lemma is proved.
Now suppose that D is a positive number. For k = 1, 2, .. ,
. (p - 1) we put
kD=P[ k: J + I"k,
where fl: is the principal remaindr of 1... D modulo p. Then, taking
the sum over all k, and recalling the identity

TIIEORY OF QUADRATIC RESIDUES
141
1  :! - . - +! (p - 1) =  (})2 - 1),
\ve get the relation
· (p-l) [ I.- D ]
(3)  (})2 - 1) [J = P   + .A. + IJ
k -1 1)
}J1Jl + ..4. + B,
where A is the sum of the nunlbers al  a2' . - . all and B the
sunl of the numbers PI' P2. . . .. Pi.. Furthel" we have
1.& i
 (1) - ai) T  Pi  1 + 2 + ,+ 1 (p - - 1) = . (jJ 2 - 1)
i=1 i=1
or
(4)
p p - A + B =  (p2 - 1).
Elinlinating B between (3) and () we obtain
 (p2 - l)(JJ - 1) (lJI- It)1) + 2 A..
Hence
(5)
p, == .1JI +  (})2 - 1) (1) - 1) (nlod 2).
\Vhen n = 2, then 1J[ -= 0, and thus
,u ==  (})2 - 1) (mod ).
Hence we ha,"e a new proof of formula (3) in Section 39_
41. The quadratic recip..ocity la".. - W e first pl'o\"e a tlleorenl
of Eisenstein:
Theoren1 8ti. Let a aut! iJ he tu:o odd intellers > 3. 1.1' (a, Ii) ;:=: 1,
aud a' = } (a - 1), 1/ = . (I) - 1), 'lct> halor
i [ b U ] + i [ a v ] =:= a' 1/.
It=l a ('=1 b
Proaf. We consider the a' (/ integers
(1)
b If - a.,,-
for it = 1,2, ° . ., a' and 1" = 1,2. , . ..1/. None of these numbers
is equal to zero. For, since a and , are relatively prime, the
equation b If = a l.' implies l' =-= II t, 1" = b t. Exactly

142
CHAPTER IY
;,'1
, )' r h " J
 (I
Ii 1
of the numbers (1) are p08itie. For. if 'll is fixed, then IJ II > {( '"
0 ' [ ' . 1 (1 11 ] .
for l" = 1" 2 " Further. exa ctl J
1/ r ]
,,1 =",' {( ,.
J. 2  IJ
, =1
of the nunlbers (1) are neg'ative. For, if c is fixed, then IJ II < a I'"
f ... - 1 ') [ l! V ] Th h 1 '.. d
01 II - . -, . . ., IJ. us t e t leorenl IS plove .
The proof nUl.J be interpreted geoD1etrically as follows: In a
two..dilnensional rectangular coordinate-system with the abscis..
sae ;).! and the ordinates 11 we draw the straight line L fron1 the
origin to the point (a, b). In the first quadrant we nlark the
lattice points (.1". .11) w]lich satisfy the conditions
1 < < ' 1 < < I '
=X={(., ==Y=.;.J.
/1
u
1/
, 1
II :: a J'
Fig, 5,

THEORY OF Q1. T ADRATIC RESIDUES
143
It is readily seen that these lattice points are distributed in the
follo,,'ing 111anner: 011 the straight lil1e L there are no lattice
points, since (a, b) = 1. Between L and the x-axis there are Sl
lattice pOil1ts; between L and the !I-axis there are 52 lattice
points. (In fig. [) it is II = 15 and b = 11.)
N O\V ,ve will apply this result to prove the quadrati(. recijJI'(j-
rit!1 lall":
'} ' J i) '"
l()Or(Jlll /.
rl)latiou
I}' jJ and q llf(' di:tiJltt odd ]JMnU'8, Ire hal'r the
() (:) = (.- 1 )11,
'll'here
h =  (jJ - 1) . . (q -- 1).
])1'00.1: Putting D = fJ in rela.tion (5) in Section 40, \ve get
p. = JI (mod 2)
and
() = {-1)-1l,
where
t {P11 [ q tt J
ill =  -- .
u=l P
Interchanging- 1J and q, \\Fe get the analogous relation
() = (-- 1)-'",
\vhere
T 1 (q-li [ p r ]
=  - .
1'=1 .'1
Hence
G) () = (- 1)J1+.Y.
Finall)., bJ putting in Theorenl 86 a = p, IJ = q. it follows that
.ill + _,r = a' 1/ = ! (j) - 1) .  (q - 1).
This proves the reciprocity la\v.

144
CHAPTER I'"
It is readily seen that this theoren1 ll1ay also be formulated
as follo\vs: If at least one of the prinles )J and q is = 1 (n10d 4),
then J} is a residue or a non-residue of fJ according as f} is a
residue or a non-residue of 1). If P and q are both ::::: - 1
(mod 4), then}) is a residue or a nOll-residue of q according as
f} is a non-residue or a residue of }).
The <luadratic reciprocity la\v was stated by Euler (1783) \vithout
proof, but ,vas first pro,?ed by Gauss (179u), who gaye not less
than eight different dellionstrations of it. Ho\v great the interest
in this theo1"elll has beeu among lllathematicians is apparent from
the fact that about fort)" other proofs, 1l10re or less diffel"ellt,
have been published since Gauss. Anlong the authors of these
proofs we Inention Cauchy " Jacobi, Kronecker.
The reciprocity law facilitates considerablJ' the determining of
Legendre's sYlubol. This is illustrated by the following example:
C ) = - C:g 1 ) = - G ) = - (:)
= -- () = - e7 3 ) = - C / ) = + 1.
Let us use the reciprocity law for deterl1lining the prinles J)
of which the number -;} is a lJuac.1ratic residue. We g'et
(  3) = ( j 1) e) = (  1) () (_ l)ilp-l) = ().
Thus \ve can state
Tllt'orenz .8. The 1lZl1nbeJ' - 3 is a quadratic. re...,.idup oj' olljJrl1Ju's
vj' the L!'ornz (j 11  1, and a quadratic U012-}Y.,..,.idue 0.1" all})rhne8
of the jorn fj Ji - 1.
Similarly, fro 111 the relation
() = ()
we deduce
Theorl'l11 8!J. TluJ ulnn/It r [) l:. a qu(uJ,'atic: reidue oj' alljJ/'iuze,'\ oj
(,ither oj' the ,j'url128 10 Ii .:t 1, a ud a quadratic- non-resid,u' (1
all ]Jrhues u.l ei the} u.f' tlte jOJ 'n1N IOn + 7.

TIIEORY OF QUADRATIC RESIDUES
145
As a consequence of Theorems 88 and 89 ,ve see that the
polynomial
x 2 + 3
bas the following prime divisors: 2, 3 and all the primes of the
forln G n + 1; a.nd, that the polynomial
4) -
;),:-- - D
has the following prime divisors: 2, 5 and all the prinles of
either of the forD}s 1011 + 1.
42. Jacohi's symbol and the generalization of- the reciprocity law.
Legendre's symbol () is defined for primes 1) only. Jacobi
introduced a more geneml symbol (  ), defined for all odd natu-
ral nunbers P in the following waJ: When
p -= JJll J 2' '1 J m
is a product of pl'iules PI' P2, . . ., Jim, distinct or not, and when
D is an integer 'P1.ime to P, then
(1)
(  ) = C) C ). . ( :J '
\vhere the factors on the right-hand side are Legendre sJmbols.
For ].J = 1 the Jacobi sJ'mbol is defined by the I.elation
(i)
(  ) = + 1.
If D is a quadratic residue of P, then () = + 1. since all fac-
tors on the rightband side of (I) have the value -1- 1. On the
other hand, if D is a quadratic non-residue of P, it is not al-
ways true that (  ) = - 1. In fact, when an even number of
factors on the right-hand side of (1) have tIle value - 1, the
product has the value + 1.
10 - 516670 TrY(Jl."e Nagell

146
CHAPTER IV
From the definitions \ve ea.silJ get t,he follo\ving rules for 011era-
tions "rith the Jacobi sJlllbol:
I. 1.1' the illtelleJ'' D and D' are )JriUle to t/if odd lJO'itire integer P,
then ll'e hal"e the relation
(3) (t) ( ') = ( n:') .
To prove it 've have only to apply the corresponding rules .for
t.he Legendre symbol.
II. l tlu 'illtcgr1., I) and D' arc }Jrinze to the odd }JDctoiitirr in-
teger P, and ,{( D = I)' (nzocl P). then 'ire hate the relation
t-l)
(  ) = ( t') .
For tIle corresponding rule holds for the Legendre symbol.
III. I.f' the integet D if'- }Jrinze to the odd ]Jo,'1itiz-e illtlgcr P and Q..
then 'U;e ha,.e the rel ([tioll
(0)
() () = ( Jf(J )'
The ])roof follows from the definition, if \ve wI"ite P and (i as
products of vrime factors.
The following I"ule is analogous to the rule given by fornlula (2)
in Section 38:
IVe J."u,. an!1 odd lJDsitife P lfe hazer the relatio11
(6)
( -- 1 )
P  (- 1) (P-1) .
For.. writing P as a product of prime factors in the form
m
P = }J1 P2 ,.. 1 J m = II (1 + pz' - 1),
i=l
we g-et
m
P == 1 +  (Pi - 1) (mod 4),
i=l

THEORY OF QUADRATIC RESIDUES
147
or
m
 (1 J - 1)  1, 1 (Pi - 1) (mod 2).
i=l
Since
( ) ) = iI ( - .1 ) ,
1 i=I}J,
\ve obtain the relation (6) by using formula (2) in Section 38.
Furthernlore, we prove the following rule analogous to the
rule given by formula (3) in Section 39:
V" For an!1 adll jJositiz"e P u:e /zal,"e the relati()1l
(7)
() = (_ 1)1 (J4-11.
For, writing P as a product of prime factors in the forD1 P =
PI ]J2' . lJrn, we get
m
p2 = 1-1 (1 + P7 - - 1).
1=1
Since every llumber lJ - 1 is divisible by 8 and every product
of two of them bJ' 6, we obviousl)"" have
m
p2 - 1 ==  U} - 1) (mod 64),
-/=1
or
7)1
 (P2 - 1) -:-   (]); - 1) (mod 2).
i=l
Since
( 9 ) ,,.,., ( 9 )
- =II  ,
J.J t-=1 Pi '
,ve obtain the relation (7) by using formula (3) in Section 39.
Finally we shall prove the following generalization of the
reciprocity law:
ThefJrp.ul .90" If P and (I are tl("O l)a.iiil"e, odd, a'nd relatil,'ely p,-im
intf!ler.. thef 'U.e hat"e the relation
(  ) (  ) = (- 1)1.,

148
CHAPTER 1\1'
1t,here
h =  (P - 1).  ( (t! - 1).
Proof. Suppose
P .=. PI P2' . pr, (.! = q] q2 1, ,
where Pl' P2 . . ., l)r, ql, Q2, . . ., q, are primes. Then
and
() = ( : ) a: ) ." (  )
(  ) = (  ) (:) .'. (  ).
( Q ) ( P ) = n ( lJ ) ( Pl ) = (-1)',
P (/ ,. PI qJ
I,)
Thus
where, by the reciprocity law (Theorem 87), f has the value
r , r'
t =  !! (pi - 1) -1 (qJ - 1) =  l (Pt - 1) · . (qj - 1).
i=l j=1 t=1 j==1
In the proof of formula (6) we showed that
r
 - (pi - 1) == ! (P - 1) (mod 2).
i=1
Then we also have the analogous congruence
It
. (q.i - 1) == 1 (Q - 1) (mod 2).
j=l
Hence
t = l(P-l). -((J - 1) (mod 2),
which proves Theorem 90.
Exan1)Jle. The number 2137 is a prime. Decide whether or
not the number 666 is a quadratic residue of this prime. 'Ve
have

THEORY OF QUADRATIC RESIDUES
149
( J6f» ) ( 2 ) ( 333 ) ( 333 ) ( 137 ) ( 139 )
2137 =. 2137 137 = 137  -333 = 333
( H33 ) ( 55 ) ( 139 ) ( 29 ) ( 55 )
= 13 = 1 39 = - 55 = - [)O = - !J
= - (:) = - ( 3 ) = - (;9) = - e:) = - ( 3 1 ) = + 1
43. The prinle di,-isors of quadratic polynomials. --. ,,:r e h11 ve
alreadJ determinpu. the prilne divisors of t.he PolYllonlial ).;2 + 1
(Theorem 58 in Section 30). Further, from Theorems 81 and 82
(Sectio11 39) it is apparent what })rinleS are the prime divisors
of the polynomials ;1..2 - 2 and x 2 + . N O\V we shan generalize
these results and show how to determille the prinle divisors of
all quadratic polJ'l1omials.
'Ve showed in Sectioll 37 that it is sufficient to consider
polynolnials of the form
(1 )
l.,2 - D.
If D is a perfect. square (2, this polynoDlial is the product of
the linear factors J - C and .i: + C: in t.his case any prime is
a 1)rin1 divisor of the polJnonlial. Suppose next that D is Jl0t
a l)erfect S(luare and let D = C 2 Dl ,vhere Dl is a square-free
l1Ulllber 7= 1. Putting  = (Y y , t.he polynomial (1) becomes
c 2 (!/2 - D 1 ).
Tllis polJDomial ob\"'iollsly has the same prime divisors as (1).
Therefore it is sufficient to consider the polynolnials (1) in which
n is a square-free int(\ger ;ft 1.
'Ve shall first prove
Tlzeol'eHi .'ll. If P i.,- a a'-qllare-JJ'ce, odd int(Jger > 1, then
(2)
s =  (l ) = 0,
n"
1ch(!l'e tlu ...:lll)Z i.... p.rtended orei' all 121l1ilber. 111 '1 n a reduced
re.'-e'idlle s!J....tenl nlodlilo J).

150
CHAPTER IV
Proof. There always exists an integcr b such that
(3)
G )=--l.
For, let p be a prime factor of P, put p' = P , and denote by t3
1)
a quadratic non-residue of p. Then we can determine an in-
teger b satisfJing the congruences
b = {J (mod 1J), b == 1 (lllOd ])');
this is possible by Theorem 40, since (1), P') = 1. This number b
satisfies the relation (3), since
( t ) = (;) (;,) = () (  ) = - 1.
If the number 11l runs through a reduced 1.esidue system modulo P,
80 does the number 1)1 b, since b is prinle to P. We therefore
obtain
s=  ( ": ) = ( ; )  (;) = - s.
Hence S = O.
Q. E. D.
Let I-' denote the numbeI' of incongruent numbers a Dlodulo })
such that (p) = + 1, and let v denote the number of incongruent
numbers b modulo P such that (  ) = - 1. Then Theorem 91
states that
(4)
It = " =  q; (IJ).
Now we pass to the deternlination of the prime divisors of
the polynomial (1), where J) is a square-free integer rf 1. It is
evident that tIle prilne 2 and every priule factor of 1) are
prime divisors of the polynomial. Therefore apart from these
primes, the problem is to clet.ernline the odd prinles ]1 for ". hich
(  )  + 1.
It is convenient to distinguish four different cases.

THEORY OF QUADRATIC RESIDUES
151
CaNe J. D === + P == 1 (mod -l); P > o.
Let (II, Q2. . . ... ll). denote the  lp (1.)) odd illtegel"s in the interval
o - "J P for which (  ) = + 1, and let hI, h 2 - . . _. b. denote the
4 Ip(P) odd integers in the same interval for which (  ) = - 1.
Then the necessary and sufficient condition for the prhlle J)
(\vhich is not a divisor of 2 D) to be a prinle divisor of the
polynomial (1) is that
1) == at (mod 2 P),
(i  1, 2 . . .. ,,).
For, it follows from Theorem 90 that
( D ) = ( E ) = ( ai ) = + 1.
1) P 1 J
On the other hand, the primes fJ. which are not prime divisors
of the polynomial (1) are characterized by the congruence condi-
tions
q == hi (mod 2 P),
(£ = 1, 2: . . ., 'J1).
".J:aHl)Jle 1. I.f J) -= 1, we find tha,t the prime divisors of the
polynomial x 2 -  1 are, apart from 2, 3 and 7, the prinles })
satisfying any oue of tILe congoruences
}J = 1, 5, 17.. 25, 37, -ll (Juod 42).
"xa)Jlplt) 2. If D = - 15, we find that the prinle divisors of
the polynoD1iai x 2 + 15 a.re, apart from 2, 3 and 5, the primes 1)
satisfying any onp of the congruel1ces
]) == 1. 17, 19. 3 (nlod 30).
Ca,,'p 11. D = + P = 3 (luod -l): P > o.
Let Gl, 112. ..., Q)I denote the . qJ (P) integers in tbt interval
o - 4,p which are = 1 (mod 4,) and for which () = +- 1. Let
b 1 , b 2 , . . " b), denote the - cp (P) integers in the interval 0 - -l J.)
which are == 3 (mod 4,) and for which (;) = - 1.

152
CHAPTER IY
TheIl, the necessary and sufficient condit-ion for the prinl p
(,y hich is not a, di'tisor of 2 D) to he a prilne divisor of tIle
polynolnial (1) is that either
JJ == ai (Inod 4- P),
t i = 1, 2, . . ., 'I'),
or
)J == bj (mod -t. P)
(j -== J, -2, . . .. 'I').
For, it follows fron1 Theorelll 90 and formula (6) in Section 42
that
() = (-l)i}J-lj (;) = (  ) = - G) =  1.
Jxaml)le H. If]) = 1 f), WP find that the prime divisors of the
pol?non1ial 3.. 2 - 1;) are. apart froln 2, 3 and 5 the prilues 1)
satisfying anyone of the cong'l'uences
p '::2 t  7, 11, 17, -13, 4-1, 53, 59 (Inod (0).
("fa.....p. l II. D = + 2 ]J ==:? (rnod ); p> o.
Let al, ([2  . . ., llq' denote the q; (]») in teg1ers in the in ter\al 0 -  P
which are = + 1 (mod 8) and for which () = -j- 1. Let hI' b 2 -
. . ., b q denot.e the q:' (P) integers in the interyal 0 - ]) ,,-hich
are =:= + 3 (mod 8) and for which () = - 1.
Theil, the necessary and sufficient condition for the prilHe p
(\\-hich is not a di,isor of J)) to be a prime divisor of the poly-
llolnial (1) is that either
1 1 == aj (Juod 8 I»), (i = t, 2. . . .. g.),
or
p == b j (mod 81»).
( . - t ') )
./- ,-....,fP.
.b'or, it follows from Theorem 90 and forn1ula (7) in Section 42
that
(  )=(_1)liPLl ()=(li)=-( ; )= + 1.
l.,t'arn}Jlc 4. If n == - 6, we find that the pritlle divisor of
the polynomial ..{,2 T () u.r(l, apart fronl 2 and 3, the priInes p
F;ntisfJing any oue of the congruences
j) == 1. i'), ", 11 (nlod 24).

THEORY OF QUADRATIC RESIDUES
153
Calp IJ T . D = -+- "'2 P  (j (IDod 8).
Let aI, 02, . . ., a q denote t.he cp (1») integers in the interval 0 - 8 P
,vhich are either = 1 01' = 3 (mod 8) and for which e ) = + 1.
Let hI, U2, - . ., h(r denote the cp (J») integers in the interval 0 -  p
which are either E 5 or = 7 (mod 8) and for which (  ) = -1.
Then. tIle necP8sary and sufficient condition fOl'1 the prinle p
(,,-hich is not a divisor of J)) to be a prime divisor of t.he polJ-
Hondal (1) is that either
p == ai (Il!O<l 8 P),
(i = 1  2. . . .  q,}),
(j = 1, 2, . . ., cp).
or
}) = b j (DIOd f< J)),
For. it. follo,vs fronl Theoreln 90 and fronl formulae (6) and (7)
In Section 42 that
() = (-l)(p-]\H{Y-l) (;) = (]j) = - () = + 1.
Exan1ph :J. If D = 6, we find that the vrime di,isors of the
polynoulial x 2 - tl are, apart fronl  and 3, the primes p satis-
f..ring anyone of the congruences
1)  1, 5, 1 9, 23 ( III 0 u. :? -l ) .
l'he results obtained may be expressed, less precisely 'I in the
follo,ving Inanner:
Lpt D ue a sfJllare-fre iutr-.'I("J1. ;.t!: 1. .A Ji10nl} the cp (-t.1 Dr) lnteger.,'
prirne tv -t.1]J I in fhe illft)}'t'al 0 - 411) I. fhere art' J-t = - (f (41 n I)
nundJ(Jr..... J'l, J'2  . . ., fa (citlz the ,follou.iny pro}Jel.('t/: J'f(,l..'1 pritne
d'il;ior uj' the jJulynonlial ,2 - D it.... congruent to a11l/ ()}Z(} q{ the
'Honuers rl, r2, . . ., r ll H10dulo .t I D 1'1 ur it is a divi.....ur oJ'  D.
44. Primes in special arithmetical progressions. - In Section 18
\ve luentioned the follo,\\Ting theorem of Dirichlet: If r anu n are
relativel.\ prime natural nunlbel"s, then there are an infinity of
pri DIes == r (nlod J1). Br applying the results of t.he preceding"
sect.ion we shall prove this theorenl ill some special cases. It

154
CHAPTER IV
follows fronl TheorelD 5H that t.he odd prime divisors of the
polynoDlial .2 + 1 ar the prinles of the form 4. n + 1. N O\v.
according to Theorenl ....5 every integ'ral polynomial \vhich is not
a constant has an infinity of prilne divisors. Thus, there are
infinitely many primes of the fornl 4. n + 1. [t follows froI))
Theorenl 88 that the prime divisors (different from 2 and 3) of
the polynoDlial :.t 2 + 3 are the prinles of tIle forul 612 + 1. Thus.
there are infinitely nlany primes of the form 6 n + 1.
:Wlore g'enerally ,ve have
ThlJureJn .92. Thei"C' aI"e i}ffinilely ,nany }Jl'inlfS of rQth of the ..l'ur11v'l
....12 + 1, 6n + 1, 8n-3, 811-1, 81l + 3, 12n-l, 12n + 5,
12 u - 5.
Pr 0 0'/: We consider the following six polynomials in x:
ji (x) =.:: 12 (:? + 1)2 + ..t.,
12 ([tl) = 8 p2 1. - 1.
J (,i"') == p2 (2.}" + 1)2 + 2.
.1:1 (.1.) = 12 p2 .!2 - 1,
.t (..) :.= p2 ({) t( + 1 f + -1-,
.16 (x) = 3 1)2 (2.1.' + 1 f + 4.,
\vherp P is an odd integer. 'Ve have, from the results In Sec-
tion 43:
1. The prinle divisors of the }Jolynonlial ii (:t') are the prilne of
either of the forms  n + 1 and  n - 3, with the eXCel)tion
of t.he primes di,iding" ].J.
:? The prime divisors of the polynoD1iai 12 (eA") are the primes of
eit.ber of the foruls  11 + 1 and 8 n - 1, with the exception
of the primes dividing P.
3. The prime divisors of the polJnolllia] 13 (.r) are the prin1es of
either of the f01"1118 8 n + 1 and 8 n + 3, ,vith tIle exception
of the primes dividing }).
-1-. The prime divisors of the poIJnomial 14 (.1') are the prinles of
either of the forlns 12 1l + 1 and 1  n - 1, with the exception
of tbe primes dividing I).

THEORY OF QUADRATIC RESIDUES
155
5. The prime divisors of the polynomial ,/5 (x) are the prin1es of
either of the forms 12 n + 1 and 12 Ii + 5, with the exception
of the primes dividing P.
6. The prime divisors of the polynoDlial f6 (:) are the primes of
either of the fornls 1212 + 1 and 1211 - 5, ","ith the exception
of the priuIes dividing p,
Let ji (,/,) be anyone of the six polynomials just defined. For
i = 1, 2, 3, let n2 =: 8; for 1,' =...., 5, 6, let 11l = 12. Then, the
prinIe divisors of .ft (:l) are the prilnes p (not dividing P) which
are either = 1 or === r (mod 11l), where r is a certaill nunIber
prime to )11 and not = 1 (mod 'ni).
Now assume that there are only a finite number of primes
== 1" (niod J1l). and denote by ].J the product of these primes. If
P has this value, the number ii (.r) cannot, for an integral value
of ../, be divisible by al1)"p prime = r (IlI0d 112). For fz. (x) is con-
gruent to one of the numbers - 1, 2 or -1 ]llodulo P. Therefore,
as a consequence of the properties of the priD1e divisors ofji (:r)
just nIentioned, we see that t.he number if (x) is the product of
primes =a 1 (mod 1n). But, this is impossible. since such a product
is itself == 1 (lnod 1n), It is, however, easy to \"erif)" that
..Ii (x) = r (DIOd 1n),
for all i. Hence, the hypothesis that the nun1ber of primes == ,.
(Ill ad }it) is finite is false, and Theorem 92 is proved.
\'T e finish by proving
Th('o,.r1J7 9H. There are iJttinitl'l.l1 'Inany pri1nes 0.( the .forJ}1 812 + 1.
}Jroof It follows froD1 Theoreln 8.... that t.he odd priD1e divisors
of the polynomial x 4 + 1 are the ]1rimes of the form 8 1J + 1.
...t\.ssulue that there are only a finite nUIlIber of primes == 1 (ulod 8)
and denote by P the product of these primes. TheIl, the nUDlber
( P.'J)' + 1 ,vonld not be divisible by a.ny prilu  1 (Jllod 8).
But, this contradicts the fact that every prilne factor of this
nun1ber must be === 1 (mod 8).

CHAPTER V
-\RITH)IETICL\L PROPERTIES OF THE
I{OOTS, OF UNIT\
45. The roots of unity. - According to the rules valid for
cOlnplex numbers "'e have
(COS qJ + i sin (p)7 == COS 12 Cf + i sin n (f
for all integers 12. (Moivre's formula.) Hence, we conclude that
the algebraic equation
.n - 1 -- 0
 ---
has the roots
(1)
 -1 J .) \
- ::r , I . . · - :It ,,, ( 0 1 ') 1 )
e lll = cos + l SIn -., 7n =  ,.oJ,.... Ii - .
U Ii
It is apparent from thei)" }Josition in the cOlllplex plane that the
numbers (1) are all distinct. For, if C is the circle with radius 1
and centre at the origin, the nUDlbers (1) form the T'ert.ices of a
regular polygon ,vith }J sideR inRcribed in (" so that one ertex
lies on the positive real axis.
The " numbers (1) are called tIle 'Jilt root8 o..f unit!l.
The llulnber
2 1t tn ..L . . 2 n 'n
cos - - I t SIn
il 1J
does not ellange if U is replaced by 111 + 11 f, ,vhere f is any'
inteo'er.
o
The nUlllber + 1 is always alllong the roots (1), the number
- 1. howevel" only if n is even.
The }>l"oduct. of two lith roots of unit.y is itself an 12th root of
unity.

THE ROOTS OF UNITY
157
If 8m dellotes one of the Ilumbers (1) which has the property
that all the Ilumbers
(2)
1:'0 £1 ,,2 ,.)1-1
c;. m ' 11l' f;, lI1' . · ., f;, '111.
are distinct, we sa)" that em i a jJ1"inzitit o e nth root of unity.
TheIl the Ilulnbers (2) represent all the uth roots of unity.
'Ve 110W prove the following theorem:
_-1 neee.'$r'aty and .uf1ieient condition ./01. em to lJe a pr';nitil"e nth
root of unit!1 is that the 'int(Jger 1n be jJ1.i'lne to u.
Prooj: Suppose that ,n aI1d n have the comnlon divisor d > 1.
Then not all the nUlnbers (2) can be distinct; for by (1) we have
n
e fl = 1 = eO °
Jlt m
On the other hand, suppose that (In, 12) = 1. TheIl the numbers
(2) are distinct; for if
e r - c"
1/1, - m'
we should have
2:r&1n(r-s) " 27&1n(1'-.,,)
cos -- + 'I SIn = 1.
II j}
But r - s is no Dlultiple of 12, since I r - N I < Il.
Thus the number of primitive nth roots of unity is equal to
the number of positive integers < Ii and prime to n, and conse-
quently equal to q; (12). The l1uDlber el is a primitive Jth root of
unity. When 11 is a prime, each nth root of unity is primitive,
except 80 = 1.
From the preceding result follows at once:
If e -is a pri11ziti'l:e nth ,'oot oj" unit!1 'lchich sati6'fic the al.fJclJrai('
eq 11 (I t i 011
Z...V - 1 = 0,
thi' positiloe inte.qer N 'nuu,.t be a flullfiple of 1l.
For 13 -= 2 the roots of unity are + 1 and - 1, of which the
latter is primitive. For -n = 3 there are two primitive roots,
namely

158
CHAPTER V
E1 =  ( - 1 + i J 3), and E2 -;: Ei =  (- 1 - i 1 a),
,vhich are the roots of the equation e 2 + 8 + 1 = O. For n == -!
there are two primitive l"OOtS, namely + i.
46. The cyclotomic polynomial. - The polynomial of degree (f.(n)
{I)
Fn (x) = II (x - ea)
a
the pl.oduct extending over all primitive 11th roots of unit.., is
called the c.t/cloto'lJzio polYJl()'l)zial of inde3.' rl.
Let 1J1, P2, . . ., }J r denote tIle distinct prime factors of 1/;
further, put
I10 = XU - 1
and for 1 < "  r
Il
l2)
II.. = II (.//, PI.' · · PI.. - 1),
the pl"oduct extellding over all the " indices i k which satisf)"
tIle conditions
1 < £1 < i 2 <, < i" < F.
Then
've haye the identit)..
 ILII 2 . .
]1n (..r) = - ----- ,
III I1: i ..
where all the IT.. with an ven index,.. occur in the numerator
and all ,vith an odd index in the denominator. In fact, let
(3)
'> b -) b
_7& .._x
Eb = COS - + -Z SIn-
11 Jl
be all a.rbitrary nth root of unity, and suppose (b, n) = d. .A neces-
sary and sufficient condition fOl" the polynoDlial
n
Pi 1 1J i.. . . . Pi,. 1
:to - '-
to be divisible by x - eb is, in consequence of the results ill
Sectioll 45. that the degree of the polynomial be divisible bJ

THE ROOTS OF Ul'iITY
159
n
d' this iDlplies that d must be divisible bJ everyone of the
primes Pi l , l)i. . . .. pi".
Denote b)T p, the number of distinct prime factors of d. If
p, > 0, clearlJ the product II" is divisible by
(:r - 8b)(:')
and by 110 higher power of J' - 8/,; for () is the number of
cOlllbinatiolls of It elements taken ,.. at a tinle. 110 is diyisible
bJ :i; - Bb, and not by the square of this linear function. Conse-
quently, the numerator in (;3) is diyisible by a })ower of x - Bb,
\v hose ex pOllellt is equal to
1 + () + () +
and the denolllillator ill (3) is divisible by a power of x - Bb,
wllose exponent is equal to
e) + (:) + .
 O\V we have
1 - (i) + () - () +
= (1 - 1 )l -= 0 .
Hence. if d > 1 and p, > O the right-hand side of (3) is not
divisible by x - Eb. On the other hand" if d = 1 and p, = 0, the
rig"ht.halld side of (3) is divisible by x - Bb alld by no hig'her
po\ver of this linear fUllction. Since, in this case, Eb is a primi-
tive nth root of unity, "'t' have establislled the identity (3).
From (3) it follows that p (,) is an illtegral polynomial in x.
For bot,h the nUlllerator IIo II 2 . . and the denominator IT I TIs . , .
are integral polynonlials, in which the highest power of .1.' has
the coefficient 1. CarrJing out the division in the usual manner.
've obtain a quotient which is an integral polynomial in x.
Fronl (3) we easil)? deduce the identity

160
CHAPTER Y
(4)
1 4' ( ' j 'p )
Pnp (X) = ;n (:}.) ,
provided that p is a prill1e which does not divide n.
On the other hand.. if p divides 1"2, we clearly have
(5)
F71p (x) = ]f"'1 (x p ).
A pplying the formulae (3), (-1.).. (5) we calculate the follo,vi!lg'
special cyclotonlic polynolnials
Fl (,;,.) -= x-I, }"2 (x) =-= ,i" + 1., 1'3 (.1') -= ,2 - X 4- 1, I" (J_') = x 2 + 1,
F5lr) = x 4 + .c 3 + .v 2 + :1' + 1. 1f6 (x) = j.:2 - t + J, 1'8 (.r) == ,4 + I,
]1'9 (,J') ==.1'6 t .l,3 + 1, j"io (.v) ..t,4 -.t 3 - .i,2 -.r - J., ]'i2 {x)=x 4 -,r 2 + 1,
1 1 '15 (.1')  .1. 8 - :1.,7 + .l'S -- .t' + .1.. 3 -:.c + 1, 1 f '20 (X)=,{,8 _,}.6 + );4 -.1.: 2 ..; 1,
F 21 (x) == .2,.12 - XlI + .1,9 - xP- \" l.6 -- ;};4 + t.3 - X + 1.
If p is a prime, ,ve have
(6)
,),1) - 1
Fp (x) -= - ·
.!' -- 1
For 11 > 1 the constant term in ]/n (.r) is equal to 1. To pl'ove
it we have only to put .J = 0 in (3).
By putting x = 1 in (3), we get the following result for n > 1 :
(i) F (1) = f p, when n is a power of the prime p,
n ll, when }2 has at least two distinct prime factors.
47. Irreducibility of the cyclotomic polynomial. - A polynomial
f(at.) in x with rational coefficients is said to be rrdftcible when
there exist two polynomials in x, not constants. with rational
coefficients, such that
j'(x) = ,l (x) h (x).
Othel"wise the polynomial f(x) is said to be irreducible.
\Ve prove the following lemma:
.Lenzma 1. Let f(x} and 9 (x) be tll'O PolY1lo1nials u:ith l'ational
coe.((icient8. If 9 (.r) is irl'eduribl(l, and if f(x) and 9 (x) lzat'e
a l-Ol1znzon zero, then f(x) i.'/ dit.1.'.'ible by 9 (,.t).

THE ROOTS OF UNITY
161
Proof Let a be the common zero. The greatest common divi-
sor d (x) of f(J:) and 9 (x) cannot be a constant, since it has the
factor x-a. Since !I (x) is irreducible, it has no other divisors
than constants and divisors of the fornl a!1 (l'), ""here (( is a
rational nunlber  O. Hence d (x) = a 9 (x,) and therefore }'(;1,.) is
divisible by lJ (x).
A consequence of this result is that an irreducible polJllonlial
can never have any zero in commOll \vith a polynolllini of lower
degree; here the coefficients are supposed to be rational.
'Ve next prove
Lenzma 2. If the integral polynonzial
f(x) = xt + £:1 X!L-l + . + Cf.
is dit'il:J'ihle by the poly'no1nial ll'itlt l'ational coe[th.i('nt.",
g (:1..) = x 7n + hI x m - 1 + ... + b m ,
the.e coefficients are necessal"ily integer:s,
PrDDf. 'Ve may suppose that
f(x) = 9 (x) h (.:c),
where the polynomial h (x) has rational coefficients. Let lJI be
the least. natural number such that lJI g (.1') is an integral poly-
nomial, and let X be the least natural number such that. ....\' Ii (J')
is a.n integral polynomial. The polynomials .]1 9 (.l') and .,- /z. (x)
are then primitive polynomials. Hence, according to Theorenl -1-1,
the product .111:Y U (.1:) h (x) is also a primitive POlY110111ial. But.
since lJI J.,r 9 (x) h () = M }l f(x), ,ve 1l1USt have 111 = ]{ = 1. Thus
Lemma 2 is proved,
Le11una 3. Let
g (x) = X 'fl + al Xm - 1 + ." + am
be an integtal polynol1zialll"itll the zeJ'o. Xl. .2 \ , , n X m , and let
G (1?) = x m + 11 x m -1 + .,. + ...1 '11
be the polyno1n-ial ,('hosf zeros arc- the n'lun"blJ}'s
11- 516670 Tl'!I!1JJt }."ragell

162
CHAPTER Y
- ,J} l 'l) J ,p
,/ l' , 2. . . ., l1& '
u:here p i,' (I, ]>rinu'. J'/u<u the C(Jr:{li('jf12t...' ..4 1 , .....J2 ...) ..,,1m are
i}2 te.fler,', a l1d all the d ':tli<}"f}tl"eS
404} - (I], ...-:1 2 _. ({2. . . .  ...t m - alii
are diri,,'ilJlc 1J!1 ]).
])rooj: By the nHlin theOl'eUl on s)"lnllletric functions we know
that everJ 8)"111 lllet.ric integ-1'al poly llomial in l"l  .1'2, . . ., J'm is an
in teg-er.
\\T e now a.pply the polyuoll1ial theorenl for calculating the ex-
.
preSSIon
( I' - 1 J' a ) p - (  ')", .[ " ')'1. ) 1)
'- l' -   11 J::..' fA I., ·
,vhere the SUIll extends over all indices oj satisfying the following'
conditions: 1 < i 1 < i 2 < " < i J . < lJl. Obviously ever" PolYllonlial
coeffici en t
p!
,
k 1 ! '2! ... k 1 !
where 1..- 1 + k 2 + ... + l-) = p.. is divisible by J), if it is > 1. Hence
we obtaiIl an equation of the forln
( - 1 I' 1I J ) l' = (- 1)' _11 .,- JI AS (,1'1' J' 2' . . . , a'",),
where S ('('I, .1'2, . . .  .I"IIl) is a sY1111uetric integral polJ T nonlial of
the numbers Xl, ..1'2' . . ., :rill and eonsequentlJ an integer_ Since,
by Theoreln 35 a; == a , . (ll1od 1)), it follows that aI' == 064 " (nIod p)
fOl' all ". Q. E. D.
We shall proye the following theorem:
Thl' cyclofo1nir )Julyuu11lial i. iri'edu('ible.
\Ve prove it indirectly. and suppose that F7& (x) is l'educible having
the decomposition
(1)
}T71 (x) = /1 (,i).f2 (,1") -. .t:. (x)
where J (x), .1'2 (,r). . . ., J. (.r) are irreducible distinct PolYllolnlals
\vith rational coefficients. in which the highest power of x has
the coefficient 1. By LeuIIIla 2 the pol.vnomials are integral.

THE ROOTS OF UNITY
163
\\T e first show that these polynomials are all of the same degree.
Let 'YJ be tt root of the equation j (,.) = O. In consequencp of the
properties of IJrimitive roots of unit). (Section 45) ther exists a
natural number  such that 'YJk is a root of the equation 12 (,}.) = O.
rrhe number k is clearly prime to '1l. 'Ve now forln the p01Jl1o-
luial .lI (x). ,vhose zeros a.re the kth powers of the zeros of 11 (.,e).
Since the equation 9 (.r) = 0 has a root in common with the equa-
tion /2 (.,f) = 0, and since the }Jol.rnoDlial 12 (..t') is irreducible. it
follows frolll Lelnnla 1 that everJ root of 12 (:.I") = 0 is a root of
!I {,I') = O. Therefore the degree t21 of }i (l') cannot be less than
the degree n2 of 12 (x). In just the same way ,ve prove that
Jl2 > 111' Hence, ,ve have 111 = 122' and all the irreducible factors
jo (.t o ) on the rig'ht-hand side of the identity (1) have the same
degTee. The roots of 12 (x) = 0 are the kth powers of the roots
of .li (x) = O.
Let JI be a number greater than n and greater than t.he ab-
solute value of allY of the coefficients ill the polynoDlial differences
ii (.r) - jj (x)
for all i and j. -i F j. If T denotes the product of all primes
< J.1I, except the priIues which divide k, we put
Q = l' n + l'.
Then the Qth po,ve1"S of the roots of ji (.x.) = 0 are obviously the
roots of J2 C.t") = O. Let us put
l.J = ql q2 ' . . gt:,
"There the numbers qi are primes which, by our hypothesis on Q,
Illust he > ]JI. Let hi (..r) be a polynoDlial ,,,,hose zeros are the
ylth po\vers of the zeros of Ji (x). Starting from hi (.1..) we forDl
11 ne". polynomial 11 2 (J') whose zeros are the Q 2 t.ll po,vers of the
zeros of hi (x). Cont.inuing in this way, we obtain a setluel1ce of
pol ynolD ials
()
.Ii (('), hI (.). 11 2 (.1'), h3 (J:'), . - .  lt (.),
in which the highest power of x is supposed to have the coef-
ficient 1. All polynomials are of the sallIe degree as ii (x), anu

164
CHAPTER Y
we have h$ (.it) = j ("r). The)" are all irreducible, since 12 (:) is
il'reduciLle. lIel1ce e,er)" pol)"nomial hi (.1") coincides with SOlue
of the polynolnials jj (x). In the sequence () not all the polyuo-
Ininls can coincide with ji (x), since h l1 (.r) = 12 (x)  11 (x). Let
h r (.r) = .f; (.1:) be th first pOljIIOl11ial in the sequence which is
different fro111 .Ii (:1"). Then the zerOR of Jj (x) are the qrth powers
of the zeros of .1; (x). Hence ]))" Leul1na 3, all the coefficients
in the polynoluial difference
Ji (l') - jj (.c)
11lUst be divisible by the prinle q,.. But, since qr > M this is
contrary to our hypothesis on the nUlllbel'IJI. Consequently, the
polYl1onlial cannot be l"educible. Q. E. D.
48. The prnne divisors of the cyclotomic polynomial. -- The
cyclotomic polynoluials ha\Te the property in common ,vitll the
polynomials of the second degree that their prime divisors are
characterized by certain congruence conditions. We shall first
establish the follo\viug' 11Iain result:
Tlu:orenl 94. 1j" q i"t (I ]Jl'hnc It"hirh dues 120t diz.ide 12, lce hal:e:
1. The neces811J'jJ and 81{t}icient eonditiou jor the c()}'.gruenc()
(1)
iT}, (:).:) = 0 (ulod q)
to be soh'able is that q == 1 (nlod 1).
"t. 1.( q == 1 (U10tl nL the solutioJ/s 0./' c012grueuce (1) are the nll1n-
Ul:J'8 l,.hic-!l l)elul2.f1 to the e."!'po12ent Ii '1J2odulo q. Thus the '12Unl-
/Ji'J or illcongru('nf sulu.ti01lS 11tudulo fJ il q; (n).
/1. (x i,\: a ,'?olutio}2 oj CuUllr'llfllCe (1), the nU'lnher Fu (or) i.' di,.i8-
ible bll ci'({cfl!1 thi> sa1ne pUIf'rr oJ" q as x n - J .
[>I'O({t: Since }'u (0) = 1. a solution ,J' of cong'ruellce (1) canDot
Le === 0 (UIO<.l q). If Fn (x) i8 divisible L.r fj, at least one of the
factors in the llUluerator 011 the rig'ht-hand side of relatioll (3) ill
Section 46 is divisible by fJ. Hence the Ilumber XII - 1 is divis-
ible Ly q. If we suppose that the solution x belongs to the
exponent Ii Inodulo fJ, the nUInber It must be a divisor of Il.

THE ROOTS OF UNITY
165
n
Further suppose > ], and denot.e by })1, 1)2' . . ., 1)lIl the dis
It
tinct l)rilne factors of 12. If q divides the nU111ber
It
(2)
n
rd - 1,
where d is a product of different. IJrime factors of 12, t1Ie number
11
d must be a multiple of It and thus a nlultiplc of d. Hence_
every prilne divisor of rl ll1Ust belong to the set of primes 111 
1'2, - . ., pm.
Now suppose that the numher ).'" - 1 is tlivisil»le by fJ and by
110 higher power of q. Thus
.£!L = 1 -t. t fj\
where t is not divisible by q. Raisillg each side of this equation
to the kth power, we have
xl: t = 1 + /.. t q S + q2, i 1 = 1 + i 2 qt: ,
where i 1 and t 2 are i11tgcrs. If J.- is not divisihle hy fJ, ncither
is the number i 2 divisible by q. '!'hus the lluu1ber :klt - 1 is
divisible exactl.y by the saDIe IJower of q a.s the nunl ber xl" - 1.
In the number (2) we have ;j = k/-l. where 7.- is an integer not
divisible by q, since q does not di,ide 1/.
Finally we can conclude: The product 110 II 2 ll4 . the nu
merator on the righthand side of relat.ion (3) ill Section 46}
is divisible by a po\ver of q whose exponent is equal to
. (1 + (;1) 1- (I) + ) .
and the product III ll3 (the deno1ninator 011 the rig'ht.haud
side of relatioll (3) in Section 46) iR divisible by a power of fJ
whose expollent is equal to
. (( 'n ) . ( 'lZ )
.l;j 1 T a.
-:- .. ).

166
CHAPTER V
But we have
1 - ('I) + (;,) - (I) + ... = (1 - 1)m = o.
Hence, if n > 1, the nUl1Iber Fn (x) is not di,iRible by '1. If
It
P == 1/, the number F n (.1') is divisible b)" the sanIe power of q as
x n - 1. Thus Theorem 9-t. is proved. It also holds for 12 = 1.
\Ve further establish the supplenlentary result:
l'h('orenl !Ji;. S'uppose that q is a lJriJnr. facto]'" o.f 11., and 1111f
Il = fJft lit, ,,"here 111 itl,( not diri.,.ible by fJ. Then U'C harf:
1. Tlu: necf8sary and Iultirient coudition for the c012grllP}U 1 f
(iJ)
FlI (.r) = 0 (mod (1)
to he .w)lcable i8 that q = t (mod 1l1)'
. 1j" q == 1 (mod 121)' the 801ufi01l8 of c(jllgruencp (3) are the nU'11l-
bf1".1,( ifhieh belong tu the f>xpouenf J21 ,uodulo f[. 1'hlf.\' the 12ll1n-
ber of incongruent ,.....Ollltio}u l1l0dulo q i, p (121)'
s. Ij'.r il n .'toZufion of (aongrlleuce (3). tlu' ull'JlIbe1' F 1Z (.J") j,.... diz'i.'-'-
ible by q and not by q2, jn'ul:ided that n > 2.
l>roof. Suppose first that n is not a power of 2. Then for
q = 2 we must have 111 > 2. Appl.ing fornlulae (4) aud (5) in
Section 46, we get
(4)
F ( ) = FUI (x"U) .
n X a ]
F fll (x'l - )
If Fn (x) is divisible b. q, the nUDlber
(f))
F"l (.r q ''')
i clearl.v divisible by q. Then by Theorenl 9.t we see that the
nUDlber x fJCt 111Ust belong to the exponent 111 DIOdulo q. Hence :)"
also belongs to the exponent 121 modulo q. For q = 2 this im-
plies 17 1 = 1. Since by hypothesis 11 is not a power of 2, q must
be odd.
If "J." = 1, we have 'Ill = 1 and b. (7) in Section 46. I."" (1) = fj.
If ('=-1, we have Jll=2 and l?n(-l)=F,,(l)=(j.

THE ROOTS OF UNITY
167
Suppose next that ;!' =;ef + t. If J' belongs to the eX110nent 111,
the number (5) is di isible hy q and, according to Theorem 9.t
(last part), by the sallIe po,ver of fj as the n un1 bel"
( x Q C!)"1 J - 1 = x n - 1,
which is =?: O. In that case the nuuIber
(6)
F ( .r{C- 1 )
III .J.:.
is also divisible by q and by the sallIe po,ver of q as the number
n
(:rfl a - 1 )'11 __ 1 = :r-;j - 1,
which is rf O. Suppose that the last nUlnber is di\yisible by q'
and not by fJ+ 1, or
n
x fJ = 1 + f q'\
,v here t is not divisible by fJ. Raising each side of this equa-
tion to the qth po\er, 'e hav'e
;l,n = 1 7 q t'1' + () ,2 q2" + ... = 1 + t 1 q"+1,
where t 1 is not divisible by q since q >. Hence ,,,,e conclude:
If the number (f.)) is divisible exactly b.y '1 ' \ the number (5) is
divisible exactly bJ' r'«+ 1. Fro In (4) it then follows that the nunl-
ber }'n (x) is divisible bJ q and Hot by q2.
Finally, ,vhen u (> 2.) is a power of 2, '''e lut\"e Fn (:r) = x 271l + 1,
where 111 = :. Hence P" (.r) is ne,er divisible bJ .t,
This proves Theorenl 95 for nl1 values of II.
In particular. ,ve see that cougoruellce (3) is solvable only if q
is the goreatest prime factor of n. For, according to the first
part of Theoren1 Do, we nlust ha 'fe fJ > }il.
Exan1}Jle 1. For}l = 20 we have
F ( ,,,. ) = ",8 - .1 ,6 -:- }'4 - _2 +- 1
20 ...... ...... ....'\. 0 .
The congruence
F 20 t. r ) == 0 (ll1od 41)

168
CHAPTER V
has the eight roots
x = + 2 + o. + 8, + 20 (mod 41),
since 41 == 1 (1Dod 20).
Since [) = 1 (DI0d ..), the congruence
F 20 (,'t.) == 0 (mod 5)
has the solutions
.C = + 2 (mod 5).
EXa1n}Jle 2. It follows fronl TheOl"enl 95 that none of the
congruences
or
F 15 (.r) = 0 (mod 5)
F 15 (.r) = 0 (mod 3)
is solvable.
Fronl Theorem 94 we obtain
Theore,n .rJa. 1".1' n ;8 a 11atural 11ll1nber, thc1°e arc i1tfinitely flla11'Y
prhJleS lfhith. are == 1 (mod 12).
PrDDf Suppose that there are only a finite number of primes
== 1 (mod n) and deuote by P the product of these prilnes. Let
us put x = 12 P y into Fn (,1'), JJ being an integer. Then, by Thea.
renI 94, every IH"ime factor of F n (12 P.l/) must be = 1 (mod 11).
But this is impossible, since
Fn (n P!I) == Fn (0) == 1 (mod 11 P).
Thus Theorenl H6 is established, by indirect proof.
49. A theorem of Bauer on the prime divisors of certain poly-
nomials.
Throrc'I1l .f)i'. Let 11l be a natural nZl1llbeJ" > 3, and let
(1)
n
I(x) =  a1;x n - k
k=-O
be all £1lteg1Yll jJolynonzial ,,.h'ic]z lza8 at lea.t 012e ,.pal zer6.
TheIl ,,((-1") ha, htinitelll 111al1.lJ p)Oi)ne dirisors u:h1..ch a1.e not
== 1 (nlod 111).

THE ROOTS OF UNITY
169
Proof. vVe can suppose that f (:I') has no Dlultiple zero. Further
,ve suppose that 00 is positive. There al"e real values of x such
that f(,x) is negatiye. In fact, let e be a real root of the equa-
tion j(x) = O. If j(,r) > 0 for all real ..r, the function j(x) has a
minimum ,alue for :J." = e. Thus ,ve ba,e .l' (e) = O. But this is
impossible since ..1'(,r). by hypothesis, llas 110 Inultiple zero. There-
fore there exists a. fraction ;, ,vhere t and T are integers (T posi-
tive) such that f( ;.) < O. Then the polynomial
) )J
. , ,  ./' " ,
9 t;r.) = 1 n .I ( T = 2.: ak 1 k xn-k
1;-=0
is negative for x = t.
Let us put
=-11
1 ( " ) = _ .fI ( - (/ I ,t: + t)
1 x. g(f)
x r/ (t) _ 'It/) 1'2 . (/' (t) +. 1- (- 1 )"-1 (g ,t))"-l 00 X".
II ' . 21
In this integral POlYllOlllial the coefficient of J....n is positive. There-
fore, 11 (.1:) is positive for all values of ,x exceeding a certain
value ,ro. Since the cOlIsta n t terlll i:; - 1, the polynomial h (x)
has at least one pl"iUle divisor which is not = 1 (mod )12). Sup-
pose now that II (..r) has ouly a finite l1unlber of Pl"ime divieol"s
which are not = 1 (nlod 111), and denote by P the product of
these primes. Then, for all 'H;].J :t' > ;l'o the polynomial 7z (nl Px)
has only prime divisors = 1 (mod 111). Thus, ,ve must have
lz (In P .:t) = 1 (1110t1 '112). This is iUlpossible, howev-er, since from
the expression for h (,X') \"e see t.hat h (11/ P x) == - 1 (mod l1l),
Hence, it follo,vs that h (x) has infinitely many prilne divisors
,vhich are not == 1 (ulod 111). But the polynomials f(.r), g (x) and
h (x) obviously have the same prin1e divisors, possibly with the
exception of the finite set of prilues which divide the numbers
g (t) and T. This proves rrheOl"el11 )7.
By means of ideal theor.r it is possible to establish the fol-
lo,ving result: If 111 is un integer > 3, every integral polynomial
(which is not a constant) hLs infinitely Dlany prinle divisors
which are = 1 (lnod 1n).

170
CIIAPTER V
50. On the primes of the form ny - 1. - For any natura.l
number n we define t.he t\VO polYllolnials ["'11 (.e) and JTll (ar) by
the equation
(x + i)ll = l"', (x) + i T" n (at"),
where 'i is the inlaginar)" unit. Thus we have
(1)
l - ( ... ) _ (,r; + i)Jl + (.r - i)n
n ,1 - 2 '
(2)
_ (.r 1- i)n - (.r - i)n
1 7& (".) = ) . ·
:.. I
If v is a positive divisor of n, the pol:vnolllial '''''1 (.r) is divisible
by the pol.rnomial 1"', (x). For, putting n = It v, Te ha v'e
U ll (::) -t- i J"J& (.r) = (.1.' + i J')" = ((J" ex) -1- i J"', ...c:)'\
or
(3) V n (x) = r.o (.d [(,) (U, (X))...-1 - (;) (U..(X))."-3 (V,(X))2 + ].
Lel1znla 1. A In.hne q lrhich i,\: == - 1 (mod -1) cannot, .for the .al1le
'integral f."aZul! uj'" .'t, di,.'ide both l-m (at..) ((nd JTm (x).
For if q divides both nUlllbers, it also divides the number
( U'Tn (X))2 + (V,n (X)}2 = (.'c 2 + 1)111.
But the number L2 + 1 is not c1iyisible by an)" prinle == - 1
mod 4).
Lenlna 2. Let tJ be a pri1ne = - 1 (Illod J) llhiloh il:1 not a di,.i,....or
of no I.l v is an.1J positiloe divisor 0.( 11, the Jlltuzbers
( ..I. )
J - ( ) 1.,1 (x)
"x and Y" (x )
Canl1()t be diz"isibZe bll q for the .al}U: intp!lraZ raZue oj' .r.
In fa.ct, put n = Ii J', and suppose that q is a di¥isor of both
numbers (4). It. follows from (3) that
IT (.1.")
,  ( ) = ,u (Ul' (',)lt-l (Dlod 1TJ' ,r).
l' ,1'.

THE ROOTS OF UNITY
171
Thus the right.hand side of this congruence must be divisible
by q. But by Lemll1a 1, q does 110t di vide 1.1 ¥ (x), and since q is
not a divisor of n, p. cannot be divisible by q.
Lennna.:1. If q i.,. a ]Jri1lle == - 1 (mod '+)t the nunzher 1Tq+l (x) i.,;
dil'i,,'ib1e by q .for all £nte,qral t.all,e8 (( x.
In fact we have
r,/+l (x) = (I]  1) :1."/ _ (If ; I) X,/-2 + ... _ (q : I):r,
where all the bino111ial coefficiel1ts, except the first and the last,
are divisible by q. Hence by Fermat.s theorem
Jq+l (..t') == (q + 1).1;q - (q + J) x == x? - X = 0 (mod q).
Len1Jna 4. Let fJ. be a. priolc = - 1 (mod 4) lfitlz the fo1lou'ing
]JrOper(ll: ]f'or a certain integral ,.alue 0.( x the prinzf! q di rides
1n (x) but none oj' the nll1nbers T". (x), It'ltere " is any po,itil:e
cih'isor of' n 1"88 than 1/. Then lrf: Iza.e
..
tJ = - 1 (mod u).
Proof Put (n, q + 1) = v. There exist t,,"o positive integers 11
ana 1" such that
II (q + 1) - r 12 = ".
This readil.y follo\"\"'8 froln Theorem 17 (Section 10). Now we haye
[l.:,.(,/') .r- i ''', (.1')] [l.Tt1n (..t') -1 i T"rn (.r)] = (..t + i),,+,.n
= (.)' + i) Il {f/ + 1 i = (or,,! ,,':"1, (X) + i J ru . q + 1 i (..-r),
or
(5)
(-J' {,,') 1 r l 'Il (x) + J"p",. (x) [rn (.r) = J"P"UI?fl\ (x).
B.y Lenlma. 3 the 11umber Jrq+l (x) is divisible bJ' q. Therefore,
since 1"' f /+ 1 (.r) divides v't(q+l) (x), the latter Dum ber is also divis-
ible by q.
Since b.v our hypothesis J- n (at) is divisible by q, 1"P"l'71 (x) is
also divisible by q. Then it. follows from (5) that the nUDlber
J""1 (.1") UJ'n (.l.) is di'f'isible by q. Since t by Lemma 1. l."" (x) cannot

l ')
l
CHAPTER V
be divisible by q, we see that q is a divisor of }rt. (x). But, hy
our hypothesis, this is possible only ,vhen l' = 11. Hence we have
q == - 1 (nl0d:I1).
For }}l > 1 the equation
J ... ( . ) = (:c + i)m - (x - i)m = 0
m .r: . ) .
l
has the roots
. (J -{- 1
.r = I ,
(J-I
where e is an nzth root of unity  1. 1'hese roots are real, since
they are also givell by
l":t
,x=cot-, (..= 1. 2,..", )11- 1).
111
For IH > 2 let us put
l/J 71 , (x) = (x - i)'I' (m) Flit ( X +  ) "
"T -2
This pol?nomial is il1tegral since, for 111 > 1.
z'l ;71.) Em (-) = Pm (...).
It is of degree q (HZ), a11d its zeros are obviouslJ' the numbers
an
cot -,
In
where II runs through a reduced residue systenl nl0dulo 7n.
If 12 alltl " are nattlral nU1nbers and " a positive divisor of In
\vhich is < 12, the polynomials JT., (x) and (/)/l (x) have no common
zeros. Both polynonlials are divisors of the polynomial Jr" (x),
and therefore we have the equation
.n.. ( ) .,. ( ) J7'71 (x)
""''' x · JJ y. = c. V (. .)'
,. .l.
(6)
where c is an integer rf 0, al1d whel'e rJ'" (lo') is an integral poly-
nODlial in x. Since (/)n (.1') lIas real zeros. it follows from Theo-
rem 97 for 111 = .1, that this polyuon1ial has infinitely many prime
divisors q == - 1 (nlod .1). Let q be one of these prinle divisors
of (/In (.v) we further suppose that q is a divisor neither of n
nor of the constant c\ Let .)'1 he all iuteger such that (/jn (;2") is

THE ROOTS OF t:'NITY
173
( ) J Tn (x). I d ... bl
divisible by q. Then, by 6, the number 1"". (x) IS a so !VISI e
by q. Therefore, by LeUl111a 2, the nllmber r"" (.1:) cannot be divis-
ible by q, and finallJ, by Lenlma 4, we nlust have q = - 1
(mod 'J). Thus we can state
Theorcnz .')8. If Jl is a Ilatural Jlzunber, there are i 1finitel!l 11la12Y
prz"111es 'll"hz"ch are = - 1 t1110d 1l).
(1)
51. Some trigonometrical pl'oducts. - If n is a l1aturnl number
> 1, it is easily shown that
1l-I l
rr 2 sin t" = II.
k== 1 1
For, if we put x = 1, it follows from the ideutity
x 1t - 1 t rr l-l ( 2 k:r .. "2 "A: J"t )
= ..t - COS - I SIn
x-I 1:-=1 Jl Il
(2)
that
71-1 ( 1 . ( J. 1. .
p , . A= :it e. n .. A" :r
n = n - 2 I SIn - ) cos - + I 8111 - ) .
J:= 1 Ii U II
Thus
7&-1 k
n = ( - i)n-l (cos l.:r; J2 - 1, + i sin   12 - 1 ) II :2 sin · J"t
k= 1 'i
71 -1,.
N1'&
= II 2 sin - .
k=-l n
Since sin (12 - k):t = :'Iin k n , it follows from (1) for any odd 12 that
12 tl
(3)
 ill-1) I.
II It ,-
2 sin .-:.. = 1'1L
1:=1 n
By putting .:C = - 1 illto (2) we obtain analogously the fornlula
(4)
 (n-1) I.
'7&
II  cos L = 1.
1:1 n

174
CHAPTER V
'Ve shall further prove the followillg formula
(5)
i in-I) (4 k - 9 ) l ll ] -
II  sin ' -!l = ( - 1) 4 · J In.
1;=1 Ii
Since the nunlbers 2, 6, 10, . . ., 4 k - 2, 0 0 0' 2 n - 4 are in-
congruent modulo n, the absolute value of the product ill (5) is
equal to the value of the product in (3), io eo.. equal to )In.
F th · · ( 4 . -- 2) n · · f 11 7. n 1 d
ur eI", sInce sIn 12 IS neg'atlve or a /L. > -i. +  an
< . 12, the number of neg-ative factors 011 the left.hancl side of
(5) is
[ Jl 1 ]
r = J ( n - 1 ) - - -t. ·
2 4 2
By considering the number 12 Illodulo 4 it is easilJ seell that
,. = [l Thus formula (5) is valid for all odd n.
B:r analogous reasoning we also deduce the forl11ula
(6)
i (n -1) . (8 l. - 4) n 1 I -
II 2 sIn = (- 1):: (n - ]) · 1 n.
1.: = 1 11
For sin (8 k - 4) n is negative when k lies in one of the inten"als
Il
 (12 + 4)-1(n + 2) and . (3 Ji + 4}-ino Thus the number of
negative factors on the left-11alld side of (6) is
1'= [ n: 2 ] _ [ Ii  .t ] + HII-I)- [ 3118+ .t ]-
It is easily shown that r is even when II == 1 (mod 4), and t11at
,. is odd when n = 3 (mod 4).
52. A polynomial identity of Gauss. - Let us put
( 1)
(1 - xl.) (1 - .£"-J) .. (1 - x't-k+l)
F (;e, h, k) = (1 _ :1') (1 _ xl!) . . . (1 - .rL") ,

THE ROOTS OF rNITY
175
where It and ,; are natural numbers and k < h. In particular
,ve have
1 - :1;"
F{x, Ii. J) = 1 _ ..c and F(x, h, h) = 1.
If k >  Ii and < h, it follo\vs fron1 the identit)"
F (.1\ It, Ii - k)
{l - xl'} ... (t - x1: 1) (I - xl') (1 - Xl:-l) ... (1 - J:.h-k+1 L_
= (1 - .1) , ., (1 - Xhk) · (1 - xh- I :+ 1 ) (1 - x't-l:+2) -.. (I - xl.:)
that 've haye
(:?)
F (.), 11, h - !to) = F (..1", 11, k).
If we define
F (.(', h. 0) = F (..r, It, Ii) = 1,
the relatioll (:?) is also valid for h = It..
"1 e ha \"e
. ]-x h
F lx, lz, k + 1) = ] _ :T 't - k - 1 · F (..r, II - 1, l. + 1)
1 - J.1' -1.:-] - ;.t" + :1'" - 1;- 1
= 1 __ /,-l., --1 · F (oX, It - 1, /... + 1)
=F(.J..:It-l,k+ 1)
1 - xl' + 1 (1 - ..c" -1) .. (1 - {!:1l-1.:-1)
-+- X"-1:-1 . - --- .
. J - X. l1 - k - 1 (1 - x) ... (I - xl.: t 1 )
Hence
(3) F (.r, It, I.: -t- 1) = F (.1.0, h - 1, k + 1) + :Lh-k-l F (x, 1z - 1, le).
This recursive fornlula sho,vs that F (.i:, It, k) is an integral poly-
nOluial in x of degree (h - k) k. From fOrJlIUla (I) \ve see t.hat
the highest po,ver of x hns the coefficient 1.
Let us define a uew pol)"nolllini .({x, It) by the equation
(-l)
"
.f{."t, 11) =  (- 1)1: F (.i., h, lo).
1:=0

176
CHAPTER ,:-
Then we obtain by (3)
{(x, h) = 1 + (- 1)h
11-1
+ !, (- 1)1: [F (-c, h - 1, .) ;- J;/;-I: F (.1.', lz - I, fL' - 1)]
k=l
and
1,-1
j"(:7', h) =  (- 1 )':-1 (1 - .),1,-1:) F (Ir, It - 1, I,: - 1).
1:=1
By (1) we have
1 - X'I-I:
1 _ .l"lt -1 · F (.r, 7z - 1  k - 1) = F (x, h - :?, l - 1),
thus
f(.t, h) = ',1 ( _ l) k-l F ( .. I _ '> 1. - 1)
1 _ Ih-l  .1... fl ...., ,... .
;1. 1;=1
Finally, by (4) ,ve obtain the recul'sive f01'1mula
(5)
j" (:c, It) = (1 - X'I-I) · f (x, Ii - 2).
Since f (x, 1) = 0, it follows for eyer.\" odd Ii that
j"(X, h) = O.
On the other hand, if h is even, we have
f (x, h) = (1 - .1:" - 1) (1 - l:" - 3 ) .. (1 - t:3 J · f' (:J:, 2).
Now
}'(.I.', 2) = 1 - (1 + al") + 1 = 1 - r.
Hence, for all even h, we have established the polynomial identity
(6)
.I" (..r, Ii) :: (1 - ..t") (1 - J...3) . . . (1 - I" - J ),
or
1 _ 1 - :l + (I - al') (1 - X'i-I) _ (1 - 1I').(1- ;2;"-1) (1- Xh-2) + ...
1 - J: (1 - x) (1 - .1_.2) (1 - x) (1 - .1;2) (1 - &t" 3 )
= (1 - x) (1 - x" 3 ) (1 - xh -1)

THE ROOTS OF UNITY
177
53. The Gaussian sums. - In his investigations on the con..
struction of l.egular }Jolygons Gauss was led to the problem of
determining the sumB of the follo\ving type:
(1)
n-l ( 2 2 )
2 1'& 1'1l 8 .. 2 1'& 'In s
p(rn, 12) =  COB + t SIn ,
8=0 n n
where 111 and '12 are integers, n > O. After much effort be at last
established the follo,ving result:
Theore'll1 99. If Ii is a ll11tu1.al Jllunber, u:e haz'IJ
[ (1 + i) Jl n for 11, == 0 (mod 4),
Jill for n == 1 (lnod 4),
p ( 1 , 'J)  I
o Jot n = 2 (nl0d 4),
il' n for 1l = 3 (mod ).
Proof Let us put
2 It)
1'& ,...1'&
e = cos -- +  SIn -.
J2 12
For 12 = 2 (mod 4) the theorem is trivial, since
n 2
8 2 +8)1+ -
8(8+1 n)2 = e 4 = _ e s2 .
Thus, one half of the terms in the SUlll (1) are cancelled out by
the other half.
Suppose next that 13 is odd, and put  (n - 1) = '. Let n be
an integer prime to 1l, and put em = 1]. In the polynomial iden-
tity (6) in Section 52 ,ve then put h = J2 - 1 and J: = 17- 2 . Since
1 - e 2k - t71
4) 1,
= - e-"
1 - e- 2 k
for every integer t, we obtain the following relation
1 + 'YJ2 + 'YJ6 -t- 11 12 + .., + '1 n tn-II = (1 - .'j-2) (1 - '1- 6 ) . . . (1 - 1]-2"+4)
or
n-l -l
(2)  1]k{k+l) = 17- 1 - 3 - a - II · - i)£-2 II (1}2k+l - 11- 2k - 1 ).
k=O k=O
12 - 516670 TrygtJe Nagell

178
CHAPTER V
Since
17"'-k,2 = 'I}'a+1;(l;-+ 1)
we have
"
'l}12  1] k(k+l) = 1 -I- 1] -+ 'l}4 + 1]9 + ... + 1'J,.2.
k=O
Further
1/ n - k ) tn-I: t-I +) = 'I/.:!-1:+)'2 = 1]1.:2+,2+ (1I-1)k = 1]il,-t/;jl
and thus
 v
'l}12  1]'n-k)(n-l:t 1) = ll,2  1}1.:!1:-1' =1](J + 1 2 + 1]'1,+2 2 + . , , + 'I}('1-1)2.
k-=l 1:-=1
Since
1 + 3 + [) + ... + (u - 2) = J.2,
it follows fronl (2) that
1 + 'f} + 1]' + '1,9 + ... + 'l}ln-l;2 == ("-1}-1) (17 3 - 1]-3) . . , (1]11-2_ 1 ]- n+2).
Here the left.hal1d side is b). definition equal to Cf (lil, u), alld
t.herefore we have
(3)
i (n-1) ( t 1 "» )
( ) II '. 'T /1: -.- 1JL j&
q: nz, U = 2 1 SIll ·
1:=1 n
For 1n = 1 this product has by forllluia (5) ill Section 51, the
value
i b - 11 (- 1)[1]. J.
Hence we see that {f' (1: n) has t.he f'alue J" for 12 := 1 (Inod 4)
and the value ilJ; for n ==  (lnod 4). Only t.he case ']1 == 0 (1110d 4)
.
remaIns.
\Vhen In and 1l are relativply }1l"hl1e natural nunlbers and h an
integer, we shall ])rOf'e the lenlnla:
(4)
q;(hlil, n). q.'(hll, In) = ({'{It, 1nn).
In fact, by puttillg
E(x) = COB 2:r;x + i sin 2 nx,

THE ROOTS OF UNITY
179
\ve have
(I .) I .)
i. 111 s'" I .n loW
q;(hnl,u).rp(hlLnl)==E +- )
,'I I 11 111 ,
_ "" J .' ( It t JJZ S + n t J2 ) _ ftl  I J .' ( ' " A 2 ) _ . (I ) ,
-- 4 - /' 4 - g. t. HZ n ,
-- HZ Il -- 1H n
':,1 !,-=-o
for bJ' Theorenl 33 the 11ulllbel-s iil8 -i- n t run through a c0111plete
residue systeul Inodulo JJZ n when  and t run throug'h a COin.
plete residue S).steln 1l10c1ulo '111 and l11odulo 12 respectively.
From (-t) "e obtain for It = 1 and nz = :!::l, if 11 is ocld:
(5)
q; (1, :,!,j 12) = cr (n, 2.i) . cp t:?,J, 11).
If P is even, we clearl). g"et
(6)
1: - 1 ) .J J. ') ), -1 /. .)
q' (:?,':, u) =  J:' ( = ) =  I' ( -.C \ = cp (I, II).
1..-=u Jt /, -=0 " l
If P is odd, we ha¥e
(7)
" - 1 ..) :j 1.- 2 \ Ii - 1 _) /., ').
(}"' (2 Ii , 11) =   ( ,_._ ) =L J:' ( -==;;-) = cp (::? it)
1,=0 J 1.......0 '
By fOrll1ula (6) in Sect.ion 51 ,"\"'e 1ICt
(H)
lp t. u) = (- i)lln-lj · l/;-
Further
(9)
(FiJI. -t)  :!(l
,.], \.
a.nu
(10) q> (1/, S) =.t. (cos Je.t -t- i sin Jell ) = VH (1 + .i) i l (n-l!.
Finally, for 1Jt == :!i J and fJ : -t  ,ve ha \?e
,11 _ 1 !!." _ I
] J. ( 'JJk2 ) =  E ( u (21..- r Ij2 ) +:! R ( -tJl/. 2 ) .
1:==0 IU 1:=0 1U k=-O 111,

180
CHAPTER V
In the first SUlll on tbe l'ig'ht-hand side the numbers (2 k + 1)2
are = 1 (ulod 8), and, if p. = ', the value of this sun1 is obviously

.t. '2: 1 E ( '2 (8 t + 1) ) = ..t.l'f (  ) lE (  ) = 0,
t-=O 111 IIi t::.O Ii
since Ii > 1. Fro III this \ve conclude that
(11 )
( '111 )
cp (n. ill) = '2 cp n, -:t '
\Vhell J1 i:; odJ, U11cl In is a })o,ver of 2 ,vhich is > 8.
Finally,: b)? repeated use of formulae (u), (6), (11) and (9) we
obtaiIl, if p is eVeJl and > 2:
1 f "') .:J ,--:--
Ip (1, 2;J n) = tp( 1, n) If' (n, 4) / . = (1 + .;) . :N 12 ,
and by repeated use of forll1ulae (5), (7), (8), (11) and (10), if {J
is odd and > 3:
1 / 2
Ip (1, 2,i 1/) = Ip (2, n) Ip (11, 8)  = (1 + i) "JI2P;1.
Thus Theorelll 99 is completely proved.
Exercises
90. Let 1l be a ua.tural nunlber, let r denote the number of
distinct odd prin1e factors of 1l, and let p be the exponent
of the big'best power of  which divides '11. If a is an in-
teger prilue to 12, and if .J,.'" denotes the number of incongruent
roots of the congruence
:l,2 -== II (IllOd Jl),
l'ro\"e that
1. .'T = 2 r for fJ = U or
;') ,'!' = .)1'+ 1 £01 " fJ = ",).
-.., - ,
3 ,.. = ')1'-;-2 f ' i-l > 3
.., .- 01 1.1 = .
= 1.
,
This result is also true if II iR a po\ver of 2 and }' = o.
Sug-gestion : Use TheorelH .t",.

THE ROOTS OF UXITY
181
91. Let 11, 1. and {J have the saIne significance as in Exercise 90.
If 4.n denotes the number of incongt.uent quadratic residues
modulo n, prove that
1. An = 9' ;2} for (J = 0 or = 1 ;
_ 9' (n) _ ') .
2. 4.n - 2r+l for fJ - - ,
_ 9' (13) . >
3. .ri n - ")r+2 fOI P  3.
-
This result is also true if n is a po,ver of 2 and ,. = o.
4. The number 12 (> 2) has a primitive root if and only if
the number of quadratic residues is equal to t.he number of
quadratic non-residues. In all otber cases the latter nUDlber
is at least thrice the first nunl ber.
9. Prove Theorem 88 by direct appliciLtion of Gauss's lemma
(Theorem 85).
93. When 1) is an odd prinle, determine the number of qua,dratic
residues ,. in the interyal 0 - }J ,vllich have t.he prOl)ert.
that r + 1 is also a quadratic residue.
94. Prove the relation
a-I [ Ib ]
  =  (a - 1) (b - 1) + } (d -1)
k=l a
where a and b are natural numbers and d = (a, b).
95. Prove the relation
 [ .b ] +  [ 'ea ] = r  ] . [ * ] + [  ] 2
1:=1 a k=l 11 - - ;;J
where a and b are natural nunlbers ancl d = (a, b).
96. Let '}}l, be a natural nunlber, and let a be a l)ositive llunlber
such that none of the numbers ka (k = 1, , . . ., 111) are in-
tegers. If n = [nz a], prove the relatioll
m ft [ k ]
 ['a] +   =11112.

18
CHAPTER Y
9R.
Show that t.he polynomial ,1';' + 1 is lleyer a prime function
to an)" prilne modulus. (Compare Sectiol1 29.)
\Vhen }J is an odd prime, we define tIle Legendre symbol
also in the case in which the 111.lnlerator t is divisible by p
bJ puttin
( ) -
... L
(;) = O.
If a and 1J are integprs. and if a is 110t divisible bJ9 p,
provc that
1>-1 ( . b)
 a:l' =0.
.(=-0 1
)f. Let'}J be all odd prime, and let f (Lr) = (l &t 2 + b [ + c be an
integral POlJl10111ial of the second degree, where the coef.
ficient a is not divisible by p. Put L1 = iJ2 - 4 a c. As in
the preceding exercise we put () = 0 if I is divisible by p.
Proye that
 ( j(:) ) = _ ( 1I )
IQ P P
if J is l10t divisible by 1), and further that
IS ( .f( ) ) -== (jJ - 1) (  )
..'=0 1 J
if .J is c1i,isible bJ }). (J acobsthal.)
100. Let p be an odd prinle, and denote by 'I1Z the Dunlber of
quadratic l1onresidues Dlodulo p in the interal 0 - . p.
Show that
11/ = .2: 1 ([ 2 2 ] _ :! [ 1..2 ]) .
k= J}J P
If }J is of the form -l- II + 1, we have already (in Section 38)
ShOWIl that '112 = 1 (}) - 1).
101. Let}J be a pri111e of the form 4 u -i- 3. How many quadratic
residues modulo 1) in the interval 0 - 1) a.re even? Express
this number as a function of 1n defined in Exercise 100.

THE ROOTS OF UNITY
183
102. How many of the quadratic residues nlodulo p in the in-
terval 0 -}) are even, when p is a priDle of the form
4n+l?
103. If p is an odd prinle, prove the fOl"Dlula
! :]j-l! [ k2 ]
 ,. = 2 1 4 P (}J2 - 1) -}J  -,
1:= 1 1)
wllere the first SUlll extends oyer all quadrat.ic residues 1.
modulo }) in the interval 0 - p. Sho,v that this sum has
the value 11) (p - 1). if P is of the form 4 n + 1.
104. If}) is a prime of the fornl 4 u + 3, })rove the formulae
IT n · :it ,. IT '" ·  te: J r-
 S111- =  SIU ---:- = 11 1 ,
r P 8 1)
whel.e the first product extends over all quadratic residues
modulo p in the interval 0 - }) and the second product
over all quadratic non-residues in the same iuterval.
Find also the value of the product
lip-II k 2 f
II 2 sin ,
1:=1 P
where f is an integer which is not divisible by p. The pro-
duct depends on the number 111 in Exercise 100.
105. Prove the follo\ving theoreln: If p is a prime of the form
8 11 + 1, there is in the interyal 0 - J;P at least one prime
q which is a quadratic non-residue of }).
Suggestion: Use Thue's theorem.
106. Prove the following theorenl: If }) is a prime of the form
,--
8 n + 5, the).e is in the interval 0 - 1- :?}J at least Olle odd
prime fJ which is a quac1ratic non-residue of p.
Suggestion: Suppose it. is true that et'ery prilne = 1 (nlod 4)
may be written as the sum of tvo illtegral squares. (Com-
pare Section 54.)
107. Prove the following theorem: If p is a prinle > 3 of the
form 4 11 + 3, there is in the interval 0 - ( J P + 1) at least

184
CHAPTER V
one odd priole q which is a quadratic Don-residue of p and
of the form 4 '111 + 3.
Suggestion: Put a = [J lp] and consider one of the numbers
J)-a 2 , (a + 1)2- 1 ) or (a + 2)2_p.
108. Prove the following theorem: If p is a prime > 17 of the
form 4 Ii + 1, there is in the interval 0 - Vp at least one
odd prime which is a quadratic residue of p.
Suggestion: Suppose it is true that every prime = 1 (nlod 4)
lllay be written as the sum of two integral squares. (Com-
pare Section 54.)
109. Prove the following theorem: If p is a prime of the forDl
8 n + 7, there is in the interval 0 - (2 J I}) -1) at least
one odd prinle q \vhich is a quadratic residue of p.
Suggestion: Consider the numbers p + ll, where llo is a
root of the COJlgruence
tt 2 == - p (mod 2"),
alld
Iz = [ Og p ] + 1.
log 4
110. Let P and Q be two odd and relatively prime integers> 1,
and let p denote the number of integers in the sequence
1 . Q, 2. Q, 3. Q, . . ., 1 (P - 1). Q,
whose principal remainders modulo Pare >.1 P.
Show that for Jacobi's symbol we have the following relation
() = (- 1)1'.
This result is a generalization of Gauss's lemma.
Suggestion: Put R (r) = x - [x + l], and let sign x denote
+ 1 or - 1 according as x is positive or negative. Begin
by proving th relation
i (P-l) ( 1; Q )
( - l)u = sign II R  ·
h=l P

THE ROOTS OF UNITY
185
111. If a and bare t\VO llatural nUDlbers, b odd, show that for
Jacobi's symbol we have the following rules:
( a ) ( (,(, ) .
2 a + b = b If a == 0 or = 1 (mod 4),
aild
( 2 a a b ) = - () if a = 2 or == 3 (mod 4).
112. Let a, band c be natural nUlnbers and (a, b) = 1; suppose
that b is odd alld <4ac. Show that for Jacobi's symbol
we have the follo\ving rule:
( 4 a: - b ) = ().
113. If x and yare integers and y2 > 1, show that none of the
following four quotients is an integer:
4x 2 + 1
y2+2 '
4x 2 + 1
Q ,
1/-2
x 2 -2
2 y 2 + 3'
x 2 + 2
3y2 + 4
114. If p is a prime, show that the solutions of the congruence
Fp-l (x) = 0 (mod p)
are the primitive roots of 1). Fn (x) is the cyclotomic poly-
nomial of index 12.
115. Show that the sum of the 9' (n) primitive ntil roots of unity
is equal to p, (11) (Mobius's function).
116. If the natural number 11 has at most two distinct odd prime
factors, show that the coefficients of the cyclotomic poly..
nomial Fn(x) cannot have other values than 0, + 1 and -1.
11 7 A P\lt
F n (.1:, 1/) = II (-c - BY),
e
the product extending over all the primitive nth roots of
unity. For what. values of 11 is the equation

186
CIIAPTER V
Fn(X'Y)=lJ
sol vahle in integers x and 'Y, if 1) is a prinle factor of n '?
Find all the solutions x and y in these cases.
118. Let 1n and 11 be integers; suppose n > 2 and (1n n) = 1; put
IT = 112 Sill n am ,
J1
the product extending o\"'er all integers a in the interval
o -  11 \vhich are l)rime to 11.
Prove the follo\ving prOI)ositions:
II = 1, if n is neither of the form lJ a nor of the forln 2]Ja.,
where }) is a prinle.
II = (j) G r- 1 . lIP, if /I is a power of the odd prime p.
II = (- 1)1 \lIz-1) (fJ-I), if n is t\vice a power of the odd
prime p.
II = ( 2 ) (- 1) L n : »l-1  · ).'"2, if 11 is a 1) 0 w cr 0 f 2.
112
] 19. Let 1n and n be integers; suppose n > 2 and (1H, II) = 1.
Prove that
J ( 1n ) ( :? ) m-l. ·
1 a n] 1. - ,If 1l IS 0 d d,
- II 2 sin fL-: 111 = I II
].I; k=l l ( ;,, ) (- l)t n im-l" if n is even.
120. Let i' be an arbitrary positiY"e number. Show that there
are illfinitely man)T primes }J such that the least positive
primiti ve root of }) is > i'.
Suggestion: Use the theorem that for any natural number
}1, there are infinitelJ nlany primes == 1 (nlod n).
12]. Show that there are infinitely lllany primes p such that the
exponential congruence

THE ROOTS OF UNITY
187
2 q == 1 (mod p)
has a solution q which is a prime..
122. Let 12 be a positive odd integer, and let '111 be an integer
prime to n. If q; (nl., t2) denotes the function defined in Sec-
tion 53, prove the formula
( ) ( nl ) .:a 1 / -
lp ))1, Ii = 12 Zl. 12 ,
\vhere " =  (u - 1).

CI-IPTER V I
D [ 0 P II L\. N T r NEE QUA 'f ION S 0 F THE
SEC01D DEGREE
54. TIle representation of integers as SlInlS of integral squares.
'Ve shall use Thue's theoreU1 (Section 36) for proving
ThcorrJlt 100. 1. Erery lJI-inlf. p lfhich i.,' = 1 (mod 4) can be ex-
}Jre.....sed ill the .fornl }J = ..2 -1- J/ 2 , lfhl'rp x and yare natural
liunzber8. .lfo other odd lJrinlp....' !lal'e thi..... pro})erty.
2. l",.er!1 }}rinl(J 1) lchich i.')' == 1 (lllOd 6) call be e"pre8sed -in tilt?
forJ1z 1) = J... 2 -+- 3 y2, 1fherf' .{' and '!I a1'e natul-al nU111bel..,a. "}to
otll(r jn"hnes hare this lJrupert!J.
.!j. J'fl"frll })rinu- J) u:hich i, = 1 or == 3 (.111od R) (au br expressed
i 11 Ih .f'al-nl 1) = .1"2 + 2 .'/2, it'here ,I' a}ld .'I are uatlirallllll1lber...
,£,r o other jl1.hneb' hat.c' thi.",' pr01Jrrt!l.
J. El'e1411 lJ/.il1ze }J u'hich is = 1, -= fJ or =. 11 (lllOd 14) Ca1"1 be
expressed in the jorlll 1) = .t. 2 -;- 7 ?,2. lfhere x and !I are natural
Illllnbrrs. No otllel. }n-inU'8 hare thi..... lJfO}Jer(ll.
:J. El.e1.!1 pI.i111C p 'l!'hic/z i...., = 5 (),. = 11 (mod 24-) can be expres,'pd
in the .(o}'nz p = 2 x 2 + 3 y2, u'here ,I" and yare /latla.al 12UIIl-
ber.,-. ro ather- })rhne8 hare this }JJ'O})e1.ty.
.A supplement to this result is
Tlu 1 0reln 101. Ij' c and d al.e [lit"en natural n!i1nber(-.:, tlze1"e i..::
at nlost one repl.e....'etltatilJ}l of the 1-)rhne p in the form p =
e :c 2 + ([ '!I 2 , 'If'her-e x and .'1 arp natural 11 11111 bers.
].)ro()j: Let us consider the con gruence
(1)
e 2 + d = 0 (mod jJ).
where d = 1, 2, 3 or 7" and \Vllere ]J is an odd prime. From the
results in Cba.pter IV we have: For d = 1 congruence (1) is

DIOPHANTINE EQUATIONS OF THE SECOND DEGREE 189
solvable if and only if j) == 1 (n10d); for d = 2 it is solvable if
and only if p == 1 or == 3 (n10d ); for d == 3 it is solvable if and
only if 1) == 1 (nlod (i) apart froD1 p = 3; for d = 7 it is sol¥able
if and only if p =- 1,  9 or = 11 (mod 14) apart £ron1 j) = 7.
If z is a solution of congoruence (1), and if the modulus is a
prime IJ. we have by Thue's tbeoren1
 == + ,o (mod p),
11
where x and .II are natural 11umbers < l,o j) : we can suppose that
(x, y) = 1. Congruence (1) becol11es
J.:2 ;- d !}2 :-:: 0 (n1od})),
and therefore
" I C)
J.0 + l .'/- = 't1l p.
where m is a natural number < d.
Hence for d = 1 we get 'Ii? = 1 and
.l:2 + !/2 = jJ.
For d = 2 we get ,n = 1 or '112 = 2, thus either x 2 + :?!J2 =}) or
x 2 + 2 y2 = 2 p. BtY putting x = 2 Xl in the latter equation we
obtain
:'II 0) + C)
 .ti l/ = p.
For d = 3 we get }JI = 1, :? or 3, thus either {.2 + 3 y2 = !I..
.1,2 + 3 y2 = 2 Jl or x. 2 -1- 3 .1/ 2 = 3 p. rfhe second of these equations
is clearly impossible 1110dulo 4-, since 1) T 2. By putting x = 301'"1
in the last equation ve goet
3 xi + y2 = }).
For d = 7 we get 111 = 1, , 3. 4. 5, 6 or 7. If J1i is eve11 both
x and y must be odd, and therefore the number ('2 + 7 y2 is
divisible by 8. Hence the equation ,2 + .. !/2 = In p is in11lossibl
for In = 2, 4- and 00 Since - 7 is u, qua.dratic non-residue of tbe
prin1es 3 and 5, tbe values i17 = a aJ1d 111 = [) are a.lso impossible.
[f '12 = 7 we get, by putting oL' = 7 t.l'
_ .J, C)
. 'i ;- y = jJ.
Thus the first four parts of Theorelll J 00 are proved.

190
CHAPTER VI
Consider next the congruence
(2)
2 e 2 + 3 == 0 (nlod 1)),
where p is a p-rilue > 3. It is easily seen that this congruence
is solvable if and onlJ if p == 1, 5, 7 or 11 (nlod 24).
If z is a solution of congruence (2), \ve have by Thue's theorem
t
 == + e. (nlod p),
U
where x and yare natural numbers < 11 p ; ,ve can suppose that
(x, y) = 1. Congruence (2) becomes
nil), 3 II) 0 ( d )
 x'" -r yw == Ino p
and therefore
 :1.,2 + 3 g2 = 112 1),
where '12 =:. 1, :!, 3 or 4. If In =, :r is odd and 11 even and
!J = 2 111; th us
eV 2 + 6 yi = 1).
But this equation is possible ol1ly for }I = + 1 (mod 8). If l1l = 3,
x is di visi ble by 3 and J.' = 3 .1'1; th us
6 J;2 + 1 12 = 1 )
1.'1 ,
which implies that p == + 1 (mod 8). If 111 = 4, both x and y
vere even; but (x, y) = 1. Hence we ha,"'e
2 .£,2 + 3 .'/2 = }),
and tllis equation is possible if and only if p == + 3 (mod 8).
This proves the last part of Theorem 100.
'Ve now proceed to the proof of Tlleorem 101. Suppose that
we have the two representatiol1S of the prinle p
(3)
.) 1 ')
p = r ..t W + t. I''''
and
(4)
p = C u 2 + d lo2,

DIOPHANTIlSE EQUATIOS OF THE SECOND DEGREE 191
,vhel.e ;r y, 'tl and l.' a.re natural nUlubers. Eliminating d from
these equations, we have
J} (11 2 - 1"2) = r (,,2 11 2 - r 2 x 2 )
and, since c < }I,
(5)
uy = + t" x (mod p).
Mlllt.iplying together equations (3) o.nd (4), we get
(u)
])2 = (c x u + d!1 t.)2 -f- c d (Zl 11 + l" :t)2,
where the upper or lower sign may be chosen arbitral"ily. If ,ve
suppose that 'i1]1 = l.X, we D1USt ba\e fl = eX and l' = y, since
(.c, .II) = (u, r) = 1. If ll.ll  l" x, it follo,,'s fron1 (5) and (6) that
Ill!1 + 1".1.: 1= p, c = d = 1 and C.Xll + d y. = U; this is possible
only for ;2" = ,. and 'Y = u. Thus Theoreln 101 is proved.
It i easy to \erifv the identit,..
., . A'
((12 + db 2 )(a 2 + d{J2) = (a a - dlJfJ}2 + d(afJ + lJa)2.
By Ineans of this and ihe first three parts of Theoreln 100 we
obtain the following results:
1. E,"ery integer ,,,hich is the product. of priu1es = 1 (mod 4)
or t\vice such 0. product can be expressed as the SUD1 of two
integral squares.
2. Every integer which is the product of prin1es = 1 (mod 6)
can be expressed as the sun1 of an integral square and thrice
an integral sqllare.
3. Ever)" integer ,vhi(.h is the product of pl"iInes == 1 or = 3
(1110U 8) can be expressed as the sum of an integ'ral square
and t,vice an integ'ral square.
'l'hese three results ,vere stated bJ Fer111at; but the proof was
givel1 by Euler.
55. Bachet's theorem. - The following identity of Eulel. is
easily verified:

192 CHAPTER 
(a 2 + b 2 + C 2 + ([2) (a 2 + {12 -I- ;,2 -t. c5 2 }
(1) = (a a + bp + c;' + d)2 + (afJ - ba - c + d;,)2
+ (a;, + b-ea-dp)2 + (a-b;, + cfJ-da)2.
\Ve shall use it for proving Bathel's theOr(Jln:
Theoretn 10,2. El"ery nat.u,.al '1111lnber can be e.i:pre....u:..(,d as the .ll'fll
of fout i12tg1"al .qlta1"e8.
Proof. In consequence of Euler's identity (1) it is sufficient to
prove the theoreUl for primes. The following proof is due to
Lagrange.
Lemrna 1. lj'}) is an odd pril1ze, there l::£isf foul" intege1". Xl, X2,
Xa alld x 4 such that
 + :l,2 + X,2 + x 2 = J1l j J
1 234 '
'tl,here m 'is a natural u'll1nbe,. < jl.
Proof. The numbers
(2)
x 2 (0 < X <  Ji - 1))
clearly are incongruent n1odulo }). This is also true for the
numbers
(3)
- 1 - y2 (0 < 11 < } (lJ - 1 ').
The two sets (2) and (3) consist of J) + 1 Ilulnbers. Thus, at least
one of the numbers (2) must be congruent to one of the num-
bers (3) modulo p, and therefore
.1'2 + y2 + 1 = 1) 1n,
where m is a natural number. Fiually we have
(£2 + '/12 + 1 1 ( p2 }J2 )
'In = · < - + - + 1 < 1),
1) J J 4 4-
which proves the lemma.
Lernnza 2. E"el":ll pl"iuze p can be e..(:pressed as the Sll171 oj" fo",.
in tpgl.al sq'lla're..

DIOPHANTINE EQFATIOS OF THE SECOND DEGREE 193
Proof. TIle leIl1ma is true for J) = . 8u ppose that 1) is odd,
and that 711 IS the least natural llulnber snch that
(-t-)
C). C). t) + 2
,.£'i -r .1'2 -T ..t 3 . :J -1 = lU }),
whel"e Xl' :1'2, Xs dnu x'" are integ'ers. According to Lel11Il1a 1 we
have 111 < 1). Len11na  is proved if ,,-e can sho\v that HZ = 1.
Equation (4) ma)' be \Vl'jtten
( ;; ) ( 1 + X2 ) 2 + ( Xl - ,[12 ) 2 ( 'lIS -f- X 4 ) 2 -L. ( .r3 - ':('4 ) 2 = .1.
v ..) ..) + ..) , ..) 2 111 p.
- - - 
If 71l IS even, the follo\viug three caRes ,viII be uistillguisllell:
1. The numbers J'l" :1'2, Xs and J'4 are all e\"ell; :!. the nUI11bers
'1, .1"2, Xs and X4 are all odd; 3. two of the nunlbers, say :2"1
and X2 are even and both the others are odd. In all these cases
the numbers i (.1'1 + J"2),  (a.'1 - .l'2)'  (;.t's + '4) anu  (.rs - ,r 4 )
are integers. But it then fullo\vs frol11 equation (b) that, t.he
number 7n does not satisfy the prescribed l11il1iInUln condition.
Hence )11 is odd. N O\V su ppose 11t ::;,... 3, and for every :I'/; (l = 1,
, 3, 4) in (4) let us choose au integer l/l: such tbat
!11: == J.,'k (n1od lIi)
and Illk I <  11l. TheIl \ve g'et
C) + C) + t) + Q 0) + O).J- '), 2 U
Yi Y"2 !I; .'1"4 = Xi "'2 -, ,)'; T J 4 =
(1110d 112)
and therefore
(6)
2 9. C) Q
'll + "" + lr" + IJ" = III ,..
. 1 .. 2 .. 3 ,14
If r=O, we \vould have !/1=.tI2=!la=!/4=O; the nunlbers £1'
;r2, Xs and ';"4 would all be difisible by 112, and \ve \"ould have
jJ = 111 ( ( ; r + (': r + c;: r i (;: f) ·
But. 1n is not a divisor of }J, since 1 < nz < 1). Hence the integer
r is positive.
Since Lillo' I < ! ')11, \ve have from (u)
2 C) C) I)
J- 71l -l- 1 In" -I 1,n" + :1 Ili" > }U I' .
and thus
,. < Iii.
13 - li16G';O 7"rY(1I'e }.'ragell

194
CHAPTER YI
If we nlultiply together e(!uatiol1s (4) and (6) member by mem-
ber and use the identity (1), we obtain
(7) 'In r · 111p = (Xl .".11 + X2 Y2 + X3!J3 + X, ']14)2
+ (.£1 Y2 - X2 Yl + X3 ]1, - X4lJ3)2 + (Xl U3 - X3 YI + X, U2 - X2 'lJ4)2
+ (Xl!J4 - XI. YI + X2 U3 - X3 Y2)2.
All the four squares on the right-hand side are divisible by tn 2 ;
for we have
4 4
 Xk'Y" ==   = 0 (nlod'ln)
1:=1 1:=1
and further
Xk '!Ii - Xi YJ: = Xk Xa - Xi Xl: = 0 (mod n).
Hence, dividing both sides in (7) by '1)1,2, we deduce
. p - 2 + ...2 + Z2 + &2
, - "'1 "2 3 "'4'
where Zl, 'Z'2, 2'3 and Z4 are integers. But, since 0 < r < 1n, this
is contrary to our definition of 111. Hence, the only remaining
possibility is that n = 1, and Lemma 2 is proved.
Thus, Bachet's theorenl is true for all primes. Finally, using
Euler's identity we see that this theorem is true for all natural
nunlbers.
Lemma 2 also gives an algorithm for determining a representa-
tion of a given prime as the sum of four integral squares.
There exist natural numbers n which cannot be expressed in
the form
2 .) 2
12 = Xl + .1'2 + :3
when Xl' X2 and Xs are integers. It is easily seen that all in-
tegers of the form
41: (8 112 + 7),
wllere k and 111 are integers > 0, have this property. Conversely,
it may be shown that 110 other positi'f'e integers have this pro-
perty; but the proof, ,vbich depends on the theory of ternary
quadratic forms, is rather difficult.
It follows from Lebesgue's iuentitJ

DIOPHANTINE EQU_.\.TIONS OF THE SECOND DEGREE 195
(a 2 + b 2 - (.2 -1- d 2 )2 = (a 2 t- 2 _ (.2 _ d 2 )2
+ (2ac + 2bd)2 + (2ad-2bc)2
that every integral square nlay be written as the sum of three
integral squares.
56. The Diophantine equation x 2 - D y2 = 1. - "\Ve first prove
the following
Lenuna. I)" D i. any natural llu1'nber folzich £.-; not a lJelfect sqlllre,
there are iltillit(!ly tna12.l1 pairs a!' natural '1lu,nbers x and :'1
'u'hich ..,'atisfy tlu! 2.nequ,alify
(1) I x 2 - D y21 < 1 + 2 J'D.
Proof The number lID is irrational (Theorem 19), and by
Theorem 20 there are infinitely Inany pairs of positive integers
x and 11 such that
x _ JI D <!.
11 y2
We have furthel'
x J ' n I - It' 1 rn n ... / D 1  1 ' n
ii + I - !I - r II + it. r < '!i + :.. .
Thus
Ix 2 -Dy 2 1 = I.t -!,J iJl.lx + !JJ lJ l < !;2 + 2LlJ < 1 +2 JI D .
Hence the inequality (1) has an infinity of integral solutions x
and '!I. Q. E. D.
Theorenl lOB. I,l D is a llatlu.al 'lll1nbl'r u:hieh is not a J)e}iect
..q!lare, there i' at lea8t 01le l)air of natural 1ul-tnbers x and 11
lL,h'ich :1atisj}1 the Dioljhantine equation
(2)
.1.. 2 - D 1/ 2 = 1.
o.
Proof. It follows from the lemma that there exists at least
one integer k, different from zero, such that
aC 2 - D y2 = k

196
CHAPTER ,. [
for infinitely nIRny pair8 of integers x and y. Among these pairs
', y there Blust exit at lea:it two pairs :tl, !/l ana .l'2, Y2 which
so. tisf.y the cou gruence conditions
(3)
.'1 = "2 (IlIOd I It I) and '!II == Y2 (tl1od I k I ).
In fact, the l'ell1ain(lel's luod ulo I k I of the four n Uln bers Xl, .r 2,
1/1 and 1I2 nItL) be c-olubined in a finite 1Iuluber (= ,(.4) of "rays.
Hence, \ve can 8U I'I'0se that
(4)
.1,2 - ]) /12 = .l,2 - D / 12 = .
J. .' 1  .'2 '
\vhel'e .l'I: j/I. .1'2 and !12 satisfy the conditions (3). Now ""e have
 .  V -
(.i'l - tit J D) (,1.'2 I- Y2 J IJ) = Xl. 1 '2 - !It !12 D -I- (.i'l Y2 - X2 !/1) D.
B? (3) and (-1-) ,ve get
,1'1 X2 - !/t Y2 D == x'i - !/ D == 0 (nIod. 11.'1)
anu
,fI !12 - ,t'21/I == :(1 !/l - Xl111 == 0 (Iuod I 'I ).
1'herefore
Xl'2 -.1111/2 IJ = I,:u
and
Xl 1/2 - ,1'2 III -= 7..' l",
"There u anu 'Z' are integers. Hence
anu
(Xt - /11 11 D ) ('2 + 'Y2 J ID) =- k (II -t- r ll))
(J'1 + III ¥D) (.c2 -- Y2 JID) = J (11 - f J.' D ).
MultiplJing' tog'ether the t,¥o equations nleluber by Inember we get
(.{'i - D !Ii) (.I' - ]) 1I) = k 2 = l-2 (u 2 -- D 1'2j.
Hence, we have
u 2 - D 1... 2 = 1.
Here v  O. For if r = 0, we would have ;,('1!/2 = X2.'/1. 11  + 1
and
(,/'1 - .'11 V jJ} lJ'2 t .'12 J lJ} (.(:2 - !l2 1/ D) = + 1- (.l'2 - .112 J 'j)).

DIOPHAl'TINE EQUATIONS OF THE SECOND DEGREE 197
Theretore after division bv /.-
01
.1"1 -.'11 V n = + (X2 - !12 V D )
,vhich ilnplies ,rl == + :f2 ancl .111 = + .'12' But \ve can choose
I Xl I  )..:t:21. Tllus Theorem 103 is proved.
The theorelu was statecl in 1657 by Fcrnlat ,vithout proof.
'fhe first c0J11plete p1-00f \ras given by Lagrange in 17()8. A bout
50 years later it was discovered that Indian nlathematicians
even before GOO A. D. possessed au algorithm for solving equa.-
tiol1 (2); but they had no proof that their Inethod alwaJs gives
a solution of the proble11l. The proof just given is clue to Di-
richlet.
Commonly equation () is called Pel!".": equation; but this is
unjustified, since Pell did uot J1lake any independent contribution
to the theory of this equation.
Let J) and k be two integers. If ."!' = 11 anu .'I = t' are integ'ers
which satisfy the Diophantine equation
(5)
x 2 -- D 11 2 = k
.. ,
we say, for silnplicity, that the nunlber
11 + l' llJ
is a ,D7utiuJ1 of e(luation (5).
The two solutions II + 'v J iJ and II' t- ,.' J ]) are equal if u = u'
and l" = r'. The first solution is greater than the second if
It + t. y n > u' + / Y D .
Let us consider all the solutions J. + 1/ JI [J of t.he equation
(6)
.l2 - D !12 = 1
with positive Ii' and II. Alnong these there is a least soll1tion
.rl f- Yl J I D , in \v hich ,rl and :'/1 have t.heir least (positive) values.
1'he number £1 -r !Jl JI D is called the .fll111lcnneutal solution of the
equation (6).
A c01l1pleineut t.o Theorem 103 is
Theore11Z 1 (j,I. l' [J i. a l1atural 1211 JU!J(>f lfh it'/z i..... uut a prrject
'''-'1ual'e, the lJjolJhantil1p. equaiiun (H) has i1tinit('ly 112any f.f)lu-

198
CHAPTER VI
(7)
tions .x: + 11 l/ D . .A.ll 801utio118 u,.£th positil:e x and y a1-e ob-
tained by flu) fOrl1lu,la
,- ,-
X ll + .'In J 1J = (Ll'l + Y1 1'D)n,
ll'lzere Xl + 'Y1 JI D i..... the junrlal1ze12fal solution of (6), lllzere 12
'-UJl8 th'rough all 12at1l1al Jltlnlbe'l"', a'nd 1l'lzerc
r .. = 0')..11 -I- '"' ( Jl ) X 'l-2. 1, 2 k D k
.(. 7& ..4' 1 .  _) 1. ] .. 1 '
1:= 1 - A
I II = ,..., ( Ii ) 1"1- 21:-+ J Y 2 k-l D '-l
,!1 n  ,,) , _ 1 · 1 1 .
k = ] - II
(8)
Proq/: Clearly it follows from (7) that
Xu - !/11 VD = (."i'l - Yl JI [) }1l.
Then, multiplying together the corresponding members of this
equation and equation (7), we have
2- D 2 - }
XU .iJ,l, - .
Hence Xn + fIn 1 '1) is a solution of (6).
Suppose no'v that it + t' JI D \vere a solution with positive 'It
and f which is not obtainable by formula (7). Then a natural
number fJ would exist such that
(,,{'1 + Y1 J' D )' < 1I -1- r J ' D < (Xl + !/1 JIJ))l1+ 1
and thus
x" + Un JIJj <: 11 + t' VD < (.l'n + 'Un 1 / D) (.r1 -j- 111 JI D ),
Hence, multiplying by the positive number X 71 - !In vD, we would
have
(9) 1 < (-u + l' vD) (..r n ..- Un VD) < Xl -1- !/1 J<D.
If we put
(u + l J'D) (r71 - 'Y" VD) = x 1- .lJ Jm,
where x = PI t, - I' Hu D and 1I = l X n - U !In, \ve would also have

DIOPHANTINE EQUATIONS OF THE SECOND DEGREE 199
(u - l: 1/])) (Xn + Yn J I f) ) = X - Y 1/])
and, multiplJing together the last two equations,
1 = (u 2 - D 1..2) (x; - Dy;,) = x 2 - Dy2.
Hence the number x + y Jm would be a solution of equation (6).
Then, by (9), we would have
x + y llJ5 > 1,
and, on the other hand
,---: 1
O<x-!lJ.D= ,- < 1.
,r + y l J)
It is obvious from these three inequalities that x and y must
be positive. Now it follows from (9) that
r=- ,-
x + y I/.D < Xl + '!II l' D.
But this inequality is impossible, since Xl + 1/1 li D is the fundamental
solution. Thus Theorenl 104 is proved.
When D is given, the fundamental solution Xl + Yl lI D may be
found by trial. In the expression 1 + D '!I 2 we put successively
Y = 1 2, 3, 4, etc., until it becomes a perfect square. Howver,
the practical utility of this Inethod is quite limited, as is seen
from the following example: For D = 94 tbe fundamental solu-
tion is
2543295 + 22106-11 / 94 .
It was shown by Euler how it is possible to determine the funda-
mental solution by 111eanS of the expansion of J/ J) into a simple
continued fraction; this method is much quicker than the method
by trial.
For certain values of D the fundamental solution maT be in-
.,
dicated immediately. For instance, the equation
x 2 - (u 2 - 1) y2 = 1,
where 11 is an integer > 1. The fundamental solution is u + J/ .,,2 -1 ;
for here we have '!II = 1.

200
CHAPTER \"1
More generally ,ve haY'e
TheoreHl 1();)" Let 1) be a natural nlu1lbe1' 'lL'hirh .,'. not a jJelfec.f
SqUQ1YI. /.(, aui! 1] are natural nlunbe1"S b'atitf.iling the inequ.alit.11
(10)
E > } 11 2 - 1.
aud I)' a = E
I} J]) is n 8olutioll qf the (.quatiou
x 2 - D 1 12 = 1
.' ,
(1 ] )
thela the ullndJl,. a is the .luudal12r'uf(ll solution (l this (.quatiou.
P,'OO.f. The theoreln is bvions for '1 = 1. Suppose now that
1] > I, ana that Xl + 111 1/ [J is the fundamental solution of (11).
Suppose further that 1  '!il < 1]. 'fhen we ,vould have
00)
xi-l E 2 -1
D=--=- - ,
y 'I} 2
and
.l.21 ] 2 - ,/22 = 1 } 2 - Y 2 = d> O.
1 · ] . 1
Henc
.l'1'] + !/1=dl' :i"111-YI;=d 2 ,
W}lere d l and ,/"}, are natural nUlubers such that d 1 d 2 = d. Thus
2 2
l: _ ell -"2  d - 1 _ fJ - .'/1 - 1 < 1 2 1
"'- --- - -1 ] -
:! .II] = .'I1  .'/1 = 2 ·
But., since this is contrar)" to (10). we must have 1] = !II as as-
serted in TheorelD 100.
We ]lave the following corollar. to this theorenl:
Let !/I (lud u. be uafllral ulunlJp1"s. 1.1' tfe pu.t
J) = It (ll !Ii + 2),
the nunlber
1  It yi + III -V D
is thr fuudrt1l1rlltal ,'ollttio}1 qf equation (11).
By letting II vary, we obtaiJl an infinit.,. of values D for ,vhich
1/1 has the saIne yaluf.'.

DIOPHANTINE EQllA.TIONS OF THE SECOND DEGREE 201
57. The Diophantine equation x 2 - Dy2 = -1. - 'Vhile the
equation
(1)
,['2 _ n y2 = 1
is al\vaJ.s solvable if the natural nUll1ber D is not a perfect
square, the equa.tion
(2)
;2 - I) 1]2 .:= - 1
is solvable only for certain ,.all1es of 1). A necessary condition
for the sol\'"abilit.. of this equation in integers E and 11 is ob-
viousl)- that all odel prill1e fu,ctors of ]) be of thp forln .t 11 + 1;
furtherlllore, if D is even. it cannot be divisible bJ.t. I-lowever,
these conditions are not sufficient, as will be shown below by
an exanlple (IJ = 3--1).
If eq uat ion (2) is sol\rable for a givpn integer I) ana if
'1 + 111 J'l) is the least solution witb positive ,; and 1]. we say
that El + 1]1 JID is the funda1Jzental 8uluti(J1Z of t.he eCtuation.
The square of any solution of (2) is obviously a solution of (1).
\Ve pro\'"e tIle follo\1fing tbeOl"eu1:
1'1zeore'J1l J(jri. Let J) /J( a lIatura7 I1l(7111)(lr 'rlzieh i.,,' not a prrfect
8quatt'. SIlPJJOxe that ('quotioJl t,:.?) is soll"al)le, aJld that l -f- 1]1 l/ n
it"! it..... .lul1rla11leutal .(J!lltjon. l'lten tlip nu}n1j(JI'
() J,1 + .'II YlJ = (E 1 + 1]1 J/ [)) 2 = .; -t J) 1}i + 2 ;1 )]1 JI ])
18 the .rlll1danzrlltal .o'tltiol1 0.( equation (I).
]11l,.the1", ':( l1"t' put
()
11 + '/]n V D == (1 + 1]1 Y D) n
lcith
(5)
I t - /:11 + ",' ( II ) l:1I- 1: 2': [)1.'
 1Z - ., 1  ') 1.  1 "h :
1:=1 - 11
l 1 } =  ( U ) l;n -21:+1 1 } 21.'-1])1;-1
" .) /, . - 1 ':II 1 1 '
1:= 1 ....,
.lotl1fltl a () [Ii rrs :

202
CHAPTER YI
1. 41l the t101uti011t1 It'ith }Jositit'e E and 'fJ oj" equation (2) ll.lzen n
runs tlzrough all po'itil"(: odd integers.
2. 411 the solutio1l8 u.'ith }Jo8itz'1:e x = 'n and '!J = 'Y}n of equation
(1) 'l(,he)2 n jOUllS throllgh a1llJOsitit"e eren integers.
Proof. We get fron1 (-1)
t - 'YJ "l i n - ( t _" l / ]) - ) n
1I n Y - 1 .,1 ,
and multiplying this equation by equation (-1),
E; - ])1}; == (- 1)71.
Hence, En + 1}n }I D is a solution of (2) or of (1) according as the
exponent n is odd or even.
TIle nunlber - l + 'I}l V D is clearly positive and < 1; it is
the largest of the solutions of (2) wllich have a negative E and
a positive 'I}. Suppose now that the fundanlental solutions of
equations (1) and (2) are not related by the formula (3). Then
we must have
,-- ,-.
1 < j'l + .l/1 1 D < ('1 + 1}1 J' D)2.
Hence on n1ultiplication bJ' the positive number - '1 + '1]1 J/[j
(6)
- E 1 + 'I}l vD < '0  1}0 1 //) < '1 + 1]1 J'Jj,
\vhere the numbers
'0 = - El Lr1 + 1}ll/1 D, t}o = '71 Xl - '1.111
satisfJ equation (2). It is apparent fl.om the properties of the
,- 
solutions '1 + 'I}ll1J and -';1 + 1}11 J) that we can have none
of the following cases: 1. o > 0 and '1}0 > 0; 2. o < 0 and 'fJo > 0;
3. Eo < 0 and 1}0 < O. If o > 0 and '70 < 0, we would ha'ge
171 .111 1 ) > 1.r1 and ;1 !/1 > }}1 :r1' whence 'fJ1 l!1i D > YJI E1 .l'i, which
is ill1POssible, since yi D - ).i = - 10 1'he llulnbers '0 and 'fJo
cannot be = 0 since they satisfy equation (2). Thus relation (3)
is true. As a consequence of this relation and of Theorem 104:
we obtain immediately the proof of the last part of Theorenl 106.

DIOPHANTINE EQUATIONS OF THE SECOND DEGREE 203
It remains to prove the second part of our theorenl. Suppose
that tt + l.' 1 / D were a solution of (2) with positive u and l.' which
is not obtainable by formula (4). Then a llaturaillumber nl would
exist 811Ch that
('1 + fl1 1 / D} 2m-1 < 'll + 'I: 1 1) < ('1 + fl1 JI D) 2 m+l .
Multiplying this relation by the positive number
(E 1 - 171 1 / D) 2m = '2m - 172m J I D
we would have
(7)
- '1 + 1}1 l'D < '0 + '1}0 J D < '1 + 1}1 J :'D ,
where the numbers
'0 = II '2m - l" 1}2111 D, t}O = l" E2 m - '"172m
satisfy equation (2). But we just proved that the inequalities (7)
are possible on I)"' for "10 = 0; this implies zt = E2m and v = fl2m.
The numbers '2m, fl2m, however, satisfy equation (1) and not
equation (2). Thus Theorenl 106 is completely proved.
Equation (2) is clearly solvable for D = 2 with the solutions
'1 = fl1 = 1. More generally we have
Tlzeorellz 107. If P i. a }Jri'11le = 1 (mod -1), the DiolJhalltine equation
(8)
,2 _ p'l}2 = - 1
£s oll..able in £'I2feger.f! , and 'I}.
Proof. Let Xl + 1/11'P be the fundamental solution of the
equation
..t,2 - P 1/ 2 = 1.
Then
(9)
xi - 1 = 1)!/.
Here Xl cannot be even, for in this case we should have - 1 == P
(mod 4). If Xl is odd, the numbers Ll'l - 1 aild Xl + 1 have the
greatest common divisor 2. Therefore it. follows from (9) that
.l + 1 = 2 E 2 , Xl + 1 = 2 }) 'YJ2,

204
CHAPTER VI
where E and 1} are natural nUJubers and /11 = 2 ,'I}. By eI11111na-
tion of "CI ,ve et
+ 1 = 2 - l' 1]2.
Since 'J} < 1/1  '\TC cannot have the upper sign. Thus t.he lower
sign ll1Ust be taken, and 'l'heoreIIl 107 is proved.. According to
Theoren1 1 06,  + 'J] Jll is the funda111entul solution of equation (8).
"r e shall sho""F that the ('qnation
(10)
(:2 __ - , , / 2 -- - 1
';, ..-r,-
has no solution. 'rhf' func1nnHntul solution of the equatioJ\
.1'2 - B-1: .11 2 -=- 1
is, as is easily ShO\\FU, 35 -i- (> y' -+ . If equation (10) ,vere sol\"able
and had t.he. fundulllelltal solution l + J}l J / 34- , ''fe woulu. have,
by Theorem lOG,
I) .. "'0).  f .')
iJ = i -r i)-t 'II] 
() = :! EI I}l .
But this systeul of equa.tions bas no intE'g'ral SOl11tiol1s 1 and 'i}l,
and thus equation (10) is not. solvable.
58. The Diophantine equation u 2 - D v 2 == C. - Let [) he a
natural nUlllber \v hich is not a perfect square.. and consider the
Diophantine equation
(1)
u2 _ [),.2 = (',
,vhere (' is an integ'er ;;6 O. l1ppose that the eqnatioli:,; solv-
able, and that II + c J ' J) is a solution of it. If:r -t- !I J ]) is al1.
solution of the equation
(:3)
.2 _ J) y2 = 1,
the number
lit. -+ r J li ) (}: -t- .7/1/])) -= U ,I: -t- 1".'11) - (II !I - 1".1') J J)
is also a solution of (1). This solution is said to be tls8ociafc-d
with the solution It + ,. -vn. The set of all solutions associated
with each other £01"111S a c.la.,',,' q!' .,'ol11tions of (1). B. Theorem
10-1- every class cont.aiJ1s an infinity of solutions.
 .

DIOPIIANTrNE EQrATIOKR OF THE SECOD DEGRRE 205
It is possible to dpcitle ,vhether the t,"Vo given solutions
1£ + f J/]j and Il' + 1:' J 1) belong- to the saTne class or not. III
fact, it is easy to se that the llecessar)'" alld sufficient condition
for these t.\VO solutions to Le associated \vith each oth.er is tha.t
the t,vo uuu1hers
, , 1
It 'It - V l )
(J
aua
, ,
i: 'It - 'It 1:
( J
be int.eg'ers.
If K is the class consisting of the solutiollS 'lii + fi 1 J) i = 1,
i, :). . . .  it is evident that the solutiol1S Hi - I"i 1/]): i = 1. 2, H, . . .,
a Iso contitute a clas:;, which BIa)" be denoted by K. The classes
K and K are said to be (;()lIju.Qaf(', of each other. (onjugate
classes are in g'el1eral distinct, but InaJ SOl11etiules coincide; in
the latter case ,ve speak of 111nln.guO'll8 classes.
AU10ng all the solutions it 2" I" JI [) in a given class K ,ve no,v
choose a solution u* + ,.*' J J) in the following ,vay: Let 1.'* be
the least 110n-negoati\e '\"'alue of v which occurs in K. If K is not
alnbig'uous, then the nUluber 11* is also uniquely uetern1illed; for
the solution -- u* + v. V D belong's to the cOlljug'ate class K. If
K is anIbignous, W get a uniquelJ deterll1ilJed u* by prescribing
that ,,* ? o. The solution lie -t- 1:* Y D c1efinell in this way is said
to be t.he funr/aJiteutal .Ollltiol1 (t' the c'/a,s'.\(.
In the fUlldanlental solution the number Ill* I has the least
value ,vhich is possible for I u I 'v hen II + ,. Y 1) belong's to K.
The case 'll* = 0 can occur only 'v hen the class is ambiguous,
and similarly for the case f* = O.
If C == + 1, clearl)' there is only one class, and then it is
an1biguous.
Suppose no\v that the nuulber (1 in (1) IS positive, and put
(' = ...,.... "T e prove
1'1heore1Jl 10$. Ij' 1(, + L'"J 1) z"
K oj' flu' equation
flu' finllll(}uenfal ,O[.ittiOIi of f1zr (./as.4.:
& .
(3)
u2 - ]),.2 -=- \-.
(inti I)' .1:1 -+ :'/1 1' 1) is fhe jilJalautenlal .()lllfi(Jn (equafiou ()'I
,f" haz'e Ihe iuequlllifif'8

206
CHAPTER YI
(-t)
(5)
o < v < 'Yl · Jtg,
- - }-/2 ('"( 1 + 1)
o < lul < J.r  (XI + 1) N.
Proof. If inequalities (4) and (5) are true for a class K, they
are also true for the conjugate class K . Thus we can suppose
that fl is posit.ive.
It is plain that
(6) li,}'1 - D V III = II Xl - V(ft2 - ]{) (.2"i - 1) > O.
Consider the solution
(It + t' JI ])) (XI -111}''])) = U.1'1 1 - D l'YI + (XI t' - 'UI u) W,
which belongs to the same class as 'll + v JIJ). Since u + t: VD
is the fundamental solution of the class, and since by (6)
U Xl - D t. YI is positive, we must ha.ve
(7)
'll .1'1 - D. 'UI > ll.
From this inequality it follows that
U 2 (XI - 1)2 > D 2 r 2 yi = (u 2 - N) (xi - 1)
or
and finally
Xl - ] > 1 _ .:.,r
Xl + 1 - '1l2
1,2 < !(XI + l)N.
This proves inequality (5); and it is easily seen that (5) im-
plies (4).
Suppose next that the number C in (1) is negative, and put
0= -N. We prove
TheOl.ern 108 a. If u + I' } ' D i. the jll ndarnental sol1.ltion of the
c.Zas8 K tJf the eqllatitJll
(8)
,u,2 - [J 1,.2 = - ")..,.,
and if' XI + YI J I1) it" the JZI12£lanlental solution of eq!l,ation (2),
u.e hat:e the i nequaZities

DIOPHANTINE EQUATIONS OF THE SECOND DEGREE 207
(9)
o < t' < 1/1 · J: }{,
V2(Xl-l)
(10)
r-
o < III I < l' " ('l - 1) .,r.
]:)J'oof If inequalities (9) and (10) are true for a class K, they
are also true for the conjugate class K. Thus \ve can suppose
that u > O.
\Ve clearly have
(Xl tl = (Yi + t) (,,2 -I- S) > !r. u2 ,
or
(11)
Xll" - Yl U > o.
Consider the solution
(It + v J: D ) (£1 - /11 J' D ) = It.rl - D t'!/l + (£ll" - .lll u) J' ]) 
which belongs to the same class as l' + t' J'D. Since II + I: l'D
is the fundamental solution of the class, and since by (11)
Xl t' - 1/1 U is positive, we must have
(12)
Xl V - Yl U  Z:.
From this inequality it follows that
D t,2 (Xl - 1)2 > D .lI 1£2,
or
1 _,T > ,"t'l + 1
+2---'
'It - Xl - 1
and finally
-u 2 <  (Xl - 1) _'T.
This proves inequality (10), and it is easily seen that (10) im-
plies (9).
From Theorems 108 and 108a we deduce at once
T1zeoJ'p))l 109. If D aud ::.r are zatuJ'al 12 lInzuers, and 1:( D is not
a }Jelfect sqlla1'(', the Diophantine equations (3) alld (8) 1za1.:e a
finite nuuzbel' oj' class(Js of 8oluti()J1s. The jiludal1le12 tal solutions

208
CHAPTER VI
(( all the classe.'$ cau lJ( jound {tjre} , It .tinite liuul1)(!r 0.( trials
U?I nzea JlS (Jf tlu ) i IIp(jual i ti(' ill 1'/1 ('OreU u..' 108 a nil loS a,
If' u* + l* J 1J is the jl(ndarnental ,f,'uilltion of' the cla."'s K, u'e
o1Jtlliu all the solutiuns 'll + ,. J/]j oj' K 1J11 tlu) jrJ1il"a
If + r J D = (1(* + 1'* W) (.1.' + y J 1)) ,
lf/u:re :J.: + y 1" D runs tlu.oll!Jh all the .olution.c.: o.,fl'quatioJl (2),
including + 1.
The Diophantine ('qllati()J2 (3), or (8), has no b'olutio12S at all
U'lit>12 -it hlts nu solution sati.ifyiug tlip inl'fJlIlllitie.'\ (-t) and (5),
or (9) and (10), J'('8peetirflll.
A supplement to Theo1'eu1s 108 and 108 a is given by
The(J1'enz 110. If}J is a jJrinu', tluJ DiojJhal1line equation
(13)
.u 2 - [J r 2 = + }J
ha at 1nob't (Jill' ;.;ullltiu/l II + l" J ]J i11 u:hil./i 11 and L' satis.f'u
the incqualitieb' (4) and to), or (9) llnd (10) retl)e(.tit'el!l, pro-
tided ,, > o.
If equati(Jil (13) is l,:olrable. it hU8 one or tlfO classes of solu-
tion,'.., accordiu{/ al!! the prilne jJ dit:id{"8  J) or not.
Proof: Suppose that 1I -+ l: J [) and 111 + rl J 1) are two solu-
tions of (13) which satisfy the conditions in t.he first part of
Theorem 110. Thus the numbers it, t., III and 1"1 are non-negative.
Eliminating D between the equations
(14)
212- J),.2=  p, ,;-I)rf= + p.
we get
.1(2 I'I - Il ,.2 = -I- )J (r - z,2),
Thus
(15)
U "1 .::: .:t 1/1 1.' (mod]))
for the upper or for the lO'"fer sign.
Further, on multiplying toget]ler equatiollS (1-1-) 111enlber bJ
1nember we have
( [ 'oJ
U 111 T ) rrd-
I - } O) 
) lit 1'1 + ill r M - - JJM.

DIOPHANTINE EQIT.ATI0NS OF THE SECOND DEGREE 209
In the equation
(16) ( II lIt +p D t' /"1 ) 2 _ JJ ( l( /.'} : tI},V ) 2 = 1
let us choose the sign so that the cOllg'ruence (15) is satisfied.
Then the two squares on the left-hand side in (16) are integers.
If 'li 1'1 + li1l"  0, we conclude froln (16) that
( 17)
IUl'l + 1(1 r I > 'YIP.
On the other hand, applying" inequalities (-l) and (5), or (9) and
(10), respective I)". "\ve obtain
I It 1'1 + lI1 l41 < .111 J) ,
which is contrary to (17). The relnainiug' case is that 1I1'1 + 1tl t' = 0,
,vhich is obviouslJ possible onl)" f01" 1I = zt1 and 1" = 1"1- Thus the
first part of Theore1u 110 is proved,
Consequently, tbere _are at 1l10st _two classes of solutions.
Suppose that 11 + 1: J 1) anll It - t" J 1J arc t\'"O solutions which
satisfy inequalities (-t) anu to), or (9) and (10), l"espectiyeIJ' l'hese
solutions are associated if and on I)"" if }J di vides the two num-
bers  It."" and u 2 + f) l'2 =  I) r 2 + i). Since 1.' cannot be di \yisible
by p, the DU1ubers :! It and  Dare divisible boY 1). But, if  [)
is divisible bJ Ji, so is  u. Thus, the ]1ecessur. aud sufficient
condition for 11 + l" J/ ]J and II - ,. J IJ to belong to the same
class is that  1) be a n1ultiple of p. This proves the second
part of the theoreln
EXllu1J)le 1. Let the equation be
(18)
,,2 - :? r 2 = 11 B.
The fundanlental solution of the equation
(19)
.1,2 - 2 .'/2 = 1
is 3 + 2 J 2. The follo\ving solutions of (1)) satisfy inequalities
(-1) and (5):
11 + J' -, - 11 + 1 :!, 13 + u J :!" - 13 + 5 JI.
I t is eas. to show that thse nurn bel's are all funda1nelltnl solu-
tions in different classes. Thus the l1ull1ber of classes is four.
14 - 5Hi6iO Tryg'c ;\"'Q!Jcll

210
CHA.PTER YI
ELcaJujJle 2. The equation
(20)
'lt 2 - 6 r 2 = - 2D
belongs to the secolld category. The fuudauleutal solution of the
equation
2 ,,') 1
x - Ij !Jw 
is 5 + 2 J/U. Fronl ine<lualities (9) and (IO) ,ye g-et the funda-
nlelltal solutions
1 r-:: C') 1 ,
f) + 3 16, - U + t:J fj.
ExanljJllJ 3. The equation
(2 J)
112 - () z.2 = - 2
has onl" one £undanlental solution
 + J 6.
For the solutions 2 + l' u and -- i + JI B belong to the sallIe
aOlbiguous class.
ExaJnple -J. Let the equation be
(22)
u 2 - 82 r 2 = 3.
The fundamel1taI solutio!l of the equation
u. 2 - 82 lJ2 = 1
is Hi3 + 18 :? From (-1-) we ge the inequnlit.," t <  V : <:).
But equation (22) has no solutio!1 .ith r -= 1, 2, 3 or --1-. rrhus it
has no solutions at nIl.
\Ve fil1ish the theol"J of etJuation (1) ,vit.ll the proofs of the
£ollo,,"ing special theorelns:
Tlu:oreJll 111. If)J i. a p1"i1ne == 1 or =' - 1 (IlIod 8), there eJ'i.t
tu:o natural ultJllbtJ' 'u liud r ."'liclt that
(23)
') ) oJ
It'" -  1:'" =- P
a1ul
(-! )
It < V 21J , zo < ) 1 i 1) .

DI0pn_\TIE EQl.'ATIOS OF THE SECOND DEGREE 211
Further.. there e,:tisf t,cu uatural JlUJi11Jrr,,;; II llnd ,. such that
u)
u2 - :! ,.2 =-:. __ ))
and
(2(i)
If < } }J, ( < J 1).
1"hroreln 11 - If )J ;,0.; II lJl'iiitC  1 (ll1od 1 ). there eJ.'ist t".u
nat Ii I'll I n 11 Hi ber,. " {( ucl r such that
(7)
C) .) C)
II"" - tJ 1:'" = }J
(lud
(28)
It < V  }J.. 1 < VI); }'-
I}' p 'l, a. pfillif == -1 (nlod 12). tlzer{J (J:J.,i....,t l,eu natllJ'al nuuz-
be J"8 II a It d l" ,,,. tl (' Ii t Ii a t
(:?9 )
.) ,).)
If'" -- i.J ,.... = - jJ
aud
(30)
II < Vi }J , I' < 1/ 1 1 -
}Jruo..( (f' Theo1'('JII 111, FOj"})::= + 1 (luod H) the cong'I'nence
,,2 === 2 tU10U p)
is sol¥able. If f! is a sointiou ,ve have b" l'hue's theorenl
. ,.t' ( _1 )
.2 :::::: T mou jJ .
!I
1-
""here J: ana !/ are natnral llnm lJers <}' 1). Ht'uce
C) ) C)
,J'- - :.. y.... == - jJ.
From this equation it follows that
# ) " ( .')
(,1; + 2 y ... -  ,(' + U)' == p-
Therefore equatiol1S (23) and (i5) are hoth sol \"a LIe. If we obser¥e
that the equation
.1,2 - :? //2 -= 1
has the fun<lamental solution 3 + :? 112, and if ,ve apply Theo-
rents 108 and l08a" ""E' find that eqnations (23) and (:?f») haye
solutions which satisfy inequalities (2-1-) and (G). (. E. D.

') 1 "
MI aJ
CHAPTER VI
PrfJof ({f Theoreln 11,-!. For}1 = + 1 (mod 12) the congruence
2 2 = 3 (nlo<.1}J)
is solvable, If. is n solution, we have by Thue's theorem
. = + .l' (mod})),
!I
,,,,here .I) and '!I arc natura] numbe}'s < Jip, Hence
X.2 - 3.'/2 = - I1Z Jj.
,,"here 112 = 1 01':1. Fronl the equation
.{.2 _ 3 !/2 = - 21),
\vhere {. and !J 1l1USt )Je odd, it follows that
( x + 31/ ) 2 3 ( X + .1/ ) 2 _
-) -  - - JI,
- -
,,"here i (x + 3.11) and -. (.t. + y) aloe illtegerR. Therefore e(luation
(:!7) is solvable for }J == 1 (JllOU 12). and equation lB) is solvable
fOl' P = - 1 (nl0d 12). If \ve observe that the equation
J':2 - 3 1/ 2 = 1
.
has t.he fundamental solution :! + 11 3 , and if \\'e apply Theo-
rems 108 and 108 a \"re find that equations (27) and (29) have
suInt-ions ,vhicb satisf)9 iue<jualities (2R) anu (aO), Q. E. D.
59. Lattice points on conics. - The )"esults obtained in Sec-
tions 54, 56. 57 and 58 Inay be interpreted RR theorellls on the
distribution of lattice points on special conics. "7 e shall now
examine the general case; let us consider the conic represented
b.v the equation
(1)
.f'lx, y) = 0,
,vhere f{x, y) is an integral polyno111ial of the second degree in
.}: and JI (Cartesial1 coordinates).
J £ equation (1) l-epl'eSents a Jlllrabula, it can be written in
the £Ol"Jl1

DIOPHANTINE EQUATIONS OF THE SECOND DEGREE 213
(2)
(a .{' + b!/)2 + C x -1- d..11 + r = 0,
where a, b, c, d, e are integ'ers such that L1 = ad - be -F o. If we
put a,,+b!I=U, ,ve get cx+d!l+l'=-U 2 . Hence
(3)
1
:r = J (b U,2 + tl Zl + b ())
ancl
()
1
!I = - J (a,,2 -i- (0 II -r a (.) .
Thus, there are lattice points on the parabola (2) if and onIJ
if the following t".o congruences are satisfied for the same value
of'll:
(5)
a 1(2 + C I( + a (' == 0 (mod I J I)
and
(6)
b 1(2 + d II + be = 0 (mod I LlI)
If these congruences are satisfied by the iuteg'er lll' ve get the
corresponding values of :l'" and y by }1utting into (3) and (4)
U = UI + J · f, where t is any integ'er. Thus "e obtain all the
lattice points on the parabola () by putting' t = 0, + 1, + 2, etc.,
into a finite number (== 1") of forlllulae of the type
(7)
,, = gz (f), !I = hi (f),
(i = 1, 2, . . ., 1"),
where {Ji (f) and hi (f) are integral I)01ynomials of the first or sec-
ond degree; at least one of the polynomials is of the f\econd
degree. The number r is the l1unlber of incongruent solutions
of the congruence s.,"steul (5), (0). Thus, there are either no lattice
points or infinitely nlany lattice })oints on a parabola.
J;;Xan1ple 1. The lattice points on the parabola
,,'l02 - 2 X!J + y2 - X .- 2 y = 0
are clearly determined by the two systems of formulae
J.... = 3 f2 + 2 t, Y = 3 t 2 - t
and
l: = 3 t 2 + 4: t + 1, 1/ = 3 t 2 + f.

214 CllAFTER VI
!:,.', t:.a 111j1!t' . rrhere are no lattice points on the parabola
2.2: 2 -3:1/-1 =0,
since the congruence
:! :J2 == 1 (n10d 3)
has no solution.
On an elli}J,,e (or circle) there are uut a finite nUIuher of lat-
tice points, ,,"hich Inay easn.v be deterIl1illecl by trial.
Consider next. 111e case of a h.'l}J('r!Jola. 'The problem aris{ls ho,v
to decide whether a g'i\en hyperbola passc-s throl1g'h any lattic
point 01' not. .A nother proLlenl is to find a. Iuet.hod for deter-
lnining- all the lattice points 011 a givell hYIJcrhola of ,,,hich \Vc
already kno\y that it pa.sses through lattice points. It is easily
seen that the eqllation of the hyperboJa can be tral1SfOrnled by
nleans of linear tra.nsforJ11atiolls \1fith integl"al coefficients into
an e(jUatiol1 of the type
(8)
.u 2 _ J),.2 == + ]..T,
where 1) and .\ are natural nUluber8. Our probJeln then reduce
to the problenl of finding all the integral solutions 'it and r of
(8) which satisfy certain congruences
(9)
'it == ll (DIOd 6), v = l' (Inod D).
where It, I and d are integers which depend on the coefficients
of the original equation (1).
If ]J is a perfect square, equation (8) has only a finite number
of SOlllt.ions, which are easily det.ernlined. In Sections 56 and 57
we have already treated the case in which J) is not a perfect
square. III this case there are either 110 solutions or infinitely
nlany. If equation (8) is solvable, its cOlnplet.e solution is gi¥en
by Ineans of the fundanlenta.l solutions (Theorem lO). Finally
we ha,e to deternline \vhich of its solut.ions satisfy cOl1g'ruence
conditions (9).
EraJllple H. Let t.he equation of the hyperbola he
( 10) U .1:2 - 1-!-..{' /1 + 7 .'/2 = - 1.

DIOPHANTINE EQrATIONS OF THE SECOND DEGREE 215
By putting 11 = 5 x - 7.'1 and r =,11, we get
(11 )
11 2 - 1  f2 = - 5
,
where the solutiollS II and " must satisfy the additional condition
(12) 'u = 5x-7!1 = -2l (lJ1od5).
Equation (11) has the fUl1dalnental solutions + 3 + J '1-!- ; its cOln-
plete solution is g'i\"en bJ the forl11ulae
(13)
.u + ,,1 1-1: = + (3 + 1 /14) (15 1- .t l'l-!-)n
and
(1)
u f- i" Vl4 = -r- ( - i1 + v14) (15 + 4 v14)n,
where n = 0, + 1, + , + 3, etc. If u is even and = 2 111, it fol-
lows from (13) that
Ii == ( - l)m · 3 (Jl10d 5), i" = (- l)m (mod 5),
and froln (1....) that
u == (-lpn.:! (TIlocl f)). 1: == (-I)m (Inod 5).
lf 12 is odd and = 2 Jll + 1, it follo\\"s from (13) that
u == (- I)W (ID 0 tl f). l' = ( - 1) 1/1 · 2 ( mod 5),
and frol11 (1-1) that
u = (- l)nz (mod 5), r = (- 1)"'. 3 (mod 5).
Thus ,ve see that condition t 12) is satisfied by taking forlIlula.
(13) and not h)- taking foro1ula (1....). COnSe(lUently we obtain
the \vhole set of solution of (10) b" the relations
Lt ==.  (u. + 7 r), .11 = l",
D
\vhere II and fare ueterlnined by fornlula (13).
In the theory of binary quadratic forms. i. e., homogel1eous
polynoJnials of the fOrB} a ..{..2 + /; oX.'I + c .11 2 , the questions examined
in Sections 54, 55.. 56, 57 and 58 are more thoroughly .discussed.

216
(;HAPTER Y I
60. Rational points in the plane and on conics. - Let K be a
gi veIl field. In this section any nUll1ber belongillg to K is said
to be a ratioual nll1Jlber. Any point (C, J/) in tIle plane with
rationa1 coordinates (Cartesial1) :t and y is said to be a ratio'nal
poiut or a point in K.
The straight line
a .)'1 + b.ll + r = 0
is cLlled rational if there exists a positive number OJ such that
the nUHlbers a OJ, b OJ and C OJ are aU rational.
N o,v let
(1) F'I(Ll', /1) = a.r 2 + b:t!1 + cy2 + d.t" + ell + f= 0
be the equation of a conic with ratioual coefficients. Suppose
that (, 1}) is a rational point on this conic, and cut the conic
with the straight line
.'I - 1] = f (L - ),
where t is a rat.ional nUll1ber. Then the second point of inter-
section is also rational, and it has the coordinates
j x= -d-aE-bl}-( 2c 'fJ+ e )t+cEt 2
, a+bt+ct 2
1 '1 = a'fJ - (2 a E + d) t - (b E + C l} + e) t 2 .
, a + b t + C t 2
(2)
'rhus we obtain all the rational points on tIle conic (1) whell
in formulae (2) the nnnlber t runs through allrationaluumbers;
further if c  O. the value t = 00 gives the rat.ional point x = E,
!I = -  (7)  + r l} + d. There are infinitely many rational points
on the curve if there is one. If the conic is a parabola, there
are al,vays rational points on it; for a parabola with rational
coefficients Inay be t.ransformed by linear transforll1ations with
rational coefficients into the forlIl .'1 2 =.:1'. ""'hen the conic is
an ellipse or a hyperbola, we have no general Dlethod of deciding
whether or not tIle curve has rational points belonging to a given
fielJ K. ()n ly for special fields (as for instance the ordinary

DI0PIIANTIE EQUATI0S OF THE ECOND DEGREE 217
rational field K (1) are we able to decide this question. (See the
following section.)
Consider, in particular, the unit circle having the equation
(3)
x 2 + 1/ 2 - 1 = O.
A rational point on this curve is (-- 1 0). Putting  = - 1 and
1} = 0 in (2), we obtaill the ,vhole set of rational points on the
curve by the formulae
(4)
1 -- [2 2 t
,t' = 1 -+ t 2 ' H = 1+(2 .
On the circulllference of the circle
(5)
{};2 + .'/2 = 3
there are cleal"l)" no rationa.l points belong"ing to the ordinary
rational fielJ K (1). For, since the number - 1 is a quadratic
non-residue of 3, the equation a 2 -f- /;2 = 3 ("2 is satisfied OIlly by
a = b = c" = o. On the other hanu, equatioll (5) is solvable in
everyone of the quadratic fields K(JI3). K (V2) and K( V - 1); in
the first field a solution is ,{' = V3, y = 0, in the second a solu-
tion is x = }, !I = 1, und in the third a solution is x  2,
y = 1" - 1.
Just as in analytic geonletry, it is convenient to pass over to
hO'1J1o!Jeueolf, coordinates and operate with a nlore g"eneral concept
of point. Let .x, 1"' and Z be tIle homogeneous coordinates for
a })oint in the plane. The point (..., Jp", Z) is said to be a rational
point in K if there exists a positive number (jJ such that the
t.hree numbers X OJ, )''If and Z co belong to K. The points (X Y,O)
are the points at infinity in the (Ll', y)-plane.
Let 1.' be the sanIe pol\"uonlial as in (1), ana put
"/2 1 ' ( X } ) ( " ( ,.. ''P' Z)
/.J  Z ' X = :r ... \., .I., '-'.
Then consider the hOll1ogeneous equation of the second degree
in X. 17' and Z:
(6)
(; (X, J" Z) = o.

218
CHAPTER VI
and suppose that its coefficients are rational nlunbers in the
ordinary rational field K (1). To eyery solution of equation (6)
in integers X., }r and X tllcre corresponds a solution of e(luation
(1) in rational ntllubers ,I' = , !/ = ;; . if Z  O. Conversel.,', if
x, /1 is a so1utioll in rational uunlbers of (1), and if J2 is a natural
number such tha.t the nUlubers Ii ,)' and n .'1 are integers equation
. 6) has the integral solution X = n:.r Jr = n.'l, X -= n.
61. TIle Diophantine equation a x 2 + b y2 + C Z2 = O. - Suppose
that the field K in ection 60 is the ordinary rational fit"\]a K (1),
anu consider once Inore the conic represented by equation (1).
\Ve shall sho,v how it can hE) uecitled whet.her or not there are
rational points on the curve \vhen this is an ellipse or 11"perbola.
By 111eanS of linear transforluations with rational coefficipnts
the equation of the conic. ill this case, can be tl.ansfortned into
the homogeneous forIll
(1)
2 . 1 '>. C) 0
a ,1..' T ) /I'" T (... = .
where the coefficients tl, hand c are integers  O. Then our
problenl is to find the conditions for the solvability jn integ'ers
x, '!J and ,? (not all zero) of tbe Diophantine equation (1). A lleces-
sary condition is cleal.ly that the coefficients a, band c be not
all positive and 110t all negative. ",Ve can suppose that the
greatest C0l1l1nOn divisor of the uunlLel.s a, band c is = 1, and
that tllese n II tn bers are nIl square-free. vVe can further precri be
that
(a. lJ) = (a, c) = (b, () == 1.
For, if we put (a, tJ) = d, (a, (') = (' and (b, c) == ./: '\ve have (d, e)
= (d. f) = (e. j') = 1.. and thel.efore it follows from (1) that :' IS
divisible by d, II by (' and , by f. Then.. by putting'
a = deal_ !J = d fh I , C == (',(('1' ,) == ..f. r 1. ,II = ('!11,  = d ?t..
we obtain the etluation
.) I '), 1 " )
a J f'... _,_ J (,,,... -1- ( . I ... ==- (
1 · l' ] ,1 IlL '-1 '
where
(all I;] e) = (al.f. Cl d) = (hI e, ('1 d) -= 1,

DIOPHANTINE EQ{TATIOS OF TIlE SECON"D DEGREE 219
If the integ"ers x =. .'I = '1], ? =, ,vhich are not all zero,
satisfy equation (1) we say that [, 1],:] is a solution of (1).
If the nun1bers . 1] and C have no COlllnon divisor> 1, [, 'f}. ,]
is said to be a }Jrojj(Jj" solution. Obviously, it is sufficient to take
into consideration ouly the proper solutions of (1). If the llunl-
bers a, u and c are s(luare-free and if [.r,!I. z] is a proper solu-
tion, we have (J\ /1) = (x: z) = C'/ z.) = 1. For, if x and.'1 had the
conlDIOll prill1e factor }J, it 'Would follow fronl (1) that c Z2 was
divisible by })2; but .c cannot be di,isible by jl, since the greatest
C01l111l0n di\'risor of J". /1 and z is 1; and (: cannot be divisible by
))2, being a square-free n ulllber.
Leg'endre proposed and proved the follo,ving criterioll for the
sol va bili tJ' of equa tiOD (1).
Throre1Jl l1H. Lrt a, b and fa ue three infpgers ,,.1zieh sati./!1 the
follou:ing condition': 1'lu'!1 arc all ;;6 0, not all })ositire and
lIof all ne,f/Cltire. l''1UJll aft all IN}llaJ'c-ji.()e, aud (a. u) = (a, c)
= (u, c) = 1. l1hen the lu'cessar.1l alld 8uj)icir'lIf conditions fDI.
the sulz:abilit.'/ o.f the lJio}J!iautilie efJ.uafio/i t 1) in inll'(Jel'c" x, p. .?'
nut all = 0:, are' that - be be a fjuadratic re.....idue o.f a, that
- a c U(' a quadratic. re.,.iduc oj' II and that - a lJ ur a quad-
rot i c rr.i d 1{(' oj' c.
])roq/: The three conditions are clearly necessary for the sol-
vahility. 'Ve shall sho,v that the)" are also sufficient.
There is no 10s8 of generality in supposing
(:?)
la) < lbl < lrl.
Then we have
(3)
I a u I < I a c I < I U ("a I.
TIle nU111ber 1= la(') is said to be the il1de.:c of equation (1).
If we suppose 1 = I a r I = 1, we obtain frOlll (3) I a 11 I = 1, and
thel"efore I a I = I u I = I r I = 1. Thus we see that only tIle equation
:r. 2 + .'/2 _ .:-2 = 0
and the equations which are obtained £ronl it by permuting x,
.lJ and z. have an index == 1. Theorenl 113 is true for all equa-
tions with an index  1  Suppose now that t.he theorem is

220
CHAPTER Y I
true for all equatiol1S with an index < I. \Ve shall then l1rOye
that it is also true for all equatioI1S witll th index = 1.
Suppose that equation (1) lIas the index I > 2. If I b I = 1(,1.
we would have I b I = I (I = 1 and therefore also I a I = 1 and
1 = 1. Thus we bav'e
and
I {( I < I b I < I c I.
la b I < I (( (.1 -= I < I b (" I.
If - (l b is a quauratic residue of f, there exist t,yO int<,gers Q
und ,. such t,ba t
(4)
1I}"2 t- h = c Q
and I f I < . I f. I H enc e
I a I 1'2 + I b I lJ
(0) 1(11 < ---17,- < {lof. l + (" <1 1 f-l<l.
If (J = 0, \ve ha Vp u = - (f ]'2, alld since b is square-free aJlt1
(a, b) = 1, it follows that lJ = - (1 = + 1 and ,. = + 1. TheIl equa-
tion (1) has the solution [1, 1, 0], and the theoreUl is proved.
Suppose next t.hat (a!  0, and let 1 be the greatest commOI1
divisor of the nUlubers a ,.2, band C (I. Then b)? (4) we have
.A =-= (a 1'2, h. c (J)  (a 1. 2 , ,)) = (a,,2, t Q)  (b, (. Q).
Since .A is a divisor of b, we have (...4, a) = (...1. c) .== 1. Thus A
is a divisor of both 1'2 and Q. Since b is square-free so is _.1.
Thus _4. is a divisor of r, and ,,,,e can put
(6)
r = ...1 a. b = ..J (1. (J = ,A. fJ. = .A.O,,2,
wher{\ ("1 is a s(Juare-frec nUlnher. It follo"s frolD (-I.) that.
(1 _ {2 f L 'J. -I . 1 (J = c.J-l 0,,2
and
(7)
(( .1 (L 2 -- rJ = (. C,,2,
where it is clear that
(a61 u. 2 .J3't = (a61 a 2 . r. (",2) == (P, t (fj,2j = 1.
Putting' 1I P = B, \\.t' g'ct _J I: = a b.

DIOPIlATIE EQrATI01\S OF TIlE ECOD DEGREE 221
Now it IS possible to sho\v that th(1 e(jUatioll
(8)
...1 Jy2 -! B }' 2 -1- (1 Z2 = 0
has the follo\ving properties: 1. rrhe l1uuILcrs ....1, B alid (} are
all  O. They are not all positi\Te and not. alllleg'ative. . They
are square-free. 3. 'Ve have (..I. B) = (...1, C') = (B, C') = 1. -t.. The
nunibel" - B (1 is a quadratic residue of A, th nUJuber - .....1 t
is a quadratic residue of B, and the llUJllber - .A B is a quadratic
resid ue of ('.
P/,()o.l: "vY' e already knO\\F that .&1 and (' are square-free. It
follo\ys fronl ....1 B = al) that (1, B) = 1 and that B is S(luare-£ree.
'urther {(\ a ...1 13) = ((t. ....1 B) = 1; hence (A., C) =- (B, (') = 1.
If (f I) = ...J B is negative. it. is elear that .J, Band (' are not
all positive or all neg'ati\e. If au is positi¥e, the nUlnbers ac
Ilnd /J (, by the hypotheses 011 ii, b una (, J11ust be negative, and
then it follo\,'g £rOllJ (-1-). ,vhell \Ve lllultipl)T by t, that
t) I C) t) 1 ( " t)
(/ C }.... + ) (" == ('- (l) -== (.-.. ,'''' .
Hence ...1 (' is negative.
Fu rther, it follows fr0111 \ 7) tl1a t - a .:1 ,3 -= - o£,t B is tl (J u adratic
resillue of ('. Fron1 the Ulle equutioll we see that f3 (" (" is a
quadratic residue of 64.. Now - {( c is  quadratic residue of b
and therefol'(- also of .il w 11ieh is a cli visor of fJ. Hence the
l)1'o(luct
( - a c) (fJ c G') = - II J1 (.2 () = - Ii r l,2
is a (juudrntic r<:-sidue of ..1.
Fronl (7) it also follows that a....4 c. (' i a lluadratic residue of p.
N O\V - a (0 i:; a quadratic residue of ,) and therefore also of p
,vhic11 is a divisor of ll. Hel1ce (- aC)({l...-!l (.1) = _1l2{"2...!(} is a
quadratic residue of p, and thus --...1 (' is a (luatlrat.ic residue
of /3.
Finally it follo".s frolli t7) that fJ l" () is a quadratic residue of a.
Therefore, since - IJ to is a lluauratic 1.esidue of a.. t.he In'oc1uet
(- h t-) (fJ e C) = -- b t3 ("2 (' = - _1 C (fJ C)2
IS al:;o. Thus -...1 (' i:; u quadl'atic residue of tI, anc1 the con-
gruence x 2 === - ....1 (' (lllOd a) i sol,at.le. 'Ve have just proyed

222
CHAPTER YI
that - 44 C is a cllutdratic residue of p. Therefore the congruence
;};2 == - 1 C llllOd fJ) is also sol\9uble. Since (a, 13) = 1 and {{ t3 = B,
it follo\\Ts froln this that the congTUel1Ce x 2 == - ..-1 (} (lllod B) is
solvable. Hence - 061 (} is a quadratic residue of B. Thus we
have proved that the coefficients _1, Band (} in equation (8)
have all the four properties indicated above.
Obviousl we ha\e the InC(IUalities
I :1 B 1 == I a II I < I. 1.1 (' I < l.tl (' I 1'2 = I (J I < I.
T11us equation (8) bas an index less than J. By hJpotbesis it
therefore has a 8olutioll [X, }-, Z]. ,,,hich Ina)'" be supposed to
he proper.
If "\ye l)U t
(9) .l = .1 a X - P }', /I == X + a a }',  = (Y y Z,
,ve g'et. by (4)) and. (7). since B = a 13:
a J.2 - 7) //2 -1- C' .2 == C {,f )J 2 (_1..."\2 + B }7"2 + () Z2) = O.
lIence:- [lor, 11. z] is a solution of (1). "T e cannot have oX = !I =  = o.
For thell ,,,,e should ha\"e froD1 (B) Z = 0, and by elilnination of X
o = (p + A II a 2 ) 1"" = c (, T ')12 } r = 0,
and since the nuuILers c., (1 and yare r 0, }7" = 0 and finally
X = 0, Thus Theoreln 113 is l)l"oed by induction, 'fhe proof
alRo provides u. Dletl10d for deter1l1inil1g' a solutiol1 of C(luation (1 )
,vhen it is sol\Tahle.
By modifying slig'htly the proof of Theorel11 113 it is possible
to obtain the e(luivulel1t }'esult:
l'heore1n 11.J a. Let a, b (l1ul (' be th1'ee integers 8l1ch that a.b(: is
,';;fjllaJ"e-frte- 'l'lu J ll tlte ]JiU1J!tautinp ('(juati(JlI (1) i.'t solcable ill
iutegPJ"s ,", 1I, .? lIot all = () ii' and onl.lI 1:( the follun'illY ..lour
cfJuditj(ju,'t are sati(fit.c7: - lJ C i8 ({ quadi'lltif' rc'idue of a,
- a (' is a quadratic J"o...iclur ())" h - alJ ;,It( a f.J.illldrlltil' fesidl{("J
(if r, aud lite C(J110J"I I CUCC
{( .r 2 + h !,2 -:- r Z2 == 0 (lHOU 8)
is 'o/'llble ZJi iuteg«rs J', '!J, ? 1I0t all ez:en.

DIOPHANTINE EQrATIONS OF THE SECOD DEGREE 223
Here the condition in Theorel11 113: .inot all the coefficients a,
b, c have the salliC sigu," has beeu replaced by a congruence
condition 11louulo 8.
It is ev'idellt that the cougrucuce
:/.,,2 -!_ .'/2 r ! 2  0 (UIOtl 8)
has 110 illteg-ra1 solutions x, !I. .? not all even. Hence Theorem
113 a holus for all equations (1) \vhich have an index = 1.
Thus to prove TheoreIu 113 a it suffice to add the following
result to the proof of Theoretu 113:
I..lP1J2rna. I..l the (.ul1!lflll'lil'e
( 10)
&) b &) .) - ( ,1 L) )
a 1'*- + .'I*- + e,'" -=- () nlOu c
is ""u7rablc in intp.'ie}'.. ::, !I, . u()f all err u. [Ilf cougruence
.J. X:! - 1: }' 2 ..:.. (! X 2 = 0 (UIO<.l 8)
(11 )
i,,' al."o sulfuvlf ill in[('!/lfS ....\. }r, X u()[ all el,'CIl.
1)1'00}: Froln the proof of Theoreul 11 B ,v(\ have the following
relat.ions bet".een the coefticieuts £l. I). c and .J, B, C:
(1 )
t 13)
,) = ...1 fJ. a f3 = B, (t u = ;1 B.
1 .., .j ['" &)
ll... (1- -.- I) = t- t; ,.....
Furtherlnore. it is easy to see t ha.t the identity
(1-1-) (- (' j,2 ta ,{'2 + ,) !/l. .- C Z 2.) -= _1 {a a C -+ jJ y)2 + 1J (1/..: 1 a - .'1")2
-! (Y (c y ?)2
holtls for all r, 'Y aUlI z.
'V" e h3. ve to uisting-nish foul' cases.
First case. II aua V are odd. c is e\""el1.
TheIl the solutions .1' and y of congrnence lIO) are odd. Hence
(If»
a  IJ == - C,2 (lDOd 8).
1 t follo,,s from (12) ana (13) that ...1, B, CL :lud j3 are odd. Hence,
Inultiplying hoth siues in (13) by a.. ,,-e ha\-e
(1o)
A + B === a c: (' J,2 (mod 8).

224.
CHAPTER Yi
Multiplyillg both sides in (15) by a..1, we ha\"'e
(17) .J + B =: - ac1?2 (lDOd 8).
If y is eyen, ,ve g-et fronl (1 u) ...1 + 11 == 0 (IUOU 8). Hence con.
gruence (11) has the solution X = }" = 1, Z = o. If y is odd,
,ve get fron"! (10) and (17): O+.4.z 2 == O tll1ocl4). Since (' i square-
free, tllis illlplies that C: and z are odd. lIeneo
c (('1 j}2 + .J .2) =-: :? t (' + 1) == () (luod 8),
If C + ..4 == 0 (ulod ), congruence (11) has the solution X = 1,
}.. = 0, Z = 1. If (' - _1 = .+ (Iliod 8). congruence t 11) ha the
solution ....\': = 1, } - =   Z = 1.
S(:(.fJnd ca,e- b anu care oull, II is ven
Then the solutiolls y aua z of cougTuence (10) are odd. It
follows froln (12) and (13) that (J aut! yare odd. 'fbu8 the
llunlber c y z is odd. Hence if ,.f'. .'1,  is u solution of (10), \ve
obtain a solution ..a\, },., Z of (11) by nleans of fornlula (1'+).
J'hird rase. ct and c are odd, b is even.
Then .:t' and " in cong'1'uence (10) are odu. If....1 is e,en_ the
nUlnber y 1 a _H..1.: is odd, aud ".e obtain L solution X, ).., Z of
(11) bJ' lueans of forulula (14). If ,.1 is oad, t3 is even, and it
follows fl'Oln (13) that y is odd. TheI1 thp nUluLt!r (. y  is odd,
and ,,?e obtain a sol utiuu of (11) l,y IneaDS uf forl11 uln (1-t).
FOil ,.tll ra,'}"e. a, b find c are all oad.
Then 1, Band pare oud. If 11 is even, Lr i oc1u aud 11..:1 a - 
is odd. If .1.' is even, !I is odd aI1d Y.11 a - x is oud. If 2 is even,
and if a is even, then .11...4 a - x is odd. In these three cases we
obtain as above a solution of (11).
FinallJ suppose z e\"en and (/. odd. TheIl, 11lultiplying both
sides of (13) by a, ve ha,e
( 18)
Franl (10) \ve have
_1 + B == a c (.,,2 (1110 t1 8 t
a + b = - (' : 2 (1110 tl  L
anu l11ultiplying L? .:1 a_
..1 + 1J == - (I (. .A .2 (11l0t1 ).

DIOPHANTINE EQLATIOS OF THE SECOND DEGREE 225
Hence .A. + IJ is divisible by 4. If...-1 + B = () (mod 8), congruence
(11) has the solution ..Y. = l' = 1, Z = o. If .-1 + B  -1 (mod 8),
it follows frOtH (18) that (J is odd. In this case congruence (11)
Juts the solution X = 1- = 1, Z = 2.
Hence our proof of the lelllDla and of Theorem 113 a is
conI plete.
A 1110l'e elegant forll1ulation of the result i given by
l'hf()I'('nl 11.j iJ. Let a, b a Jld c be three integei'8 sitch that a b c is
8f}lCa}"(:-fr(le. Then the jJiopltanliue equation (1) is solrable in
jHtegr1"S :1:, .II, , I/of all = 0 if and un!.ll if thr: cOJigruence
(1 D)
a .:1.. 2 + b y2 + C Z2 == U (1110tl ....,")
i. 8011:a1Jle J1r all integral 11loduli }."r iu iute[Je1"8 x, !/, Z sllch
that (or, y, ., 1\T) = 1.
Pr()oj: [t is evident that the conditioll is necessary. Suppose
next that congruence (lH) is solvable for all....\..,.. Letp be a prime
factor of c. Put _\T = ))2, aud consider a solution .I:, !I, ,? of (19).
If y is divisible bJ )). it follo\"fs from (19) that x is also since
(a, lJ) = 1. Since (, is s(luare-free this ill1plies that z is divisible
by]). But this is ill1possiblc since (x, H, z, p) = 1. Thus]1 cannot
be di¥isible by 1), and silnilarly we prove that x is not divisible
by lJ. Therefore it follows from the congruence a a.. 2 + b!J2 = 0
(mod ]J) that - (f 1J is a qULdratic residue of 1). lIence - a iJ is a
quadratic resiuue of c. SinliJarlJ we finu that - a c is a quadratic
residue of b, and that - b c is a quadratic residue of a. Further-
more.. by l1ypothesis congruence (19) is solvable for T = 8. Thus
all the conditions of Theorem 113 a are satisfied and equation (1)
is solvable.
Suppose that [Xl' Yl, .2'1] is a proper solution of (1). Applying
the results in Section 60 we find that the ,\"llole set of proper
solutions of (1) is given b. the forluulae
(20)
f J .c = - /I 1'1 /(2 - :? lJ.l1t If t' + lJ ,)'1 ,.2,
I J y =- II .'11 11 2 -  {f ,1"1 It f - h .Ill t 2,
+ .d.? =- (( ':'1 u 2 -1- lJ;1 ,.2 
1 G - ,j I t.Hi;1J '1'r!'tJt' .Yay"ll

226
CHAPTER \"1
where 'Il and t. are relat.ively pl'ill1e integers, and where L1 denotes
the greatest conlnlon diisor of t.he three right-hand sides.
If the equation is
(21)
;l,2 + .'1 2 - .,2 = 0,
,ve can take Xl = - 1, Yl = 0, and 2'1 = 1. Then fornlulae (O) give
L1 x = u 2 - Z.2 L11 1 = 2 li l' + LI z = 112 + lo2 ,
',1 ,
where L1 = 1 or 2 according as the product 'II to is even or odd.
In the latter case, by putting t1l = } (u + to) and Cl = - (ll - iO),
we get
..r = ::? Ul1°1'
.) ')
1/ = u- - r"'"
.. 1 l'
+ .?: = ui + l'I.
Consequelltly, we obtain the whole set of propel. solutions of
(21) ,vith odd . and even ,1/ bJ Jl1eans of the forlDulae
(22) x = u 2 - ,;2. 1/ =  U l', + z = u 2 + t;2,
where II and l" are relatively prill1e integers, the one odd and
the other evell. Conversely, every triplet of positive J.., y and z
uniquely corresponds to n pair of 110sitive 'll and lO.
The result expressed by (22) is fOUlld in Euclid's Ele'YJ1pnta.

CHAPTER VII
DIOPII_\NTIN-E EQUA TI()NS OF
HIGHER DEGltEE
62. Some Diophantine equations of the fourth degree with thl.ee
unknowns. - In the lliarg'in of his copy of Bachet"s edition of
Diophant.os's .:!1'itJunetic"a Fermat stated and proved the follo,ving'
theorel11: Tbe area of a right triangle with rational sides cannot
be the square of a rational lluDlber. 'Ve formulate this result
in the lang'uage of nUll1ber theory as follo,vs:
Thforenl 11,J. The Diophantine equation
(1)
,4 _ !}4 = Z2
ha. I/O sDlutions ill natural ilund)l}r :I', JI and ..
})"O(if. 'Ve suppose that :1', 1/ and .? are positive. If we put
x 11 '
(x. /1) = d, d = '('1> d = /h and ( = Zl, equation (1) becomes
X 4 _ Y 4 = ;2
1 (1 'l'
,v here '1 is an integer. Since (,1"1_ .'11) = 1, it follows fronl this
relation that (.r1, '1) = e'/1, .C1) = 1. Thus it is sufficient to prove
the theorenl ,vhen x, 1I and z are relati,ely priJue in pairs.
Since .'}4 + ".2 cannot be divisible by 4, x is odd. N O'Y suppose
that equation (1) has the solution ['', y, 2'], where x is the least
(positive) integer for ,vbich (1) is satisfied.
}'irst ,,"'e consider the case of an even !I. The nU111bers ,2 I- ?
and [t,2 - z then ha,"e the greatest common divisor 2. Since the
product of these nunlbers is divisible by 16, one of them is
divisible jut by 2 a.nd the other by 8. Therefore tIle nun1bcJ.s
 (.1,2 + .2') and  (.l:2 -t ?')

228
CHAPTE 'II
are relatively prhlle, and siuce tlJeir product is a biquat1rate, they
nlut thculsel,-es be hiquadrates. Hence
,};2 + ; = :2 a 4 , L1,2 + z = 8 lJ4 '1
wher4? (( and bare reluti'\"'el)" prillle natural nUlllbel's, such thnt
JI =  {( IJ: (l. is odd. By eliIllinatiug ., ,ve get the equation
,l,2 == a 4 -t- _! /;4,
\\"hich IJHlY be ,,"rittc.'u
(..c + ( 2 ) {..t' - a 2 -== -+ IJ 4 .
Since the t,vo factors on the l&ft-haud side hu,"'e tIle greatest
COHl1l1011 Ji \''isOl' 2, it follo,vs that.
.(' + a 2 = :! (.4, lot'" - a 2 =  (/4.
,,,here e and d are relutiYel prilnc natura) nUDluers such that
b = l'd On subtraction we obtain the equation
(.4 _ a 4 = 112.
Starting. fronl u, solution [,l'..'/. z] of (1) ,,,,itll an even.1/, we ha,e
thus deduced a ne,v solution [e. d, (I]. But we have
c < :! a (" d = ," <: IC,
which h COIl trnry to our h)-pothesis on ,i'.
Suppo:;e next that .'1 is od'}. Then  is even. Since the l1Ull1-
hers .{.2 - li 2 ana &.t.. 2 + .'/2 have the grpatest COUlmon divisor 2,
'\""e ueJ nee fronI (1 ) that
.. ., ) "
,.- - - 1 / - -" ,,-
. .: -- __ ..f .
  ) .
..f.,... - .'/- = :. I J- .
where II and bare relativel.v priIne uatural nUD1bers. such that
:? z -== (( h. Hence
.,  I :)
J'- =-.: {,'" '. J"- .
" 2 oJ
.1'- = (I - /;-
and b" multiplicatiou
(l.'Ir=a.l -- lJ 4 .

DIOPHANTINE EQ1!ATIONS OF HIGHER DEGREE 229
Starting' fronl a solution [.l'..'I.,] of (1) ,yith an odd .'1, ,vc have
thus found a ne\v solution [ a, b, .r l/]. But we have
a < JI a 2 + 12 = ,\
w]lich is contrary to our IJ.pothesis on x. This proves the in-
solvubility of (1) ,v}len z  ().
()ur Inethod of proof is an illustration of Fermat's fa1uons
method of illinite descent. In g'oneral, it UJay be characterized
as follows: Onc' aSSlunes that any natural nun1ber .')' has a certain
propert. E. rrhis assulu ption inlplies the (\xistence of some sJnaller
11atural llunlber ')"1 which also has the )n.opert. E. nut this leads
to a contradictiol1 since there lUllSt be a snlullest positive 111teger
'v hich has the property E. 1-1 ence Ye conclude that the aSUlll p-
tion is false for ev'ery' natural lluJuLer ,I'.
This luethod has been uscd ,,"ith great ::;uccess by Fcru1a.t and
later investigators for solving' Diophantjne problen1s.
I t is, however, i1uportant to llotice that it SODletinles occurs
that certain ,"alues of ,I: do not ilnply the (\xistence of any Xl
less than x. Iu these cases the assuInptiol1 nl0Y be tl'ue for a
finite or perhups infinite 11U111 bel" of positive integers x. By an
appropriatl) 1110c1iticatioll of tl11 reasoning- it is t.hen often possihle
to ueternline all the positi\.e integ'crs ;1" wllich La\""e the property E.
In the next section ,ve shall g"ive an exanlple of this g-eul)l"alized
Inet.hod of iniinite decent
Froln Theoreln II.! it follo,'ls at once that the equation
(2)
I " ,4 +- 11 4 = ) :; 2
.. ... ....
.
has no other solutioll in relati'\'"cly priIne Batura1 nl11111)f'l'S than
.(' = /1 = ? == 1. For after squaring it may be \\"'l'ittll
( iC4 - '1/ 4 ) 2
:4 - \,l'lIj4 = 2 ' ,
",r e use Fcrlllat's l11Cthod for proving' the follo..i}jg three pro-
positions:
.1"/if»Ore1n 11;j - The IJjophfl lit i lit' rf/llatioJ1
t H )
,),4 + H4. = :2
has /tl) .""01" (iOl1,'. ll1 natl(ral J/lI}J,IJf-;r.,' ,I'. !I (IJI!! ....

230
CHAPTER 'II
Tlzeorenl 11(j. The Diophantine cfluafion
(4)
,-...4 _ 'I' = 10) ,2
& .'/ ... .-
has no 8olutio'18 in natural uunzuers J\ 11 and .,.
TJu-JorC1l2 117. Tlz,. Dio]Jhantiue equation
(5)
x 4 _ .'1 4 = }) E'2,
tvlzere p £ a l)rhnc = 3 (olod 8), hat? no .90111tion. in natural
'1Hunbers x, '!I and 2'.
In equations (3), (-t.) and (5) \ve can obviously suppose t.hat the
natural numbers x, 11 and z are relatively prin1e in pairs.
Pl'oq( oJ" Theo}eenl 11;). In equation (3) the Ilunlber ' DlUSt
clarlY' be odd; one of the n ulllbers :r: and .II is er-en, sa). ll. Sup-
pose that (3) has the solution [x, !I, .z], wllere z has its least
positiye value. Since the greatest coronIon divisor of the factors
.? + x 2 and z - x 2 is 2, we obtain fronl
(2' + {,2) (z - .r 2 ) = Z2 _ [1'4 = .11 4
the systeln
';' + " .2 = ') ' , 4-
,.., _ . ...J L ,
- 0') - l4
2' + ;r =  ) ,
!I =  a b 
where a and bare relatively prilne positive integers, and a IS
odd. Elinlinating E we obtain
+ ,i,2 = a' - 4b',
where the upper Sig)l must be chosen since the lower sign does
not hold nlodulo 4-. Since the greatest COlnmon di\?isol. of the
factors a 2 -+- x and a 2 - ;r; is :2. ,ve get fronl this eq nation the s)'stenl
a2+.=2c4. a 2 -Jo=2d 4 , u=cd,
where rand d are relatively prime positiye integers. Hence, bJ
elhllinating &(',
a 2 .:= ('I{ -1- d 4 .
Thus the assulllption that [x,.1/, ] is a solution of (3) leads to
the existence of a ne\v solution [c. d. a]. But we have
Z = (l4 + 4 IJ' > (14 > a 2 .

DIOPHANTINE EQUATIONS OF HIGHER DEGREE
231
Since t.his is contrarJ' to our hypothesis on z, ,,'e have thus proved
that the sum of two biquac1rates cannot be a square.
Proof 0..( ThfOrC}}l 11(l. In equation (4) both &r and 11 must be
odd; z is even. Since the greatest COlnnlOl1 divisor of the nunl-
bers J...2 + .'/ 2 , .t\ + y and x - !I is 2, \ve deduce fronl (4) that
(6)
&x 2 + .'/ 2 = 2 a 2 , x + y = 2 tJ2, X - lJ = '2 c 2 ,
where a, band care natura.l nUlnhers relativel:r prime in pairs,
and z = 2 a b c. FrOIl1 the last t\VO equations in (6) we get
,.,.. = /J 2 + c 2 , lJ = L 2 - c 2 .
Introducing- these values of ,e anu /1 into tll( first equntio11 in
(u) ,ve obtain
a 2 = iJ 4: + (,4.
But according to Theorenl 115 this equation is not sol vaLle in
natural numbers. Froul this Te conclude that equation (4) bas
no integral solution for z  0 either.
])rool qf Theoreln 11 t. Suppose first that z is odu. Then either
x or .'1 is even. In the first case \ve oLtain fro III (5) - 1 = ]J
(IlIod H) and in the second case 1 ==}) (Tuud 8); but tllis is im-
possible since }J = 3 (luod S). Thus Z Hlust be even.
Now suppose that. equation to) has the solution [.1:, ll. z], ",here
x has it.s least positi\Te valul:'. t i8 necessaril'y divisible by 4.
The greatest C01l11110n divior of .£ - 1/. ,i: + .'I und :2 + y2 is .
\\T' e theIl have either the 8J'ste1l1
( ": ) ,...... 2 I 4) 0') 4) n 2
.. .i" :!: .'1 =  if, X + !I = -r )J r . x + 1I =  if .
or the svstem
.
(8) ;r + U =  P lt2 .r + ?I = 4 ,.2, ,).2 -.- .11 2 -= ;! ,r 2 ,
,vhere 'll, C und fC are positi\"'e integ-ers relatiyely prio1e in pairs;
'll and It" are odJ. By elinIination of x' and 11 we get froln (7)
(9)
u 4 + -11 12 ,.4 = u.2,
and shnilarly from ()
(10)
p2 u: 1 + 4 Z-4 = ,,.2.

232
CHAPTER VII
From the latter equation it follows that
U' +}) u 2 = 2 a', if -}JU 2 = 2lJ 4 ,
where a and bare relativel)" prin1e natural numbers. Hence
}),,2 = a' _ '.
This Diophantine equation is of the sanle tJ'pe os (5). But here
'it is odd, and we have jURt shown that Z lllUst he evpn in (5).
On t.he other hand, since u: + 2}) r 2 and U' - 2}) l.2 have t.he
greatest comn10n divisor 1, it follo,,"s from (9) that
1(" + 2}J r 2 = a 4 , If - 2}) 1: 2 = b',
,vhere a and bare relativel)" prinle positive integers. Hence, on
subtraction
a 4 - lJ' = }J (2 l")2.
Thus the assun1ption that [x.'J, ?] is a solut.ion of (5) leads to
the existence of a new solution [a, l, 2 t]. But \\"e have
a < a 2 b 2 < a 2 2 + }J z,2 = 1(2 + 2}) r 2 = .r,
,vhich is contrary to our hJPothesis on x. Hence Theorem 117
is proved.
BeUla1"/-., In TheorenJ 117 the restriction on p to b{-'  3 (nlod 8)
is essential. For in each of the cases j) == 1, [) or 7 (UlOU ) eCflut-
tion (5) Inay be solvable for certain \"alues of the prilne p. For
instance, when 11 = 41, it. has the solut.ion [5, 4, 3]. 'Vhen p = f),
it has t.he solution [3, 1, 4]. "Then 11 = 7, it has the solutioll
[4, 3, 5].
Applying sio1ilar nlethous, .Billing, Lind and other in,pstig'ators
have ShOWll the impossibilit.)"' in posit.i\"'e integers L1', 11 anu z of
a great nunlber of Diophantine equations of the type
a,);' -1 b !14 = C Z2 ,
where a, band rare int.egers.
63. The Diophantine equation 2 x 4 - y4 = z2. - The equations
il1vestigated in tbe preceding' section had no sohltions in nat.ural

DIOPHANTINE EQUATIONS OF HIGHER DEGREE
233
llulnbers, except equation () ,,'hich had only the solution x = !/
= z = 1 in relatively prinH positive integ'ers. There exist, how-
ever, equations of a inlilar t.ype ,,,,hic]l ha,e infinitely luau)"
solutions in }'Ielati\?ely prinle integ'ers. As an exalnple of this
category \ve choose the equa tioll
(1)
:! .1;' - 1I 4 = Z 2 .
,\\T e denote a solution of this e(llultioll in intgers ,I'. !I and z by
[,x, .lJ, .2']. \Ve consider only solutions with positi\.e '. '!J and Z..
Such a solution is said to he n positi,e solutiOJl. ',\7 e can ob-
viousl.v suppose that :1:', 1I and z are relati \Tely In.ilne in pairs.
It is easil. seen that they are all olid. Elluation (1) nlay be
written
.,4 = ( .'12 + 2.' ) '2 ._ ' .'1 2 - e ) ' 2 .
 \:!
Applying e(luations (2) in Section 61, ,ve obtain the following
systelu, provided that Ii' > 1 :
2 l) 1 2 C) C) I 2 '\ ,
:i: = a'" + u, .'/ =: Ie" - J --  a ),
:: .2 = ([2 - lJ 2 -  a h,
where l and b are relatively prinH i1l1eg'ers  (I. rrhere is 110
loss of gcnerality in supposing- II + b > u. It foilo'vs from the
first e<.]uatioll of this s)"stenl that a b is eeIl, TIH.'ll ,ye conclude
fron1 its second equation that b lllust be e,en. E'rom the first
equation of the systeJIl ''fe obtain in the sanH way as before
,I' = ('2 -1- lt 2 , a = ("2 - rt 2 , 1) = :! cd,
,vhel'e c and dare relativel,,\'" prinl(\ integ'prs F o. I lltroducil1g'
these 'ralues of a and b into the expression for .'1 2 . "..e have
.'/ 2 = (.'t +_ -t- (,3 cI - () t..2 (l"!, __ -! (J (/3 + (/4
and
( 0) , ) I l ') C) C) L.) C) I C)
(.. -r :. (. ( - ( ... r - .If' = , C'" ( ....
Hence, since a -1- b = (.2 + :! e d - It!. is positi\e,
('2  :? (0 tl _ r/2 + !/ = :!./'2 
')
('... I
 . 
:! (. d -- "... -;- I' = -f (J'"
T ., .f ,

234
CHAPTER VII
\vhere f and U are relati v'ely prinle integel.s rf: 0 such that.l 9 = cd.
By sumlnation we get
(2) c 2 +:lcd-£l2=J2+2g2.
Since fg = cd, we can put
f= E t" g = 'YJ u. c =  ll, d = '7l",
,v here " 'YJ, II a.nd l' are integers :F- O. Introducing these ,allIes
into equation (2) we obtain
(3)
2 (u 2 - t. 2 ) + 2  'I} II t' = 1j2 ,.2 + :! 1}2 11 2
and thus
(-I-)
YJ _ 'It V :f: 11211' - t o 4
 2 it 2 + ,.2
The numbers ..l' and .'I are expressed b)F the fornlulae
(5)
(6)
.£ = (;2 + d 2 = 2 112 + 'Jj2 zo2,
+ C]I  C) ... 2 C]I ) ') ')
_ y = J'" -  g =.; ,." - :. '1" l:".
COllsequelltl)?, starting fronl a positive solution [x, 11, .] of (1),
we llave by this procedul.e deduced a ne,y positive solution
[Iul, I r f, 1 / 2 u'-v'] of (1), where
1111 = I  I V .c - 1] 2 V 2 < JI x-I < .J:.
But ,,,"e had supposed x > 1; for :1 ' == 1 the reasoning is not valid.
Conversely, starting from a positive solution [It, t" 1f] with
'l{' > 1, we can bJ means of formulae (-1-), (5) and (0) deduce two
11e\V, positive solutions; E and 'YJ are deterlllined bJ (4) as rela-
tively prime integel.s. For II = zo = 1 we get fron1 (4) either
')
1 = 0 or  =;; since onl)' the latter value is applicahle. we ob-
tain 'YJ = 2 ana; = 3 and thus ;.. = ] 3, 1I = ] anu,; = 239. Starting
from the solution [13, 1, 3U] we g'et. by (-I-)
IJ 13 + 23g
- ,
; 33B

DIOPHANTINE EQUATIONS OF HIGHER DEGREE
235
w hence tIle two st?ts of values: '1 = - 3, 'I}l = 2 and 2 = 1] 3,
1}2 = 84. This gives two new positive solutions of (1), of which
the least is [1525, 13-&.3, 2750:!57]. By repeating this procedure
we obtaill in succession all positive solutions of (1).
Obviously the nlethod applied here can be regarded as a gen-
eralization of the Fernlatian lllethod of infinite descent. It was
first discovelwed b" Lagrange that Fermat's method can also be
used for solving conlpletel)" Diophantine equations which have
an infinity of solutions.
lVlordell ha,s shown that the Diophantine equations of the type
(7) 1(.'1:, !I) = a ,1;4 + 1J ,'1;3 !I + e X 2 .11 2 + d ,1' N 3 + e y' = "h: Z2,
\vhere the coefficient.s are integers such that the equation.f(j', 1) = 0
]l3,S JIO Inultiple root, Catl be solved by the sanle method. If
equation (7) is sol,.able in integers .i.., .'I and z, there exist a
finite numher of iuitial 8011liiolls \vith the following property.
Starting frol11 these solutions tl1e total set of solutions can be
found bv Ineans of a system of fornlulae analogous to (4), (5)
and (n). In our exa.mple the initial solut.ion was [J, 1, J]. But
equations are known ,vl1ich have ll10re than one initial solution.
(COlD pare Section 68,)
64. The quadratic fields K (V -1) , K (V = 2) and K ( V 3 ).-
,-
ft ,,,,as sho\vn in Sectio11 6 that the quadrati field K {J' 1)) con-
ists of all uU1uhers of the forlll a = a + I) J ]), where II and b
are (ordinary) rational 11t1ln bel's. Sil1ce D is rational, a and bare
uni([uely deterulined wht?n a is g'iven. The nUlnber a' -= a - hl' [J
is the ronj ligate of u.
/-
BJ the uO}"1n ..\T (u) of a == ct + b J'D we understand tIle product
of a and a'; thus
... ((1) = a a' = (a + b J Jj) (a - b 1 /) ) = a 2 - n b 2 .
If we put P = (. + d J '/) with l"atioual c and d, we obtain from
the identity
(a 2 - [) h 2 ) (c 2 - f) ,,2) = (a c + lJ d D)2 - D (a d + b r)2
the following rule for the nor1118:
.\or (u. 13) = .,.. (u.) ..\r (JJ).

236
CHAPTER '-11
In this section nllnlLer 1ueans a 1lun) bel" in the fieltI K (co). ".here
(J) denotes one of the Ilulubers 11 - I. 1/ -  ur  (- I -:- V - :-3).
The nUluber a + lJ w is said to be an lutcg'er in K (w) if a and
bare ordinarJ' integers. In the follo\viug, int.epr IUf\anS an in-
teger in K (co). 'fhe ordinary intf'g'ers are called rational iutegers.
The norn1 of a = a + IJ w is e(lUnl to a 2 + 1)2 in K tJ ' - 1 ) , equal
to a 2 + 2 [;2 in K (1/ - 2) and equal to
( 'C) C)
b" a') /J'" '> 2
a - ) +;) -t. = 1I -- a I.. + I,
in K (} (- 1 + 1/ -3 )) .:=. K ( JI 3). It. foiJo'vs f'()n' this that .\"la)
is al\\"a..rs positive in K(w) except ,viJen a = o.
If a is an integ.er, _''''(a) i a natural llUI:1Ler. It is clear that
onl.v a finite nU111ber of intgers errn hn \'C the saIne norlH. There
are positiye rational integers n such that 110 integ'er has the
norm J1. Thus the nUluher 23 calluot he the norm of an). integer
in any of the fields K ((0).
An integer ,vith the Hornl I is called a IlIIit. Recalling ,vhat
,vas said aboye ahout the IJorlUS. n.e easily prove: Iu the field
K Cv=1 ) there are the four units + 1 nlJa + i = + V - 1. In
the field K (V - 2 ) there are only t ;h' t \\"0 units ..L ]. III the
field K (JI - 3) there are the six unit.s -t 1, + e = + i (- 1 - 1/=- 3)
and + e 2 = +  (- 1 - 1 'c 3 ).
If . 'I} and x are three iuteg-ers ill K (lfJ) such that
(])  = )} %. (1J  0).
,,'e sa.v that 1} is a divisor of , or that  i di\isiLle by fl. 01"
that 'I} divides , Each illtegel"  (O) ba on1y 11. fillit.e llUUlber
of divisors; for, b" (1), we haye
.\.l) == .\T (17) 2\T(U).
Thus the natural nuntber \. tJ7) is a. ui\"ior of the natural num-
ber ...t' (). (1onsequently, there is 0111). H fin i te JJulnbcr of possihle
values for ...,r ('I}) and threfore nI o for "I.
Ever)- intf'ger (different fro1H 7.er(/1 i:; a di \T180r of its norD!.
.l\. "unit. rnar also he detinl}d afoi a divisOl" of the 1l111n her I. siuce
. '
Ee' = 1 \vhen E. is a l1uit.

])lOPHANTl:XI EQl ATTONS OF UIGIIER DEGREE
" 3 '"
.... 
T\vo int.eg'ers ure saia to be associated \vith each other ,vhen
their Cjltotient is a unit. The Ullits and the integers associated
vith a given illteg'er a are called the tri\"ial divisors of a. All
the di¥isol's of a unit are thenlselves units.
An integ'cr ,vhich has Oll]Y trivial divisors is called a prinze
\vhen it is not n uuit.
An integer , ,v hose HorUl is a. rat.ioHul prilue p, is an irra-
tionul prilne. For it follOlrs fro111  = 1}  tha t ...\" () = ... (I}) ...Y' () = p.
Therefore, one of the l1Ul111}pr .. ('Ii) ana ...:\' () is equal to 1) and
the other to 1  hence one of the lluulbers 'I} and C is a unit.
ConSellUelltly all the ai\,"j()rs of  ure triviaL and  must be a
priIue. In t.he nelJ K tJ - I}. 1 +  1 -1 is a prilne; in K t V ;J),
2 + J I 3 is U. pl'illH; i 11 I (I -  ), 1 + J -:! is a prilue.
LeUHna 1. I}'  {lud 'I are [,eu iutegers, and 1}  O. there fxi.....t tl('O
infl'gel's % {l },d A ....llch that
 -:= '} ')! .+).. ,,.ith ...\" tl  < T C,,).
c.
J,.
J)ro()J: Put  == Ll + h i'J. \\71lt:re II and u are r3. tional n unl bers.
'1}
Then there exist. t\VO rational il1tcgers .t' and /I such that
()
III - . t I < !, I b - ?/I < .
Puttin o .

,,' -!- ,II ({) -=-;,e ana
c. 
 1 . ,, - 
 - J r., - ,..,
\ve have
i.. = 1] ( - x) = 'i [u - ," --;- (b - !I (0].
If (0 -=='J ==-1 or J/ - , it follo,vs from (2) that
'li) == ..".. II}) . [(a - ;,c)2 - (,)2 (h - 1/)2J < .\' ('I]) . (t + 7) < ..\.. (1}).
If (I) = ! (- 1 .'- } J - a), it follo,vs that
.i'l(A) =-= .\-(17)' [(a - ;J.)2 - (a - ,i:)(1J - .II) t- (IJ -- 1/)21,
and fronl ()
..\" ())  }.? (17) . ({ + ! -t- 1) < -,or {tJt

238
CHAPTER '-II
This proves the lemma. We next prove
Le1111na 2. E't'e1.y priHle jt u:hich dirides the product ''fJ of fhe fleo
integers , and 1] 'is' a divisor of eith£'r , or 1].
Proof. Suppose that , is not divisible bJ j(;, Then ,ve shall
prove that I} is divisible b).:'t. In the Ret of llunlbers
(3)
a  + p 71:,
where a and fJ are arbitrary integers ill K (£0), there exit certain
lluDlbers  0 which have the least possible DorIn, snJ .'T. Then
..V < }..,. (n). For according to Lelllrna I,ve can choose the integer
fJ such that
).T (c: + fJ 71:) < T (n).
Here , + fJ n  0, since E it not divisible by. Let y = a E + fJ n
be an integer such that TlY) -= lV. Then, according to IJemlua 1,
there exist two integers " a.nd ).. SUCII that
n=Y+A
with ...\T (l) < N(y). Further we llave
l=n-"y=n-x(aE + (J)=(- "a) + (1- fJ):n:.
The number A therefore belongs to the set of numbers (3), and
its norm is < 'T(y) = .'T. By the definition of }.7' this is possible
only if A = O. Thus we have jt = "y. Here y Call110t be asso-
ciated with n since T (y) = }.T < .'T (n). Hence y is a unit.
Multiplying the equation
u.+p1t=y
by the integer 17y-l we obtain
(/. E'I] y-1 + fJ 1ll] j}-1 =-= 'I}.
Since n is a, divisor of E'I], the left-hand side of this e(luatioll
is divisible bJ n, and therefore a.lso the right-hand side -= 'I] is
divisible by n. Q. E, D.
An immediate consequence of LeI:nma 2 is

DIOPHANTINE EQUATIONS OF BIGHER DEGREE
239
Theo},(Jn 118. 111 the jield8 K CV -1 ), K (V - 2 ) and K t )1 - 3) (!l:ery
intpger 'n'hose nO]"Ju is > 1 CQ 11 'ie II Ii iquely e.£jJressed as the
product ql a .tiuit(, nZilu"hp1" of })rz'1ne8, GjJ£lrt front the order OJ
}Jrinle .factur8, and providpd that Clt.'::8()(.iatecl jn'i'l12cs are not COll-
sic! el'P.d ({,.... cl i8ti nct.
Thp proof is analogous to the proof of Theorelll -l- in Sec-
tion 4. '\Ve first prove that every integer ,vitII a 110rDI > 1 has
at least one vrilne factor, Anlong the divisors of , \vhich are
not units there are certaiu divisors "yhose norUlS are the least
possible, Let '7 denote Olle of these di\yisors. Then, apart fronl
the units, 110 divisor of , has a norn1 < T (1]). Now suppose

'J} = ':, ,
wher  and  are integers which are also divisors of ,. Hence,
if : is not a unit, we nlust have }.T ()  ),7 ('J}). But since
...\ (17) = , () ,,. (),
we also have .£\.,.() < .,,. (17). Thus ., () = -,. t?}), and  is a unit,
According to the definition of a })l'iUIP, 1} is then a prime. Con-
sequently , has at least one divisor \vhich is a priule.
By induction it is 110W easy to sho,v that , can be \vl'itten
as a product
, = nl:72 .." 7Cr
of a finite number of primes ni,
Finally, the 1111iqueness of tI1is decolllposition follows by the
same argumel1t ,vhich ,vas used for the rational field K(l) ill
Sectioll 4.
C0111bining t11is result with Theoreul ] 00 (first th1 4 ee parts) we
obtain
1/l(:ure1n 119. 1. III th(, thl1d K ( V - l) there are the folloll.ing
))J'in2P8.' l'!ze nU1n1Jer 1" + 1/. 1 and its (u:sociate8,' the rational
prilnes == 3 (nlod ..t.) and their a8$oc:iates: the integers x + JII/ -1
'If:hose 1l0fJ}l8 are '{J1.inzes == 1 (mod 4).
.. In tlu) .rielll K (Y - 2) there are the &(ollou'i ng pri7nf!s.' The nu,n-
ber.,' + 1,' - 2: the nunber,'t + 1) where p "8 a rational prinle = [) 01.

240
CIIAPTER Y II
== 7 (n10d 8); t!l£, integers x + !I JI -  lfhub'e J'orJ}1t....' are lJ1411nes
== J or == 3 (l11od 8).
.=1. In the .1;eld K (l l 3) thrre are fhe fu/lou'in.fJ }J}"inle8: l'lu nU1}1-
lJ('I. JI - 3 and ilt aS'()Cilllrs: the raliunal ])riJJ2c. ==  (ll1od 3)
{ud their a,i""'o('iaf( .....: t he int('grr.' )1 + .'I J -;3 lrho,....(: U()J'nl arr
}J1.hnes == 1 (Inod 3).
It easily follo,\s fro 111 Lemlna I that if a and pare t".o
integers, not both = 0, thy ha\"e certain COlnnlon divisors 6
with the followiug properties: The numuers  and  have no
coronIon prilue divisor; the llunlLers  are associated with each
other; all the COUlI110n divisors of a and p are divisors of 
These nUDlbers  are said to be the grl'atest C()11lJnOn diris(jJ'8
oj' a and p.
If the nunlbers  consist of units only, ,ve say that a and p
are relatirellllJrinze and \vrite (a. P) = I- The integers aI, a2. . . .. am
are relafil,'f'l!l prinle in pair.." ,vhen (ai, :) = 1 for all disinct
i and j.
The concepts of eouyrueuf Ulf1nUf'J"8, cougfuenC(', ,.esidue.,' and
,.esidut, ("la8.'jc;;: can be defined in the san1e way as in the rational
Dum ber theory. If the diffel.ence a - fJ is divisible by , ,ve say
that a is congruent to p nlodulo  and ""rite
(J. == P (ulod ).
The nU1l1bers a and /3 are then saia to uelong to the sanle residue
class luod ulo E.
Finally we pro¥e sOlne lelnnlata on numbers in the fielJ Ktl/ - 3).
As above ,ve put (! =  (- 1 + J'  3 ). The uUluber ).. = 1 - (! is
a. prin1e, since its l1orll1 is 3. It is associated ,vith the prime
JI 3.
The integers in K (J  3) fall into three residue classes modulo )"
which may be represented by the numbers 0, 1 and - 1. J n fact
a + b e = a + b (1 - )..) == II + lJ t 1110 a A),
and the nUlllber II + b has oue of the residues 0, 1 or - 1
tnodulo 3.

DIOPHANTINE EQUATIO:NS OF HIGHER DEGREE
241
,--
Lellt1ua /J. 1.( a is an iuleger 1'U K (J' - 3) ,chich is not di,,'isible
uy A, n'e haloe
a 3 -= + 1 (Juod )..4).
Proof. Since a is not divisible b. A, we have
a = + 1 (Iliod ),,),
aud therefore
a 3 = ( + 1 + A fJ)3 = + 1 + 3) f3 + 3 )..2 rj' + A 3 {33,
where f3 is an integer. Hence
a 3 + 1 == AS (f33 - (12 fJ) = ),,3 (13 3 - [1) (ulod A 4 ).
But it is easil.v seen that the nUluber p3 - fJ is al\vays divisible
by A rgardless of the reidue of p 111odulo A, Thus a 3 + 1 is
divisible by A 4 . Q. E. D.
Ltnana 4. 1)" 8 i,o.: a. unit, aud lj"  i.r.; an int(-grr i12 K(J"'=-3), the
tongruen(O(J
3 = 8 (IDod :?)
hold. onlLl .f'f)}' 8 = + I.
For 2 is a prime, and the residue classes modulo 2 are obviousl)"
represented by the four nUDlbers 0, 1, (! and e 2 .
.A. cOtl1plete theor.r of the field K tV - 1) 'vas gien by Gauss;
it is therefore called the Causian field. The field K (V - 3) was
treated by Jacobi n.nd Eisenstein.
65. TIle Diophantine equation ;3 + )3 + ;3 = 0 and analogous
,- -- -
equations. - Suppose t.hat , 'I] and : are integers in K (l - 3),
different fron1 zero. ,vhich satisfy the equation
(1)
:3 + 1]3 + 3 = O.
A prin1e which diides both ; and 'lJ D1USt, by Leluma 2 in Sec-
tioll 64, be a divisor of . Thus ,ve can suppose (. 1]) = (;, C)
= (,]. :) = 1.
If none of the llunlbers , 1] and  is divisible by A = 1 - (!,
\ve derive fronl (I) by Lemlua 3 in Section 64 that
+ 1 + 1 + 1 == 0 (mod ;,,4).
16 - 516070 'l'r!J(Jl'e .a!lcll

242
CHAPTER YIf
But clearlJ this congruence is iJupossible for all eight cOlubina..
tions of the signs. rrhus we can SUPl)ose that one of the integers
E, YJ and , sn. " is tli visi LIe L. )". Putting
 n
" = A y,
\vhere 11 > 1, and ,,,"here r is not divisible by)", equation (1)
becomes
3 -+ 1}3  ),,III 1'3 = o.
".,. e consider, ho\vever the 1110re general equation
(2)
E 3 + TJ3 + e A 31l y3 = 0,
,vhere 8 is an arbitrary unit in K ( V- 3) .. and where (, 1]) = (E, y)
= ('l}, y) = (" A) .=: {l}, ),,) = (y, A) = 1. Further ,ve snppose that the
exponent " has the least positive value such that equation (2)
l1as integ-ral solutions. Then n > :!. For, a pplJing Lenlll1a 3 in
Section 64, it follows £rol11 (:?) that
+ 1 + 1 + 8 A au == 0 (nlod A 4 ),
,v hence
j,,3 n :::: 0 (Illod ),,4)
and therefore 11 ? 2.
The three nUJ11bel'R
al = ; + l}, a2 ==  + (! 'l}.
,)
as = ; + fl" 'I}
are all divisible by A.. For uccord.ing' to (2) the product al a2 a3
is di visi ble bJ' A, anu fUl'thC'r
al = (..(2 == as (Inod ),,).
The three inteo'ers

a]
PI = -,
}"
U2 ag
132 =: - , fJ3 = -
}" )"
are relativel)" priJne in pairs. In fact, if "-e suppose that. PI and
P2 are divisible by the sanlt") pritne :t, the l1ull1bers PI - P2 = I}
aud )..fJl - PI + {32 =  \\'onIJ ulso be divisible by :t: but this is
cOl1trarJ to hJPothesis. Thus ,"'e have (PI- f32 \ = 1: in the salue
,va)" it llla)" be sho,vn that (PI' fJ3) = (fJ2 - P3') = 1.

DIOPIIANTINE EQ1TATIONS OF HIGIIER DEGREE
243
Fron! () it is obvious thnt one of the nUlnbers PI, P2' P3 Inust
be diviible by ),,3u-3 ° F].'here is no loss of gelleralitJ' in supp08ing'
that tbi nU111ber is Pl' For equation (:?) does not change if y
is replaced by re or ye 2 .
If \ve \¥rite e<luatiol1 () in the forul
PI {J2 {J3 = t;" },,3 n-:J ( - 1,)3,
it follows that
P = S ']3,£-3 "-3
1 1 1L 1'
(12 -= f2 1i, P3 = E3 : '
wbere £1, E2 and C3 are ullit, auti \vhere 1' 'Ill and ;1 are in-
tegers relative]J prime in pairs 'v hich are uot divisible by A.
N O\V we have
0) ") '). R l)
{!04/ J3 -t e P2 -r 1'1 =
and thus
e 2 E3  ,- e E2 ,' + £1 ),,: 11-3 ; -= O.
Hence
(3)
ta   ')13 +   3 Ii - 3 ,'"3 = 0
"'I I c4./] . (;5/1. l '
where EJ a.lld ES are units. Since n > 2, and since the product
;1 'Ill is not divisible by A, it follows frolll this equation that
+ 1 + E4 == 0 (nIod ,1,3).
Hence E4 -= + 1. Then equation (3) can be \vritten
(4)
 - ( + '111)3 + E 5 A 3 {u -1 \  = O.
This equation is of the sanle type as (), al1d it is solvable if
(2) is sol¥able. But since the exponeIlt of A. is less in equation
("*) than in equation (:?), this is a contradiction of our h)'pothesis
on the nUlnber 1'1. (onsequelltlo\r we conclude that equation (2)
is not solvable and state
1'/ze()J o enl 1(). The Diophautiue eqaation (1) has }1() ,olllti(j1i8 ,,1
il/tegcrs ;, 1] aud  in the .field K lV - 3 ) 1:( E TJ:  o.
By a simple moJification of the proof of this theorenl we
easily establish

244
CHAPTER YII
Thr-or(J111 1:!1. The Diophantine equation
3 _I- 1}3 + 3;3 = 0
,--
has 120 ,';:CJlufic.'J/s iu integers , 1} and C lJ2 the fit'lel K (l- 3)
i)' ,  O.
]J/,()O'/: EvidentlJ it is sufficient to slJo,y that there is no solu-
tion of the equation
(5)
3 + 1}3 -i- F j,,311+2 1'3 = ().
".11e1.C E is un arbitrary unit in K (JI - 3) , anu whel.e (, 'I}) = (, y)
= (rJ. 1') = (. i) = (1}. )J = (": j) -= ].
"T e suppose that n has the least posit.ive value suell that
eljuation (f)) has integral solutions. Appl).ing" Lenlma 3 in the
I.receding section we see that n > 1. By the saIl1e arguinent us
i1) the proof of Theol.en1 l:?U it follo,ys froIl1 (6) that
 ;- 'I] =  131&-1,"'3  +_ C! I} _. 3  + e 21 } _ t3
)" C] IL "'I · )" - l:2 lit. -0..-.£ - £3 "1 .
Finally ,ve goet the equation
t3 +- ( + 1 ) )3 -l- E A,3/l-1 .<-3 = 0
'"'1 " 1 I 5 1 '
,\.hich is of the sanl€' tJpe as (u) and is solable if (5) is solvable.
But this is contrary to our hYIJothesis 011 the IlUU1bel' n. Hence
(5) cannot be solvable.
l'heo1"eJ} l,."d. TluJ Diupltautiuf efJllativu
{tj) 3 -f 1}3 = 2 3
hat:; no 8olut'ious -in £nteger8 .;. I} and  in K (1' - 3) oth('f than
3 = 1}3 -= 8 r  r' O.
Proof: ,v e conider the l110re g"elleral equation
(7)
3 + 1}3 + :? f' :3 = (t,
,vhere e is an arbitrar'y unit in K (11 - 3). If, 1},  and E satisfy
equation (7). ,ve saJ that [, Ij. C, E] is a solution of t7). The
110rU1 N (rJ:) of the produc.t  I}  is said to be the Jzpi!lht of the

DIOPHANTINE EQU:\ TIOS OF HIGHER DEGREE
q..-
-.)
solution [, '1], , l:-]. The height is a natural number. "... e obtain
all the solutions with heig-ht 1 wheu E 3 = '1}3= + 1, 3= + 8- 1 = + 1.
N o'v suppose t.hat [C;. 1] ,e] is a solut.ion which has t.he least
possible beight > 1. Then the l1unlbel's , 1] and C Jl1ust be
reIn ti vely priDle il1 pairs. 'Let llS denote by a, {3 and r the
numbers
I:
'-I : II,
Coo .,
!!; -. (j- I}
.) ..
!!-; -. !! II
in this 01' some other order. Then
(8)
u. + {J + )J = 0, U I J I' = -  /;' 3.
( ' (1. I' ) _ ( {J )' ) __
-- . - 1
b D. .D (), .
(8) that
(9)
If <5 is a divisor of a and fJ such that (, ) = 1, it is clear that
'l'hen it follows fronl the last equa.tion in
.)
(1 _ :J P _ :J" -- .) L' ..;;
() - f'l  l' b - f2 'it. i) - - 3 '::' J ·
'.
.... --- l: I ] '100
IJ --  1 1 ":) 1 ,
where Cl, f'2 ana Ea are units. anu ,here ';l 1]1 and ] are integ'ers
relatively pritne in pairs. Adding the first three of these f[ua-
tions and recalling- that a + p + r = O. ,vt? obtain after multi-
plication by Ell
(10)
....1  ) <..')
.J-.' 1- C'  } '> -+- · F  ,) = ( )
":'1 I c-I'] , - .)-'1 '
where 84 and 85 are units. If we consider this equation modulo 
and apply LemJna 4- in Section 64, ,ve see that E.t = 7 1. Thus
we have deduced a new solution []. 'I}]. :" E5] of equation (7).
This solution has height N(1171 '1) = N(). Thus according to
our hypothesis ,ve Inust have either
(11 )
(  )
.,..  - - : .
7'" -- __' :\ I t I] . )
.L' <5 -  . '='
or
(12) x() = Si;11hZ:l) = 1.
Since  divides a - (! p and (1 - (12 f3, it also divides (1 - e) E and
(1 - (1) I). Hence, since (. I}) = 1, b is either a llnit or associated
with A = 1 - O.
\00

246
CIIAPTER VII
If relation (12) holds, : is associated ,vith D. Froln (7) and
fron1 Leluma 3 in Section 64 it is evident tbat t is divisible uv
- 
)..2 if it is divisible bJ )... Thus  and;; must be units. It then
fo11o\\"s from (9) that a and p are units. Hence A'; and ),.'} are
both the SUD1 of t\VO units. It is easy to see tbat this is pos-
sible only \vhen  and 1] are therllselves units. Hence we should
ha\?e 4'T"( YJ ) = 1 contrar) to our 11ypothesis.
Frolll (11) ,,"e obtain ....'T (1} c))  1, and thus the nU1l1 bel's ,
1] and c) are units. Froll1 (7) it t.hel1 follo\vs that  is also a unit.
Hence we sllould have }.j'" ( I} ) == 1 contrary to our llypothesis.
In particular, Tlleorelns 1 O, 121 and 1 2 also hold in the
rat.ional field K (1).
Theorenl 120 was stated for rational llull1bers by Ferluat (see
Section 68) ; th e proof was given 1,)" Euler. Theorenls 121 and
1 2 ,vere proved by Leg'endre for rational numbers.
66. Diophantine equations of the third degree with an infinity
of solutions_ - To Theorems 120, 121 and 122 in the preceding
section we add the following result:
Tlzcorern 123. IJ' a is an int(1uer > 2 'Ic/zirh is not dit'isible bll the
ruu,' of an!1 }J},lJne, the Dio}Jha12tinc equation
( 1 )
,/,3 -t- !,3 ::= a Z3
has either 120 solution or in.tinit(l'!J tnauy 8olutio1ls in relatil"el!1
lJrhlle 'integers J", !J and .!, 'zl'ith z;;6- O.
Proof. Integer means here ordinary rational integer. As usual,
denote by [.r, 1I, z] a solution of (1). As in the similar cases in
Section 65 we can Slll)pose (,r, .l/) =-= (..t, 2') = (//, z) = 1. From the
solution [.X", .'/'2'] ",,'e obt,ain a second solution [Lrl, HI' 2'1] of (1)
by the forDlulae
()
(D)
(.t)
b '/"1 = .r (,1 3 +  !/3),
b .'/1 = - !I (,;:3 + ,lI3),
() : 1 = e' (,,.3 -- N 3 ),
where b denots tbe greatest comn10n divisor of the three num-
bers on the right-hand side. This is easily verified bJ introducing

DIOPIIANTINE EQUATIONS OJ.' IIIGHER DEGREE
247
the values of Xl' .'11 and 2'] given by (), (3) and (-l) in equation (1).
The geometrical interpretation of this is as follows: If \ve draw
the tangent to the curve (I) at the point P (x, !I, 2') (hoIIlogeneous
coordinates), IJ 1 ((1' !II, Z'1) is the point in which the ta.ngellt cuts
the curve again. 1 J 1 is called the tangential point of P. PI co-
incides with P only ,,,,hen JJ is a point of inflection.
We have t"1..: rl=.'I; for other\vise .'JJ = .If = 1 and a = 2 contrary
t.o hJ'pothesis. Therefore it follo\vs from (-l) that 2'1 rf O. FroDl
the eq nation
J'. + .'I = a 2
we see that (.:r.t, l/l) = (,rl, Cl) = (!/l, 2'1) = 1.
Let q be a prinle factor of. If .1' is divisible by q, it follows
froDl (3) that .'1 4 is divisible by 'I; this is ill11)ossible, since
(x, l/) = 1. Sinlilarly ,,"e sho,v that n{ither .II nor z is divisible
by q. Thus (t5, .),.) = (, y) = (t5, ,,) = 1, alld the nunlber  is a
divisor of the three numbers
:/3 + 2 !/3,  ,/,3 + .'/3, .r3 _ y3.
d is therefore a divisor of 3.1,:.3. Hence ,,"e have either  = 1 or
d = 3.
To pro\"e Theorem 123 it is obviously sufficient to prove that
(5)
1
1£'1 < I zll =  Z (x 3 - .lr) ·
Since !I;-6 0 and t i ;  - !I, we have
.),2 + x y + !12 = 1 {:? .1" + l/)2 + .'I 2 > 1,
and since x 3 -.11 3 is divisible by b. so is .r - .'1; thus Ix -!II > 
and
1.,;3 - .'f31 > b.
From this it follows tbat I.] I > I : I, and Theorenl 123 is proved.
Each of the equations
..e 3 _;... !13 = (j Z3,
,.3 -+- ) / 3 = 7 7 3
L"- . ..f I _ ,
.l.3 + 1/ 3 = 9.z3

248
CfiAPTER VII
has infinitely Inal1)" solutions. For the first one has the solution
[17, 37, 21], the second one has the solution [2, - 1, 1] and the
third one has the solution [2, 1, 1].
:Wlore generally, the eCluation
x. 3 + !/3 = (c 3 + l)z3
has an iufinity of solutions in relativelJ prime integers if (- is
an integer  0,  - 1 and  1. In fact, this equation bas the
solution x = c, y = 1, Z = 1. Further, the equation (.3 + 1 == 2 iJ3
has no solutions ill integers c and b, if c -F - 1 and  1 (Theo-
rem 122), and finally the equation c 3 + 1 = iJ3 has no solutions
in integers c and b if c  0 and  - 1 (Theorem 120).
67. The Diophantine equation 7 + y7 + z7 = O. - \Vben , '!J
and z denote the roots of the cubic equation in X,
(1)
.x 3 - pX 2 + qX -lJq+ ,. = 0,
we fil1d by means of Newton's formulae that
(2) x 7 + !/7 + Z7 = p7 - 7 r (p4 - ))2 q + (2) + 7 lJ 1. 2 .
'Ve shall see whether we can have
(3)
.1'7 + .'/7 + z 7 = 0
when 11, q and t are rational numbers. If x, y and l' satisfy
equation (3), it follows froJn (2) that the coefficients p, q and ,.
must satisfy the following equation:
(4)
i - ( 4. C) 9 ) ... ') 0
jJ - l,. }) - lJ q 4- q... + . P 1.... = .
First we suppose 1) r o. If we put fJ. = 1,2 fJl and I" = 1 /3 rl, it
follows from (4) that
7 ri - 7 rl (1 - ql 4- 'Ii) + 1 = 0,
whence
1"1 ===  (1 - ql + fJI) + 1 1 (1 - ql + qi)2 - }.
Here the square root must be a rational nUDlber. Hence, by
putting 2 qt - 1 = , where sand t are relatively prime (rational)
integers and t > 0, we ha\e

DIopnATINE EQt!ATIOS OF HICHER DEGREE
249
(5)
.r.: 4 + 6 ,').2 t 2 - ! f'" = U 2
, ,
where u is a rational nUJDber. Since 7 u 2 is an integer. u must
also be an integer, 3.nd therefore f is di,isible by 7. "r e clistin-
guish t\VO cases: according as f is odd or even.
Case 1. t is odd. Froln (5) it follo\vs that either s or If must
be even. Equation ([») Dlay he written in the forD1
(8 2 -t a /2)2 - u 2 = 1:/ {4..
and by the sanle argull1ent as in Sections 63-66 '\,"'e conclude
that
( (1)
f 8 2 -:- 3 t 2 + II =  a a4.,
l..2 + 3 {2 + It = fJ IJ 4 ,
where a and b are odd nat.uralnunlbers such that a b = f, (a, b) = 1
and a == 0 (mod 7); a anJ p are even natural numbers sllch that
a fJ = 64. Froll1 (6) we have by additi011
C) no e ') l ') a I 4 P l 4.
li'" = - a U'" ')... + ) ',.. a + - ) ) .
:..  
(7)
This equation implies the congruence
.,,2 ;::: - 3 -  -l...  (mod 8).
(8)
If we had a = 32 and p =, '\ve \\"'ould haf'e \",2 == -  (Dlod 8)
which is impossible. If we bad a = 16 ancl p =.+, we would
have ,..,,2 == - 1 (lnod 8) which is also impossible. For a = fJ = 8
,ve would ha-re the in1possible congluence 8 2 == - 3 (Iuod 8). For
a = 4 and p = 1G we would have the inlpossible congruence
8 2 == 3 (mod ). Thus '\ve must have a = 2 and p = 32 whicll
implies 6,2 == 4 (Inocl H). Equation (7) may be written in the
form
8 2 = - 3 a 2 lJ2 + ! a 4 + 1 (j lJ4
.
or
G-! S2 - (3:? lJ 2 - 3 a 2 j2 = ! a 4 .
Hence
+ 88 + l32 b 2 - 3 112) = + l4.
+ 8' - (32 lJ2 - a a 2 ) = +  d',

250
CHAPTER VII
where c a.nd d are odd natural numbers, and d is di,.isible by 7.
But these equatiolls do not hold modulo 8.
Oase 2. If t is even, both s alld 'll are odd in equation (5).
Writing (5) in the form
{b 2 + 3 t 2 )2 - it 2 = lJ:/ f4 =  (2 t)4,
we obtain
t:;2 + 3 t 2 + II =  a 4
-- .,
.f;.2 + 3 t 2 + it = 2 b 4 ,
where a and b are natural nUlnbers such that a b = 2 t, (a:- b) == 1
and a = 0 (mod 7). Then \ve get by addition
(9)
8 2 = -  a 2 b 2 + b 4 +  a 4 .
Either a or b is even. If b ,vere even, we would have
( /J ) 2
S2 = - 3 2 - 1 (n1od ),
whence, since s is odd,
( b ) 2
2 == - 3  (mod 8),
which is impossible. Hence a is even. Putting a = '2 at we get
by (9)
1 == S2 = - 3 {Ii + 1 (nlod 8),
Thus at is divisible by 4. It follows from (9) that
2 _ ( 12 _ :: ' 1 2 ) 2 = 1 £( 4
t.: U 2 " 1 2z.. r
and by the usual a.rgument,
-+- R +. ( b 2 - :I a 2 ) = + ') r 4
...:.. c :! 1 --..,
-t- s - ( b 2 - S a 2 ) = +  el 4
- 2r 1 -. '
where c and d are natural numbers such that at = 2 ed, (c, d) = 1
a.nd d == 0 (mod 7). By subtraction we ha¥e
b 2 = 6 (2 d 2 + {(:4 - ! d 4 ).

DIOPHANTINE EQUATIONS OF IIIGIIER DEGREE
251
Since u 2 + c' is 1Iot diviible by 7, we must take the upper sign.
fJ.'hus the equation is
(10)
r 4 + 6 c 2 d 2 -  ([4 = 1;2.
This equation is of the sanle forn1 as (5). But we have
f = (11 b = '"2 b (: d > d.
Thus b)p the Jnethod of infillite desceut we have shown that the
Diophalltine equation (5) has no solut.ion in integers 8, t and ft
when t  O. So we have proved that relat.ioll (3) does not hold
for })  o.
Consider finally the case 2J = O. Then. + 1/ -f- z = 0, and equa-
tion (3) takes the form
{,{,; + !17 _ (.t: + y)7 = o.
Now we have the identity
(.t' + !1)7 - :x,7 -. 1/ 7 = 7 X 11 (..r + .'/) (x 2 + :t"!1 + y2J2.
Hence we conclude: Relation (3) holds if and only if one of the
follo,vil1g four conditions is satisfied: L'r = 0: .lI = 0; x + y = 0;
t:2 + ."Lay + !12 = O. Thus we have established the result:
Tlzeure'ln 12.J.. If" r, 'U and z al'e the 1.00ts of a cubic eq'4atioll °u'itlz
,.ational cor:(ticiel1 t8, thi> relatioll
:,,7 + y7 + Z 7 = 0
lzold8 ouill in the .lollol,'il1g cases: 1. one of the nunzbers x, !/, z
i. equal to zero ,. 2. thi> 12ul1zuer.." x, y and .? are l JrO portional
to the 1'00t..,. oj" the equation e 3 -- 1 = 0 tak(Ju in a cO'lu"ellienf
order.
This theorem, which ,vas stated alld proved by V. A Lebesgue,
is, of course, also true in the special ca.se when x, y and z are
rational numbers.
68. Fermat's last theorem. - The Diophantine equat.ion
(1 )
XU + yft = en,

'> 5 '>
 *'J
CHAPTER VII
\vhere Jl is an integer > 2. has acquired great celebrity, thallks
to the fanlous statement nlade by Fernla t, In the Inargill of his
copy of Bachet's edition of Diophalltos's 4J rit/onl":fic:a he asserts
,vithout proof that equation (1) cannot be satisfied by any integers
all different from zero. To quote his actual ,yords: ('UbllJi7 (tufe,n
t.u duos clllJ()s, auf fjlladrato-quarlratu}ll in duos quadrato-ql:adrato8,
ct g(ll1eraZifel' nullanl. in illJinitu,n ultra fjuarlJ'atunl pofcstat()11l in duos
e.jusdenl nOlninis fatY est fl.iridere,: CI{jU."'. ,.ei d('JH0J2sfratiol1f1i2 lnirahilen7
sant d(texi. HaJlc ,nal.gini,,? (>xi{Jl(ita non (-a}ifrcf.
He thus believed that he had a really reJuarkable proof of the
theoreUl; but he never communicated hi supposed proof. (T p to
now Ferlnat's "last theoreul" has only been proved for special
values of the exponent n. III the case n = .t. Ferlnat rpallJ proved
his statenlent; Jle had only to replace ,,2 by 4 ill Theorell1 11-1-.
Apart frol1l the case 71 = 4 'e can evidently suppose that the
exponent 11 in (1) is an odd prilne }J.
It follo,s fronl Theorenl 118 that Fertnat's assertion is true
for }) = 3. Fernlat's theorell1 for }J = 7 is clearl)" a special case
of Theorenl 124. About 1825 Leg.cn<lre and. Dirichlet proyed the
theorenl for }) = 5: their proofs are httse(l on the Dlethod of
infinite descent.
Kummer was the first to succeed in proving Ferluat's la,f=;t
theorem for 1) = 11, 13 and for certain other larger prinle expo-
nents. In his investigtations of the eqnatioll
,r P + /I}J = zP,
p being an odd prilue, he applied the theory of the algebraic
field K (1]). \vhere 1] is a primitive }Jt.h root of unity. In the case
p = 3 in Section 65 we have alread)" used this Inetllod. Bnt,
when ]J > 7, the factorizat.ion of integers into prinle factors in
the field K (1]) is not unique. This fact g'ave rise to great tlif..
ficulties when applying the theory to Ferlnat's equation. But
Kumnler overcame t.hem by creating tIle concept of idea7::... The
theory of ideals, which lIas turned out to be of g'reat iInportallce
for the developlnent of seV'eral pUJ'ts of mathematics, will, ho'v-
ever:, not be treated in this volume.
'Ve shall only give the luain result discovered by K umlller
(1850) :

DIOPHA1\TIE EQI.iATIONS OF HIGHER DEGREE
253
1.( ji 1-". a ]{unnucriou prinle, the [)i()}Jhauti12e equation
(2) all + fJP = I"l
is not 8ulcllhlp in inie[/(:J's a, P (llld Y in K (1]), e.l.'cept jOl. afJy = O.
.A 1l odd prilue jJ 18 culled ]l(.n2nleriaJ1 (or also reg'ular), if it
does not J.ivide an)" of the llunlerators of the first" (1) - 3) Bel'.
noulli llu!nbers:
B - t B_1 B _1 1  - I B _:t B - Ht 1 B - ';' etc 1
1 - I;" 2 - :.0" 3 - 1:!, .J4 - :;v, 5 - Ijli' 6 - i:J(), i - ,;, ·
rrhe integ'crs iu K(17) are the lluu1bers of the f01'I11
+ . 2 +
Llo al I} -! a2 Ij
- a p _2 1 }1)-2,
,,,here ai (0 < Ii S P -) are rational integ(rs. Kumnler's proof
proceeds u,nalogously to the proof given in Sectioll 65 in the
case J) = 3. 1'he first part of the proof consists in sho,viug that
one of the nUluhel"s 0., {3, y in () is divisible by 1 - 'YJ. If y is
di,isible by 1 - I}, the next step is to write the left-hand side
of () in the form ·
(a + 131 {a + 1} P) (Ct -1- 1}2 fJ) , , , (a + 'fJ/J-l (3).
....\.180 in Kun1ulcr's proof the application of the lllethod of in-
finite descent is all essential feature.
If ".e ,vrite dO\YIl the first  (H7 - 8) = -1-7 13el'noulli nUlnbers,
\ve see that all prilntS < IoU are KUlllll1erian except 37, 59
anu 1)7. KUlnmer sho\ged, ho\\e,"'er, in a special investig'atioll,
that llis theoreln is also true for these prilnes and for certa.ill
other 1l0n-KUIllmel'ian priInes.
'The investigation of e(luatioll (:!) has been continued along the
lines sug'gested by KUD11ner, unu up to no,v Fernult's last theo-
reIn has been proved for all exponents at. least np to Jl = 600.
"But it has not beeu possible to show that there exist an infinity
of l\:\ullruel'iall l'rilucs. On the other hand 'e know that there
are non-KUJTIlUeriall l'ritnes la.rger thaIl a.ny g'j'Pll nuulber.
69. Rational points on plane algebraic curves. l\-Iordell's theo-
rent. - L et K be any giVe!l field. In this section an)? nunlber
1 For the ueHnition or the Dcrnoul1i I1nmhers see any intl'odul'tion to an:llrsis.

254
CHAPTER "II
belonging to K is said to be a ratiDJlal Jl1lilzber. rrhe point ('', y. 2')
in hOlllog-elleous coordinates in the plane is aid to be a ratiDna!
point in K if [)...., '!I and z are proportional to thr(e rational nUDI-
bers. In Sectioll 60 ,ve considered tIle rational points on plane
curves of the second degree. Here \ve hall estltblish a fe'v
results about tIle rat.ional points on plane curves of the third
degree, called silnply cubic curves. I
Let the equation of an alg'ebraic CUI.ve in homog'eneol1s coor-
dinates x, lI, e be
F (,I', !J. 2') = 0,
whel"e t.he left-hand side is a honlog'eueolls pol.Yl101uial in ,I', ,II
and z with rational coefficients. FroIlI the hOll1ogeneous equation
F (:c, y, 0) = 0 "re obtain the rational points on the curve which
are at an infinite distance £1"0111 the orig'in. If the cur"fe is of
the nth ueg'ree (or order), tIle polynoln.ial is of the same degree.
The singular points of the curve llla.y he obtained by the
systeln of equations
() If' iJ E" {J ]1"f
 = 0, 0 -:-- = 0,  = o.
uX 11 uZ
There are two different categ'ories of cubic curves: 1. The cubic
curves of genus zero or unicursal cUJ",es, i.e., the Cllrves which
have a singular point. . The cubic curve of g'enus one. i.e., the
curves wIlicll have no singular point.
The follo\ving result holds for unicursa] cubic curves:
Theorenz 125. The 6'inglllar poiut un a uJ/icursal callie Clll"l"e ifitll
rational cDtficielJt8 1.8 a rational lJoiut.
PrDof. Let the equation of the cubic curve in homogeneous
coordinates be 1/(.1',.lI, z) = 0, '1fhere 1" is a ternary cubic for111
in ;)', y and z. rrhere is 110 loss of g'ellerality in supposing that
the singular point (.l'o,.'/o, zo) is at a finite distance frolH the
origin, thus 2'0  o.
\Ve cut the cubic curve ]1' = 0 by the conic (polar curve)
1 In this and the rollo".ing section some elcnlentar,Y kn()wldgf' of the theory
or plane algebl'uic curves is supposed.

DIOPHANTINE EQlTATIONS OF HIGHER DEGREE
255
d 1" iJ l" iJ 1 1 '
a-::--+v-+c-=O
iJ .£' () !J 0 z '
w here tIle paralueters a, 1) and c are rational nUl)] bers. Two of
tIle six points of intersection coincide with the sing.ular point.
It is obviouslJ possible to choose the numbers 0, band c so that
the other four points of iutersection are distinct from each other
and from the singular point. The coordinates .r: of the six points
z
of intersection are the roots of the equation of sixth degree \vitIl
rational coefficients
f(:) = 0
r
which has the double root . Then the greatest COIUnlon divisor
,to
of the t,vo polJnonlials with rational coefficieuts f(u) ana f'(u)
is u - , which must also have rational coefficients. Conse q uently,
Zo ·
the lluInber .1'0 is rational, a.lld sinlilarlJ we can prove that "0
Eo Zo
is also rational. Thus 'rheorenl 125 is proy-ed.
Every rational straight line throug.h the singular point cuts
the curv in a third POillt ,vhich is llecessal'ily rational. In this
\va" we ohviously obtain all the rational points on the curve.
There are infinitely many of theIne
1'he01.PUl 12()t. If a cubic: (.Ufl"e u.(genus oue 'fitlz rafiuuol coe..t)icie2t8
has II rationalj)uiut, by Uleans oj" a trans.I()rlnafion u.'ith 1-afioual
(.ofj}iti(:nt. -it can be trau..1'urnzed .ntu a ru.bie CUrl"e oj' !t'hith
tlte eqllafiuu is
}7 2 = X 3 - t X-B.
ifhfJ'l) ....1 and B are rational JllonbtJ'8.
]J roq /: Let .f'(J:', .lI) = 0 he the equation of tIle curve in non-
hOlllogeneous coordinates, and let ])0 be a rational point on the
cur,e. There is no loss of generality in supposing that Po is at
a finite distance from the orig'iu. The tangent to the curve at
Po cuts the curve in a third point PI w hicb is obviously also

256
CHAPTER VII
rational. (If 1)0 is n point of inflection, PI and Po coincide.)
By taking 1)1 as origin ana the tangent at 1'>0 as the !I-axis, ,ve
define a linear tranSfOl'Ination \yith ratiolla.l coefficients which
transforlDs the equation of the curve into the form
(1 )
Cfa (,(,\ ,II) - f{'2 (.r, ll) + Cfl (.l'". !I) = O.
,,,,here Pi (.r, y) denotes a hOll1ogeneous pol)"no1l1ial ill .i' aud y of
degree i ,vith ratioual coefficients. Then, cutting the curve (1)
by the straight line ,1/ = 1.I' the nUlll bel" ,):, a part f1'o111 the value
:r = O. is determiuetl hy the equation of second degree
,(,,2 (Fa t 1. l) -(- ,/" q'2 ( 1, t) + q;l ( 1, t) = O.
FroIll this \\-C ohtain the following- paralnltric representation of
the curve
(2)
- tn 2 (1 t) + 1 1 ll (f )
.,: = - 't' 2 Cf3 (1. t) ,,11 = t.J:.
where If (t) Jellotes the polYllou1iai
(1' ( 1, t) --  lf1 t 1: t) g'3 (I, i).
This polYllon1ial has rational coefficients nnd is of the third de-
gree. For its zero:; are oLviousJy the slopes of the tangents to
t.he curve which pass through the origin. except tbe tang-el1t
,/" = 0 at tbe point })o.
Now, if we put at = X -  lJ 111 the expressioIl
R (t) = a t 3 + lJ 1 2 + c t + d,
,,"here a. u. c anu tl are rational lluluhers , we ()Iet
t'"
a 2 11 (L) = X 3 - .J X - IJ
\vith rational -1 and B. B)" Ineans of relations () Wl- t.hen arrive
at the follo\ving relations bet\veen the coordilJates X, IT of a
point 011 the cure
(3)
, 2 = '.3 - 1 ,.. - B
... ....\..  \a.
and the coordinates ,(', !I of a point 011 the curve (1):

DIOPHA:STI!\E EQt'" A TIONS OF IIICHER DEGREE
" 5 '-
... .
(5)
f · - - a f{J2 (a, ..t\. - A b) - a. 2 1"'
.1... - ') ( "'9' I J_ ) ,
"J era Cl,...\. - : u
l y =  (X -  b).I"
J " } 7 = a CP2 (.L''I .II) +  a CPt (..I..,', .lI) .
 '
:c.w
I '" - 1 ({ .'I
\. - :J 1J + --.
,1."
(4)
and
The systeln (5) represents a so-called birafiunal trau.1'or1uatiou,
which transforuls the cur\"e (1) into the cure (3): the systenl (-1)
transforn1s the curve (3) into the cur\"e (1) and loepresents the
birational transformation which is inverse to (5). These trans-
formations define a Ol1eone correspondence between the points
of the curves. Since the tra.llsforlnatiollS have rational coefficients,
each rational point on one of the curves corresponds to a rational
point on the other curve; the rational poiuts at infinity are also,
of course, taken into consideration. Thus Theorenl 126 is proved.
The theorems proved in Sections 62 and 63 may be interpreted
as results concerning the distribution of rational points in K (1)
on certain algebraic cur¥es of the type
(6) a X 4 + b ",\3 + c:\2 + d \. + e = A" }"2.
If [,r, y, .!] is a solution in integers of tIle Diol)hantine equation
(7) a ;,.4 + b ,)",3 Y + C :1.. 2 1/ 2 + d x 1/ 3 + e 1/4 = k z 2 ,
the point .x = ', J" =  is a rational point 011 the curve (6).
!I N
Conversely, if (",y 1"') is a rational point on (G) and if ...V is a
natural nunlber such that ...\r...\. and }.}... are integers, then
x = ]..T X, ,'I = l{, Z' = "2 JW is an integral solution of (7),
Theorenls 11-1, 11 () and 1 17 D1ay be forlnulated in the fol-
lowiug way: On the curves
.11-\4 - 1 = k 1 72 ,
whel'e k is either 1 or 2 or a prinle = 3 (n10d H), there are, apart
from the points (1, 0) and (- 1,0), 110 ratioual points at a finite
distance.
17 - 5166i() Tl'UfltOe l\...agell

258
CHAPTER YII
It is evident fronl the result ill Sectioll 63 that we obtain all
the rational points in K (1) on the cur,.e
2 .\4 - 1 = } "2
by Uleans of the rccul"si\e forluula
()
"'2i.) ,"'2 ' 1 ' ) 2 . ( '" + } " ' ) 2
r -40 1 \ - - \ 1 -r -r.. \ 1 - 1
+ X= ('>X 2 + - 1)2_ 0 )X 2 (X '- 1')2 '
...-40 1 _.. 1\4 1:1. 1
,vllen \ve start \\'ith Xl = )'1 = 1.
In a sirnilar \va)? the theorenls in Sections 65 and 66 ma" be
interpreted as results concerning. the rational points in K (1 - 3)
OIl certain cubic curv-es of the type
X3 + )3 = (Y.
The curve
(9)
}P2 = ..3 - ...l.. - B
is of genus one if and onl)? if
J) = --1-.:13 - 7 B2 r O.
The cubic curve (9) has the follo,,'illg paranletric representation
b" means of elliptic functions: 1
-40\ =  (u; --I- .. L -! IJj,
}' -==  ,/ (It; --I- .J" --I- B),
-.J \
\y llere --1-.....1 and --I- B are the ill ,"ariants. E\"ery point on the curve
(9) correSpOI)(]S to a certain argullleu t II \yhic h is uniquely deter-
Iuilleu apart fro 111 addition of periods; for the sake of brevity
we then speak of hthe point u'" Th(\ point it = 0 is the point
of inflection at infinitJ; it has the houlog'encous coordinates
0, 1, 0 and is therefore a rational point.
] f lit, "2, . . ., U R are rational ]Joints 011 the Cl1rYe it is easil)"
seen by nleans of the addition theorCllls for the fUllction8 &J (u)
a.nd ' (u) that the point
(10)
?U1 III + n12"2 +
+ JU,e: U g
1 See for example Gourstlt, CfjlO"S (1'..1 llalYf..r, tOllle II.

DIOPIlATINE EQrATIONS OF HIGHRR DEGREE
259
is also rational when nil  '}U2, . . .. JU x are rational integ'ers. In
19:!:! l\lorllell pro\"ecl the follo\Ting' theorenl 011 the rational POiHts
in K {I) all the curve (V) ''iheu .Ll and B art} ordinary rational
nllD1bers al)d J)  0:
There are on the curre (9) a .finit( ulonver vi ratioual })oiut,,,, "1,
U2, . . H if r such that tlte j'ur}}21da
(11 )
'}Hl Ul + Jn2 U2 -
+ }lir HI"
jitl'ni",h('8 the u'lzvl( set 0.( rational ]Joiut...: on the curre ,["hen 71/1'
1112. . . ., nz, J'if/ through ull rationed integ(Jr8_
"T eil (1930) has sho\vn that this result IS also valid in any
field K () ""here  is an algebraic number.
The least possible value of the l1ulnber ,. is called the reu'A-
or the curve ill the field. If r has this value, the set HI'
U2, . . .  'lt r £01'111 a uasi' of the rational points on the curve:
the points Itl, 1(2,'.., ll>, are the gell('rato}'c..
The point u on the curve () is uid to he C,Y'C('pt/vual when
the llU111her it is cOlnulellsurable "vith sOlue period of  (u). Uther-
,\.ie the poiut u is ordinary.
If at. least one generator point ill the basis of the cur\e is
ordinary, there are an infillitJ of rational points 011 the curye.
If all the r g'ellerator points are exceptiollul, tllere are only a
finite l1ulnber of rational points 011 the curve.
'rhe follo".ing' theoreUl, which "3,S l)rovcd Ly the uutlJor ill
1935: g,j'res a luethod for clptpJ'lniuing' all tIle eXleptjol1all'atiol1al
points 011 the cur\Te \vhcn the tielcl is the ortliuary rational ficlcl
K{l):
Lt.t X and }" IJf: tlte cCJurdinalt.<; (r an (L{'("f)Jfional J'lIliuuallJoint
,n K (1) (;1/ the c-urre {) Iflle}"(1 .1 aud Bare infl}gt/\..:. 1'lzeu X and
IT life illtt!/(,I".,', and }. 2 is either {{ diri,...,v}. of D = -l...-l3 - 27 B2
Of ==- O.
E ' . (ihtLtelet f 19-10) obtained a siInilar result for algebraic fields.
Billing (1 n37) anu other in\esti'ators have deterlllil1P(1 the basis
for a great llumber of Cl1r\es ,vith an infinity of rational points
in I( (]). For illstanCl). nIl the rational points 011 the Clll'\-'e
, 2 _ \. __ ..)
J - _. ...

60
CHAPTER VII
arc goivell by the arguDlellts l'uI, whel'c III is the argunlel1t of
the point (3, 5) and \vhere k = 0, -!- 1, -t- , + 3, etc.; the POillt
'U1 is ordil1ary. l'his curve has the rallk 1.
It ,vas sho\vn by Euler that there are exactly six rational
points on the curve
}"'2 = X 3 + 1,
namely, besiJes the point. of inflection at infinity, the following
five points: (2,3), (U,l), (-1,0). (0,-1). (2,-3); the corre-
sponding argunlents are k , for l = 0, 1, 2, 3, 4, 5, \vhere liJ is
6
the least posit.ive real period of the function  (u; 0, - 4). In
this case all the rational points are exceptional, and the rank is 1.
A g'l'eat number of siulilar results have been established by
other investigators.
However, no general lllethod for deternlining a, basis of a goiven
cubic curve has been founel.
If a rational point on the curve (6) is knowll, it is easily shown
that there exists a transformation \vith rational coefficients which
tl'allSforlns the curve into a cubic curve of the type (9).
Rplnark. The definition of an ex'cejJtional J}uint can be extended
to the case of the general cubic curve (,' of g'enus one as fol-
lows. If Po is a point on (', the tangent to () at Po cuts C in
a. point P1, the so-called taJ2gential point of ],)0' Denoting by P2
the tangential point of ].)1, by P3 the taugential point of P2. etc.,
we get an infinite sequence of points
}) 0' Pl. ]J 2' P 3, · · · 
wher Pm is the tangential point of Pm-1. If, in this sequence,
there are onl)" a finite number of distinct points, Po is said to
be an r.cceptional ])oiJlf. Otherwise Po is ordinary.
70. Lattice points on plane algebraic curves. Theorems of Thue
and Siegel. - In Sections 56-59 \ve developed a theory of the
lattice points on conics. Tbere does not exist so conlplete a
theory of the lattice points on 11lane algebraic curves of higher
degree. Invpstigations of the integral solutions of Diophantine
equations of the t.vpe

DIOPHANTINE EQUATIONS OF HIGHER DEGRE]
261
(1)
.f(x, y) = 0,
where I(.l', y) is an integral polynomial in .)J and y of deg'ree
> 3, lead to problems concerning algebraic nunlbers which can-
not be solved by nleans of elementary number theory. In this
exposition we shall only lnention, \vithout proofs, something of
the main problem ill the theory of Diophantine equations of the
type (1), nalllely: 'Vhich equations ha.ve an infinity of integral
solutions, and \vhich have only a finite number of solutions
including those which have 110 solution at all'? Thanks to the
inlportant work of Thue, Siegel and Iaillet this probleul is now
completely solved.
It is eas. to indicate algebraic curves of an arbitrarJ degrfle
\\"hich pass through an infinity of lattice points. For instance,
the curves gi,"en by the parametric representatioll
:JJ = j'(t), .II = g (t),
where .((1) and .q (I) are integral polynonlials ill t, have this prop-
erty.
On the other hand, a closed curve which has 110 infinite branch
can pass through only a finite n1.lnlber of lattice points. Thus
the curves
.(.2n + y2n = 1,
where 13 is a natural number, ha.,e only the four lattice points
(1,0), (-1,0), (0,1) and (0, -1).
For certain classes of Diophantine equations it is possible by
considerations with respect to convenient nloduli to show that
the equations have no integral solutions. For instance, the curves
2 :r 3 + t = 7 yn,
where n i a na.tural number, have no lattice points since the
number 2 is not a cubic residue of 7.
Ferlnat and Euler g"ave seeral exaulples of a. cOlllplete solu-
tion of non-trivial Diol)hantine equations of higher degree. Thus
Fernlat stated that the equation
()
.J3 = :! + !12

262
CHAPTER VII
has no other integral solutions than oX' = 3, .'I = + o. To prove
it \Te Inay- apply the theory of the field K (JI - ). The two fac-
tors .'/ + J -=  and .lJ - JI -  on the right-hand side of (2) are
relatively l'riJne: and therefore, by Thcorenl 118. they n1ust both
be cubes in K ( 1/ 2). But. froIn
1I + y  -== (if + rV - 2)3,
wher(\ II and l' are rational integ'ers, it follows that
.1J == u 3 - I.) It ,. 2 
1 = ;1" 2 l" -  ,.3.
rrh(, latter equation is onl. satisfied bY' if = + 1. ,. == 1. Thu we
have {:.stnbliheJ tL(, result. of FerInat. There are a great numher
of equat.ions of the saDie t:vlJe as () which lIH1Y be solycd C(Jl1L-
pletely by sitllilar Jnethods.
N o,' let
(3)
71
I.' (.r U) = a II (,/" - i /I),
i=l
(a  0)
be a. hOl110g'eneous il'reducibt. intgrul polynonlial in ,/' aud !I of
deg-ree n > 3., \ny alg'ebruic l1uInher (Spe Section 12) \'rhieh is a
root of an il'reducil,le equation of lith dT(\e ,\\Tith ra tionnl coef-
ficients is said to be of degTee il. Thus t lie alg"ebraie llumbers
1. 2' . . ., n are of 11th (1pg-ree.
jrllrther consider tlJe Diophantine equation
{- )
If VI;) ,11) =-= (; (./', 1/),
,,'h(\re G (:1::,?J) iF; nn integral polyuolnial of deg'rce 1n less than
11 iu '-1' and;1l. If we RllppOSe 1111  I,}'). t.hp.re i a positir con-
stant Cl sl1ch that ""C' have for snffieient.lv larg'e I N I
(n)
I (](x, !/)I < Clllll lll .
It foJlo\\"s froln equation (4) that ""e D1USt ha ve, for at leat one
of the"' factors ,r - i .11 ,
/ 110 -- 71l
I.r -  ill < 1 I ; I I ?11 m = ('21 !II ;; ·

DIOPHANTINE EQr.ATION'S OF HIGHER DEGREE
263
where t2 is a pOi3itie constant.. If;; is distinct fronl ;, it fol-
10"7s that
m
1. , : - i!ll = I ( - i) Y + ,[' - ; .Y I :> ("3111 I - f21 !/I n > c..1 y I,
where ("a and (".. are pOsiti,e contants. Hence fronl (.+) and (u):
("
I /' - t /1 1 .1(", I II I ,;: 1,'" I ' 1 1/ ) "'.
. , . 0,t" I (, I . I
and finally
( (1)
c.
'-
""
,
,i' (-
<.... -- .
/:' ... I ,III If-1ft
whel"e c' is a positi \"e constant. ,\"hich depends on t he coefficients
of (.+). By supposing I,'I > 1!l1 'fe should obtain an analog-ouB
1
inequalitJ' in ,vhich  was replaced bJ -, x by y and y by L/"
Thus we conie to tIle follo,ving' conclusion: If for ever)"  the
inequality (G) anu the analogous inequality have only a finite
l1unlber of integral solutions :-/', .'1, the nUluber of integral solu-
tions of the Diophantine equation (-1-) is also finite.
\Ve are led t.o the problclIl of finding especially good approxilua-
tiOl1S to a rea] algebraic nun1ber hy ra.tional nUlnbers. An itn-
portant contribution to the discussion of this question was given
in 1908 by' Thue, n..ho estahlished the following result: If  is
an algebraic nUluber of uegree JI > 3 and if c and e are positive
nUlnbers, the inequality
(7)
f: _ ,_/'

JI
c
<
) ?/I  n+l+
110lds for 0111.,'1' a finite nUlnher of integ-ers :r and y.
'rOlJl this it follo".s th3t equation (.+) bas a finite nUlnber of
integ-rul solutions if rn < - u - 1. In particular, if 111 = 0, the
eq nation
()
[4'(,/', J/) = (f,
,vhere (' is an integer, has n. finite nunlber of solutions.
In a pa.per published ill 1920, Sif!gel succeedd in proving that
Thue's result holds good erell \\"Len the exponent  Ii + 1 + E

264
CHAPTER VII
is replaced by 2 V; ill inequality (7). Thus equation (4) has a
finite number of solutions if Ul < 12 -  1/;.
BJ nleans of Thue's result for equa.tion (8) Iaillet (1919) de-
termil1ed all the unicursal cur,es W' hich })ass through an infinity
of lattice points. A unicursal cur,e is characterized by the prop-
ert.y that its coordinates ct.:, !I can be expressed as rational func-
tions of a paran1eter I. faillet proved tbe theorelll: A unicursal
curve passes tllrol1gh an infinity of lattice points if and only if
there exists a l1araluetric representation of tIle form
j(f) g (f)
.r == (It Ct))n' !I -= Ul Ct))1t '
where n is a. natural nUlnber, and \"\?here .f (t), !I (t) and h (t) are
integral polynonlials in 1 satisfying' one of the following conditions:
1. Either h (I) = at + b \vith (a, b) = 1 or Ii (t) = 1 ; 1(/) and
g (t) are both of degree n; 2. h (f) = a 1 2 + b f + c is irreducible,
and a > 0, b 2 - 4 a (. > 0; It f) and 9 (t) are both of degree 2]1;
the form a,,2 + b" l: + C 1"2 can represent for integral values of
'it and v a certain integer k  0 sue h that k n di vides all the
coefficien ts of both f (t) and (J (t).
In a paper published in 1 U30 Siegel proved that Maillet's
curves are the oul)" curves whicb })ass through an infinity of
lattice points. The proof of this tbeorelll is very complicated; it
is based on a generalization of the theoretl1 011 illequalit)"" (7),
and on a generalizat.ion of J\Iordelrs tbeorenl (Section 69) due
to Weil. This result of Siegel siguifies that it is al\va)"s possible
to decide whether or not a gi\yen algebraic curve passes through
an infiJlity of lattice points. "\Vhel1 there are only a finite nUl11-
ber of la.ttice points, it is possible by 111eanS of the lllethods of
Thue and Siegel, to determine all l1])per linlit for this number
as a function of the coefficients of the cur,e. However, their
methods give no algorithnl for deternlinil1g all the lattice })oints
on the curve.
Such algorith111s have been founu for special classes of curves
by means of quite different nlethods.
Thus, in 1925 the author of this YOIUlne sho,ved that the prob-
lem of deterlnilling all the integra] solutions x and '!J of the
Diophantine equation

DIOPHANTINE EQUATIONS OF HIGIIER DEGREE
') 6 -
... t)
...4 ,1;8 + B !l 3 = C,
A and B being natural numbers and C = 1 or 3, can be reduced
to tbe 1)roble111 of finding the fundamental Ul1its in certain cubic
3
fields K CJ /11l ), 11l being a natural 11un1ber. There is, apart from
a special case, at JDOst one solutiol1 with X.'I  o.
In 193R Ljul1ggren showed how the con1plete solution in in-
tegers :/.. and 1/ of the Diophantine equation
.A x 4 - B 1 14 = C
.1 ,
where A and B are natural llun1bers and C = 1, 2, 4, 8 or 16,
can be obtained by deterolining funda1l1el1tal units in certain
quadratic and biquadratic fields.
By the work of Mordell and other investigators tbe Diophan-
tine equation
.'r = .1.. 3 + D
has beel1 solved cOll1pletely in integers x and !J for a great nun1-
ber of integral values of D. For instance, the cOll1plete solution
of the equation
y2 = .£.3 + ] 7
.
is given by: x=-2, !1=3; Lt'=-l, ]1=4: x=2 11=5;
.1'=4, 11=9; ,r=8 11=3, x=43, y=282; }:=5:!, y=375;
x = 523.t., !I = 378(Hl 1 (sho,vn by the aut.hor in 1 HSO).
Other types of Diophantine equations maoJ' be sol,ed by 111eans
of a general method deelopecl by Skolell1 (1 fJ;Jfj).
Exercises
123. Show that every prill1e 1) = 1 (mod 6) can be expressed in
the form
p =  (a 2 + 27 b 2 ),
where a and b are natural nun1bers. It can also be written
in tbe form
p = ..1'2 + ;'}'.'I + 1/ 2 ,
where ii' and 1I are integers.

266
CHAPTER YII
12.t.. Pro,e the theorenl: Ever)" priule 1) == 1 or == 7 (111od 24) can
be expressed in the form
1) = :1;2 + 6lJ2,
\vherc ..I.' and yare rntioual uUll1bers; no other prin1es 11ave
t his property.
12u. Pro\"e the theOrt'l11: Every prime }J = .., 13, 23 or 37 (luod ..to)
can be expressed i 11 the forln
:'\ ')  C)
jJ =  X" + u ]''',
\vhert' .1' and 1I are nutural nUll1bers; 110 other prill1es have
thjs propcrt...
] 2U. f:;ho\v that cYcr" l'rilue ]J = 1, 3. 4, 5 or 9 (Iuod 1 t) call be
"'Pl'itten in the f01"Ul
C) I " ')
P = .r'" + .. y 1"" ,J !,..,
1 -)-
- . .
where .j" and .'1 are in tegers: no other priules r= 11 have
this property.
Deterll1ine the llulnber of solutions in integers ,1' and /' of
the Diophantine equation
I}. ')
,1....  .1'''' = jI 
\\'here II is the pr()t1nct of r priInes ".hich are = 1 (mod .t.)
or t ",'ice sneh 11, product.
128. D'etcrnline the l1uInber of solu tiOllS in integ-ers .. alld y of
the Diopbantine etluation
.) , I ")
.1.'''' T J:.1I -:- /'- = }J,
where n is tl1e product of r prilues which arc = 1 (nJod u)
or thrice sllch a product-.
I !}. For what prilnes }J is the Diophant.ine equation
.J 0') , >
"IJ - P t'''' = ...
olYahle ill integers If and 1.?
rrl1(\ 5U11ue qncstion for the Diophantine equatiol1
1,2 - J) 1"2 = - 2,

DIOPHANTINE EQUATIOS OF HIGHER DEGREE
267
. Show that the Diophantine equation
;,.2 -  ]) 1I 2 = - 1
is sol va ble ill h) teg-crs .)" and .1/ when 1J is a prin1e = 5
(nlotl 8).
Let J) and fJ bp t,,'o prhnes = 3 (n1od 4). Show that the
Diophantine equation
p ):2 - f} .1,2 = + 1
is sol \Table in illtgers ,(" and !I either for the upper or for
the lower sig"n.
Prove the follo\ving' tl1coreu1s:
1. Eve1"y prinJc p == 1 or 1 D (mod 2-1) ean he expressed in
the for!!1
}J = 3 .1. 2 - 2 y2,
,y here II' and !J are natural 11 uInbers. .No other priInes . .' 3
have 'this property. ""'f can choose .l' < Vp and 1/ < }/;.
:? Every l'riIne j: ==:> or 23 (mod 2-1:) can be expressed in
the forrrl
) ,) ') 
I =.. J'. - . , 1 ' ..-
}, _ . , t : '
\\7 here ,'I" and /I art) natural nun1 bers. No oth er prill1eS =:ft :!
ha ,"e t 1 tis pro pert. . We can choose ,/' < J  jJ and y < J f: p .
11 consecutive positive integ'ers have the property thnt the
SUll of their squares is itself a sqnal"e. For what \"'alues of
/I < 25 is this possible'
Prove the follo,,'iHg' theorem of Stl)]"mer: Let .1' and !I he
nat.nral nUlnbers ''rhich atisf.\" the equation
:J" - ]) /,2 = + 1.
,\"hcrc J) is a natural lJumber ,,"hich is not a perfect s(luure.
If all prilne factors of 11 divide f), the nun1ber .r + y J })
is the fundan1ental solution of this equation.
Suppose that "1 alHl '.1 are the l(\ast positive illtg.ers which
satifJ the cc InatioH

268
CHAPTER VII
u 2 - D r 2 = + J
- ,
where D is a natural number and not a perfect squal"e.
Sho\v that all the positive solutions tt and ". of this equa-
tion ma)? be obtained by the formula
If + tel' {J = ( Ul + l.t }in ) '1
 Q'
- -
where 12 = 1, 2, 3, etc.
'Vhat connection lIas the JIUIUber ! (Ill 1- 7:1 1 7)) with the
fundaIllental solutiollS of the equations
.).:2 _ D y2 = 1 and 2 - n 17 2 = - 1 
136. Show tJlat there are au infillity of integers of tllC form
:I;2 + ]  (:1' integer)
which are divisible onl. bJ prinles == 1 (n10d 8).
Sinlilarly for the integers of tJIe form
2 ,1:2 - 1, (x integel").
137. Show that thel"e are an infinity of integers of the form
3.1.,2 + 1, (,t" integer)
which are divisible onlJ by ])l"imes == 1 (mod 12).
138. Proye the theorem: If D is an integer, the congruence
.1'2 - 1) 1/1. + 1 == ( (n1od J[)
is solvable for all 1l1oduli J1 if and only if J) can be ex-
eI
pressed in the form
1 J = u 2 + l.2
,
where u and t' are relatively prime integers.
139. Show that the Diopl1antine equation
OX = 2 Y + 3 z
has 110 other integral solutions t,han the following:
I' = 1./ .= ;' .:: 1. I':: 1 1 1 == '>  = o. (' = C) Y = -t. z = -)
. .1 ... , · ..1  . - .. .., , · .., .

DIOPHANTINE EQUATIONS OF HIGHER DEGREE
269
t '+0. Find all relatively prilne integers ;(.' ana 11 and all natul"al
numbers n such tllat. the number x"" - !ll is divisible by
no prinle > o.
1'+1. Sho\v tha.t the Diophantine equation
XV - y.r = 1
has only the following solutions in positive integers: x = 3,
11 = 2; .1' = 2, 11 = - 1.
12. Prove the theorem: If ;l', 11 anu " are natural nunlbers..
the nUIllber
-1 "t' .'lu 2 - ,I' - .1/
is l1ever a perfect square.
1-13. Show tJlat tIle Diophantine equation
,1,4 - .114 = 2 pl,2,
where }) is a prime == 5 (lllOd 8), has no integral solution
with E  o.
1-1'+. Show that the Diophantine equation
,1.,3 + y3 =  2=3
is not solvable in iutegers jn the field K (1-' - 3) if "7* o.
1-15. Show that the Diophantine equation
13 + y3 = })Z3,
\vhere 1) is a prime = 5 or 11 (ulod 18), is not solvable in
the field K(J/ 3 ) if eo  o.
1-16. Show that the s)"stelIl
x + 1 = 2 y2,
J,2 + 1 = '2 .f 2
has 01l1J the following solutions in na.tural numbers:
.i:=ll=-.=l and :.1.:=7.. ,11=2, !=u.
147. Prove the following theorem: Apal't from the point of in-
flection at iIlfinity and the point x = - 1, 'Y = 0, there is
no rational point on the cubic curve

.'''' 0
-.
CHAPTER YII
./,3  1 = ]J .I/.?.
,vhere J) or - 1) i the prod uet of t1inercll t pl'iUICS '''' hich
are == 5 (tl1ocl 12). The sallIe result holds for ]) = - 1.
(CoIn pare Sectioll 69.)
1..t8. Sho\v that the Diophantine equation
4 .....') 4 ,)
.1.. - ').r"'!l M :- II =:-
is solvable in integ'ers Ollly for xy = u.
1-19. Sho\v tllat the cubic equation
.r 3 - a; -;- 1 = 0 
whel"e a is rational, can ne-rel" have three rational roots.
150. Show tJlat the cubic equat.ion
.1 3 - tl ,I' 1- :! -= O.
,vhere a is rationaL has three rational roots only \ylun a = 3.
151. Sho\v that the cubic equation
,/ 3 - It {IJ + , .:= 0
has three rational root.s for au infinity of rational values
of a.
1 ;;,, _ :-' . Show that tIle I .. I ·
v"/ po yno1I1U't 111 ."j-
OJ .)
./'''' + ,)' - a....
where a is rational, is reducible olJl\ for a = O.
Similarly for the pol)?llo111ial
.... ) .)
.1" + ,i. - :, fl':".
If>3. Pro-re the theorelD: There are an infillit.y of ratiounl nU1l1.
bel's a such that the polynolnia1 in .1'
i . ) '"
..1' - _ ,e - a-
is reducible.

DIOPUAl'iTINE EQr A TJONR OF HIGHER DEGR}::';;
271
15-1. Sho\v that the polynonlial in .,'
:1 :)
" - l' - (/ ,
wherc a is rational. is reducible onl \" for II = 0 tlllU II = + .
. - ,.
15. Sho,,' that the 1)0]ynoll1ial in ,J:
J.. ,3 - " -- -' l ,3
. r .. .
where {( is n rational nUluber  0, is a1 \yays irr(-dnciblE'.
150. Show t.hat the }Jolynolnial in .l:
4 'J
." + .,' - {l-,
,,"here (t is a rational lluluber  0, IS al\,.ars irreducible.
157. Sho\v that the pol)'noluial ill a.
::; I ) 6')
I' -- .. / ' - ii -
. ..... r ,
w here a is a rational nU111bel"  0 n.l1d  U, is 0.1 \yaYE ir-
reducible.
1u8. Deterlnine all the s01utions of th€' Diophantine equatioll
U 1:4
.. ')
I / == _...
r t. .
in integers x, !I antI z.
159. Deterllline all the integral solutions .1' and I of the Dio-
phantine equation
.) 'I 3
,r- -1- -t = U .
160. Sho,,; that there are 110 integers ,.1\ .11 and z .hich satisfy
the conditiolls .T' 1/  0, (x, !/) = 1 und the equation
/ 4 1_ // ,1 =  3
. I. w.
1 {jl. Sho\v that there are no integ-ers .i'. ?I and ? ,vhic11 satisfy
tbe conuitions ,J'!I   0, (.l', II) = 1 anu the cquatiou
,3 _ 1/ 3 _ .4
./ - -:: .
.
1 G2. In ,,"hat quadratic fields do there exist two J1ulubers ,J' auu
]1 ,vhich satisf- the equation
'). .. .) . I
.r- I .'r" == ,) :'

C)"'')
w..
CHAPTER VII
163. If Il is an oud integer > 3, deternline all the integral solu-
tions l:, '!J, z ,vith (.{', y) = 1 of the equation
J. 02 + -) J,2 = :;.n
.., ., .(1 ..' ,
expressed as functions of t\O -varin blps -u and 1".
Sho\v tllat x = + t oul)''' for !I = + 11, : = 3 u = u.
1 (j-I-. Show that the Diophantine equation
:;;2 + &1: + 1 = 3 yn,
where n is an integer >, has 110 solutions in integers :
and y for y rf + 1.
165. A number in the field K( V 7)
1+11 - 7
II + b 2 -
i said to be a.n integer when a and b are rational integers.
Sho\v that the factorization of integers is unique in K (V - 7).
Use this to prove tl1at the Diopllantine equation
:1;2 + & + 2 = 1/ 3
lIas 110 solution in rational integ'ers x and y if y  2.
Show that the number
I) + 
x'" ,
x being a positive integer, is a power of 2 only when 
has one of the following values:
J, 3, 5, 11 or 181.
166. Let 11'(:c) be a primitive integral polJnomial of the second
degree which has distinct zeros. Denote bJ' q (x) the num-
ber of square-free numbers in the sequence
]4'(1), F(2), J4.(3). ..., 1 4 . (&-r),
x being a positive integer. Sho'v that thel'e exists a posi-
tive constant 1c such that
q () > I. x.

DIOPHANTINE EQUATIONS OF HICHER DECREE
273
167. Prove the follonring theorenl: The Diopllantine equation
,('4 _ Y ( = ..1 7 2
.. ... .-,
where .A. is a positive integer, has either 110 solution or
infinitely nlany solutions in relatif'ely priule positive in.
tegers :1.:, 11 and z.
Suggest.ion. If [, 1], ] is a solution of the equation, show
that another solution is gi\"en by the formulae
:r = 4 - 'J}4 +  2 IJ2,
I II - 1:4 _ _4 -  1: 2  7 2
../ -  ' I -"',
z = -!  1]  { + 1}4).
168. Let p be a prin1e which is not == + 1 (Inod 16). Sho\v that
1) cannot be expressed a.s tIle difference of t\VO rational
biquadrates, apart fronl t.he case p = [) = ()4 - ()4.
169. Show that every natural 11ulnber 'Jl can be represented in
the fornl
0') .'\ 2 3 0) ( . 2
12 = .r'" j-  11 + - z'" -!- ) it ,
where :1',!/, . and It are integers,
Suggestion. Apply Bachet's theorenl and the identity
2.l1 2 + 3 ,,2 + 6 u. 2 = ('I -j- .? + 1l)2 + (- !I + " + It j2 -j- (z - :! it)2.
170. A 1jarfifiu1l of n. natural nuruuer II is a, representation of n
as t.he SUlll of any positi\-e integ'ra.l parts. Denote bJ (T (n)
the number of partitious of Ji into unequal part.s and by
U 1 (12) the number of partitions of Il into odd parts ,vhich
are eq unl 01' unequal.
Prove Euler's fornluln. (valid for I :1'1 < 1)
O'J
II t I + ;£':) = 00
k:el IT (1 - J.. 2h 1"t)
I, .:- U
I
and show by 111eaUS of it that
[?(n) = ["1(11)-
IS - 516670 Trygt't }."Pagell

274
CHAPTER VII
17]. If a, b, c are integers in the field K (}I - 1) such that a b c
is square..free, find the necessary and sufficient condition
for the solvabilit,. of the Diophantine equation
a .1. 2 + b 1;2 + c Z 2 = 0
..
in integers t', y, z (not all = 0) of the field K (}/ - 1).
Same question for tIle fields K (}I - ) and K (} / - 3).

CHAPTER VIII
rrHE PRIJ\lE NUl\IBER 1'HEOREl\J
71. Lemmata on tile order of I.I1ngnitude of some finite sums. -
Let x be all iuteg-ral or continuous variable ,,'hicIt tends to in-
finity. If 9 (x) is a positive function of :J.', \ve dellote by
o (g (x))
any function of [{ \vhich lU1S tIle l)l"operty that the quotient
fJ (g ex:)
- !1 (x)
tenus tu zero \\"hen x  00.
Thus :r = u t{':!), sin ,I' = () tl x), log.£ = (J (.r). Tlleoreln H (Sec-
tion 17) luay be enunciated: n: (x) = 0 (x). The priIne nU1ubel'
theoreUl (ectiol1 16) D13J' he formulated as follo\\'8:
( . .1 ( X )
IT Il') = 1 + u 1 ·
og I' og "C
}i'urthernlore, ,V'e denote bv
() (g (x))
any fUIlction of .l ,yhich has the property that the fjl10tielJt
I () Cq ex)) I
!I (x)
is less than a positive cOl1stant wIlen x  00.
Thus 2.1.' -1- J/ ;t' -= U (x). sin x= 0 (1), log x= 0 ( ]Ix ). Forulula (II)
in Section 17 Inny be ?rittell
(1 )
 log J)
£..J -= 10g'.1' -i- UtI).
- 1 )
IJZ

276
CHAPTER VIII
LeHi11Zll 1. l"hcJ'f ' ()..I.'il.r.l (I, l)O'it i re absol ute C01&".ta lit " sllrh that
'l'e hare
(2)
2: I = log !I + I' + 0 ( ! ) ,
nyJ y
tho BUnt e..t(:ndiug oreJ" all positil"e iutege}'' 11 < y.
lienulr1.. The nUlnher I' is known as Euler's constant.
Proqt: Let " be the least integer > 11. If we put
d ll = 1 _ log ( 1 + 1 ) ,
u II
it is }) lain that
:-,1 ( 1 :-1 1 ':.;1
(3) log z =:  log 1 + - ) =  l -  <5n.
1& = 1 II It = 1 t ,& = 1
It is kno,vn fronl the elenlentary tJleor)" of logarit.hms that
(-1 )
1
o < /l < .;-;;.
 }2'"
I t follows from this that tIle infinite series
QO
, 6
n
n-I
is converg-ellt and has a positive value ". Further we have
)\ <5n <  )' ]. < ! )' ( 1 _ 1 ) = 1 .
:: 2 n u- I 11-1 1/ (z--l)
- Thus we conclude froln (3) that
z- 1 ] 0
 - = log' ,. + " + - ,
11 ?
1& := 1
where 0 is n function of ,c such that 101 is less than a positive
constant. This fornlula clearly leads to (2).
Le1nna t.
Tlu J l'e exi.t,.-: a 1/ absolut() rOludant C. sllch that It'e hat.e
(5)
"" lon' Jl I) ( log II )
  =  (log 1/)- -I- e + () --. ,
ny n y
It"lzrrf' the .unz ext(Jlld1 orer all positire intp.lJers II < !I.

THE PRIME NUl\IBER TllEOREl\1
277
Proof. Let z be the least integer > !I. Clearly
:,-1
(log Z)2 =  [(log ')1. + 11)2 - (log 11)],
11=1
1
and since log' (n -I- 1)'= log- II + - b ll , we have
n
"-I "-1 (
2 -  log' )/ .., I 1 1.)
! (log 2') ==-  - --  c)n1og 12 + 1l - ') 2 - to) c); ) .
n-l n 7&-] 11 _Ji -
By means of () we see that the latter SUIll on the right-hand
side tends to a finite liInit c when z -> 00.
Further ,ve have
 ( )
 . 1 1 1 2
 6 11 log- Jl T - bit - :)2 -  6"
1r=% )l.;.l U -
, log' 1/
< ? ---
 ),2
II.=::
00
lo z J .. log x I . _ log z  log' 2'
< 2 + 9 (.1.,. - :! + + .
Z .1'" Z t' Z
..
Thus we conclude that
.: - 1
 1 0 11 =  (log ,)2  t + () ( 10: Z ).
" - 1
where c is an absolute constant. This formula at once gives (5).
Lenl1na .. I.t" 't' (n) dpuoff?8 the 1l11112b('r of positi,.c diti80r."1 o.l1'l,
"1£'(> har:e
 't' (11) C) ( 100" 1/ )
.£.J - =" (log y)! + :!" log.11 + Y" -:! t ,J.. () - ,
'liS,!! n vy
u'/zere the Slun extend..'! Ol:er all lJositizoe intpger8 n < ]1, and
'll'hel-e " and c are the Sa'l12e absolutp C012stantb' a8 in Leuunata
1 and 2.
Proo..f Since T(n) is equal to the number of pairs of natural
numbers a and b such that ab = 1i, we have
'Y. !J!D = 2:  ,
1I n ab

278
CHAPTER VIII
where the sun1 011 the right-hand side extend over all natural
11un1 bers II and b such that a b < II. Now let 51 denote the part
of this sUln in ,vbich a < 1 'y" let 'J2 denote tIle l)al't in which
h  J .11, and let 8 3 denote the pal"t in which both a < J'1/ and
II <:. J lI. Tben the alue of the required sum is clearly 6 1 1 + ,s'}.- ;:J.
1- 11
Putting- . = 1 .'1 and f.=.' . \ye have by Lelnmata 1 and 2
a ·
' _ , 1 " 1 _ " 1 [1 N , 0 ( a )]
# J -  - L.J -  - og +" T "-
«:abt b aza 'a ,Y
'-., 1 '" log- a ,., 1 . ( 1 ) '"'
.=. 100" 1 1 '  - - " "- .+ 'V / -j- () - 7 1
.   F""'-J 
a: a flZ a ll: a 11 a.:
= [log.1l + y] [ log z + y + 0 G) J -  (log z)! - (01- U ro,; Z ) + () C)
.. (1 ) 1). II 1 .J, ) ( log y )
= N og 11 - -,-  y o!J + ,.- - Co T { ir,; ·
I t is plain that S! = 8 1 . Further we have by Lenll11ata 1 and '2
Ss = (&l r = ((log z, ,. 1- 0 G ) r
:-= 1 {log !I):!. I' log l/
Hence
2 () ( log' !/ )
,., -Vy-.
1
 a b = 8 1 + Si - 8 3
== :, ( 100' 1/ ) 2 + 3 'V 100' '!I -i  IL':! -  (' - 1 ( 100" 7/ )  - 'V 100' 11 - "\/ I 0 ( IOY )
oj 0 · F 0 I J  · F::- · I 1/1/
( ) C) ,,\ .,  - ( log II )
=  log ?I - + ,. log 11 + ,.- -  c + () JI '
\vhicIl pro¥es LenlDla 3.
72. Lemmata on the Mobius function and some related functions.
The Iobius function !-l (11) was defined in Section 9. For any
integer h > 0, we define an arithmetical functioll by the equation

THE PRIME NUMBER THEOREl\[
279
9'h (11) = !L (d) (log d)h,
d
where the sum extends over all positive di"\"isors d of the natural
number n. (log d)O means the llul1lber 1.
LC1111na 4. If the 12atu1.al 11 lunbe1. n £s divis£blc by nlore than h
d(fferent pril1zes, 1('e harc
9'h (12) = o.
Proof This is true for h = 0 according to Theorem 14, and
so we cau suppose It > 1. \\T e use mathemat.ical induction and
suppose that Lemma 4 is true for all the functions 9'e(n) when
e < h -1. If we suppose n = ])" 1)Z, where a > 1, and where the
integer J12 is not divisible by the prinle p, we obtain
lplt (u) =  f.L (d) (log d)h =   !t (lI I d 2 ) (log ell + log d)\
d (11 (I;:
where the outer SUlll on tIle rigllt-hand side extends over all
positive divisors d 1 of 11l and the inner 8UDI over all positi"\"e
divisors d! of })". Then
CPl. (11) =  ()  fL (d 1 ) (log d 1 )"  It (d 2 ) (log d 2 )1.-.
=  (  ) <px (nz) f{h-s (pa).
O ....
....:.
Since 1l has nlore than It different pri DIe factors, 11l lIas n10re
than h - 1 different prinle factors. Therefore by hypothesis
cps (nl) = 0 for 8 = 0, 1, . . ., h - 1. The remaining ternl 9"& (1)1) <Po (pa)
is also equal to zero, since its last factor is zero. Thus the lemnla
is proved.
Lel1l1na i). TJrhen x Z"S a )Josi ti re 121U1i be}", uoe j)1( t
( X ) 2
A (d) = It (d). log d
and
f(u) = A(d),
d

280
ClIAPTER Vlll
l('/ze,-e the SllU2 extends 01:e1' alll)()sitirp dit'is01-S d of the positit"c
integer u. Then u:e haz:e
f(1) = (log X)2;
,t'(jJ a ) = - (log 1))2 + 2 (log .r:) (log ]')
u'heu 1) i8 a JH"i1ne a nd a (Ill illtge'  1;
.f(jJ a fj{) = :! (log })) (log q)
U:hl'J1 p and q a,'c d(tfe,.ent )1rhJ2(,,", and a aud pare iutegers > 1;
f( 12) = 0 iflzen '11 is d h:isible 1)/1 tll1"ee or 'Jnore dt1erent p1"i,nes.
The proof follows imn1ediately from the definitions and Lelnma 4-
(for h = 0, 1 and 2).
LeU21na 6, For et'cry natural J21t1nbe1' "J.' l,£"e hat.e
, J:L (d) < 1,
 -
rl=1 d
Proof It follows from Tlleorem 1.1 that
.1'
1 =   f-t (d),
u=1 rI
where the inner sum exte11ds o,.er all !)ositit'e divisors d of the
positive integer ll. Hence, since the nUluber of multiples of
d ;;:' -1 and < x is equal to [J]'
 il (d) [ . ] = 1.
(/=1 d
Consequently
x  It (d) - 1
d=1 d
.., "- ( d ) ( .r: - r .r ]) < , ( :?: - [  ]) < "[' - 1
...;.J r' d rI - _-f-J = 1 r/ d - · J
"-=-1 u
Thus
.r " ,II (d)
 d < l+.c-l=;r..\
,1=1
which proves the lemma.

TIn: PRIME NPMBER TIIEOREM
281
LenHll a i".
]..01" ('l'Cl.!/ posifivc x tle hat'c
(1)
 !l (rI) log x = 0 (1 ),
d:r. d d
'1f.'here the stun e.rfends ore}" all lJosifi'L'c intp(}Pl.. d <. x.
Prou.f. Applying Lemnla 1 we find that tIle left-hand side of
the assertion is equal to
 p. (d) (  ! - i' -I 0 1  ) ,
dJ.' d 71t n
J..'
where t = d alid where I 011 is less t.han a positive constant C 1 .
For d u = 111 tllis becollles
 1   p, (d)  ( ) 01
2.J -  p () - l'  -. - +  p d - 
m ill d ,/.t d (I% X
where d runs tllrough all positive divisors of 712. By' TheoreIll 14-
the first SUIn has the value 1, and by Lemlna 6, j nst pro'fed,
the second sum has an absolute value < y. The absolute value
of the third sum is at most
ft " 1 -
-  - C 1 ·
,/ d%
This gives the desired result.
Le11Ulza 8. For ez'el'l/ uafll'al 'Illl1nber n ft,'e hat'e
.
(2)
p.(d)T(j) = 1,
the Sllln pxtf!12dil1g orcl' all positire d:iz'isol"s d o..f n.
Proof. Since T () =  1, the sum extending over all positive
n
divisors  of d' the left-hand side of (2) becolnes
 p, (d)  1 =   It (1)'
,1 f1  cJa

282
CHAPTER YIII
where the inner SUUl ou the right-haud side extends over all
positive didsOl'S 0 1 of . By Theorem 14, this inner sum is equal
to zero \\"'hen () -;.d. n and equal to 1 when d = Jl. Thus the right-
hand side is equal to 1. Q. E. D.
Lc'rnnza .'). I.'vr el"f'1"!1 })(Jsitire .L' Ice hare
(;3)
(1)( ) 0')
--, ( :"1: ..
 log- =210gx+ ()(1),
tl:a;c d d
n.here the Slun extends Ol"er all liosilite integcr.-.: d  x.
Proof. Applying Lemma 3 for !I = ft, the left-hand side of
fornlula (3) 1113)" be \vritten
,, " /1. (d) [  T: (n) .) I ,'}' 2 T ' 2 c]
 2.J - ,  - - - j' og' - / - y -
fll...,( l "clx n (
i-  It (d) ( 0 ( d )  100' :r ) ,
d:i:x; {l ,{'  cl
where 101 is less t.ha.n a positive constant c. For all sufficientlJ
large x the absolute valne of the last sum is snlaller than
t ",' 1 ( d ) t ( 1; ) :1- ( X, ) (J  )
c d -  =.i'-.l() ll-t =x- 1 0 z-:-d: =0(1).
dSx x dz
- 1
Further, by l)utting k = 11 d, we have
s = :?  fL (d)  T (n) =:! 1, t !l (d) T ( I.: ) ,
(t d kx n k,1; I.. (L d
where the inner sum 011 the right-band side extends over all
positive divisors d of k. Hence, by means of Lemn1ata 8 and 1,
 =  log at" + 0 ( 1 ).
Finally, applying' Len1n1ata 7 and 6 we see that the left-hand
side of foru1ula (3) is equal to
2 log .v + () (1).
Thus Lemn1a 9 is proved.

TIlE PRIME NUMBER TIIEOREl\1
283
73. Further lemmata. Proof of Selberg's formula. - By Ineans
of Theorem 30 we prove
Lr11Z1na 1 (). The BUIll lJein[1 extended or:er all pritnes p < x, n.e hare
(1)
 (log lJ) ( log . ) = () (;t" log .1').
]> :i. ;r }J
[)}"()o(. I f !J = I x , the sum on the left-hand sjde js e( J ual to
· ogx
 (log p) ( log L' ) +  (log p) ( log  )
py p pp 1
< (log J')  log p + (log log x)  log P
py pJ
= (log (") () ( 1 x . ) + (log log "..) f} (.:c).
og J.
.Applying Theorenl 30 ,ve see that this function has the o,"der
of magnitude
U (x log log .'('),
which is somewhat better than (1),
Lennna 11. The sunl bi?in[l cxff'nded orer all }Jrl11le lJOll"erS })a < ,:,
,,'here a is a natural J2UHlber, lty) hat'c
(:!)
.log]1 = 0 (.r).
Proof. The sum on the left-hand side is obviously equal to
n

o (.r) + o (V x) + 1?('J/ x) +.
1:
/-
+ 0 (1 :r),
lvheI.e k is the greatest integer such that 21.: < x; therefol"e
k < log x Th . · t 1 t
= log 2 · IS sum IS at mOB equa 0
o (Ie) + !, 0 l J I x) .
From Theorem 30 it follo,vs that it lIas the order of magnitude
log x r:
U (x) + . 0 (1 x) = () (x).
log 2

284
CHAPTER \"111
Le1nn1Q 12. I.l I(n) i/j' the .fi.l'l2rt-ion df'.(illed in Le1,una tj, u:e hat'e
/(n) = (log x) {J (.r) + 2 J' {J () log p + 0 (x log x),
'u.he}'e the SU.I11 OJl the left-ha.nd s-ide (',rt(}nd,,' Ot'e1. all }Jositil"e
'inte(Jf1.s n < .r, and u"lzere the .'lin on the l.ight-hand 81.de
,-
f>.r:fends oter all l)rhnes 1) < .'1 = 1 ,L'"
Pl'oo.f. It follows fronl Iemnla 5 that
(3) j(1l)
7&  J'
= (log X)2 +  {2 \log x) Clog-1 J . .- log 1) 2) + :!  (log p) (log q).
Here the first sun1 on the righthand side extends over all plime
l)owers l)u < ,}, a being a. natural number; the second SUln on the
right-hand side extends oyer all prinle powers per. and f)1 such
that ]J" fjl3 < X and p < f}, where a and p are natural nunlbers.
In the first sun1 011 the right-hand side we first consider the
terms with a > 2. If we denote by g (.r) the number of prime
powers J)" < x with a  2, we see that the contribution of these
ternlS to the SUIJl is at Ulost equal to
:\ k
2 (log .I')!' 9 (j;) < 2 (log X)2 (JIX + Vx + ,., ; J/J')
when k is the greatest integer such that 2 k < J..... Thus the con-
trib1.1tion does not exceed
2 (log X)2. k .IX <. 2 (log X)2 · :; V x = 0 (x log :.r).

Consider next the terms with a = 1 in the first SUUl on the right.
hand side. The contribution of these terms is equal to
 (2 (log x) .Jog p'. - 'log 1) 2)
p.r
= (log.1')  log p +  (log p) ( log  ) = (log x) () (x) + () (,"log Ij'),
px p P
according to Lemma 10. Thus the first sum on the right-haud
side in formula (3) is equal to
(4)
(log :t) -0 (x) + () (x log x).

THE PRIl\lE NUl\IBER TBEOREl\-1
285
Finallr consider the seco11d SUlll 011 the right-lland side. Apl)l)"ing
Lemma 11 for x ; instead of for x, we see that the contribution
lJ."J
of the ternlS with {J > :? and a > 1 has the order of n1agnitude
 ( X ) . "" loa q
logq.O R =U(x)£..j 0., =0(.1:);
q' q.
for the infinite series  log' q extending over all primes q is
 ql
clearly convergent, si11ce (J > 2. Thus the second SUln on the
right-hand side is equal to
(u)
  llog 1)) (log q) + 0 (./.),
the SUIll extending over all primes p and q, such that p q < x
and p < q. If \ve put y = J/', the latter SUIll is equal to
 (log p) (log q) -  (log p)2
p pp
=  (log })) (log q) +  (log'}J) (log' q)
py qy
pq pqz
-  (log'p) (logq) -  t1og' p).
py py
f/:fJy
According to Theorem 30 the last two Sluns have at D10st the
order of magnitude
(1J ,Yx))! = () (J o )
and
(log 1;;) () (V;;) = 0 ( Yx log r.)
respecti\Tel)". Hence ,ve conclude that expression (5) is equal to
 (log ]J) 0 (  ) +  (log q) {J. (  ) + () (.J') = 2  (log 1)) {} ( .) + 0 (.r).
pray p qy q jJ!I P,
Introducing expressions (-1) and (5) into forn1ula (8), 'e finally
obtain Lenlma 12.

286
CIIAPTER VIII
'Ve finish with the proof of Selberg's basic forluula.
Tlteoreln 127. Putting .11 = }I x, 'll"e hate
{) ( x ) 100'' + 2 ')  .f) ( .. ) 100" J } - 2 x Iou. x = 0 ( X 100. ,t').
  I ') e  e
}J  !I
Proof: According to Lemlna 12 the left-band side is equal to
 1(12) - 2 oX log J.' + () (x log .A:).
n.r
According to the definition of the function I(n) we have
s = .I'(1i) =   ). (£1).
1I.c ,,:r d
\V}lere the inner sum extends over nIl positive divisors d of Jl. Hellce
s =  ;(d) [  l] =  i (d) ( 1- Ed ) '
tl;az l d .c (
where 0 < Ed < 1. By putting z= (lO;x)! we ha.ve
 i A (d) I <  ( log  1 .. ) 2 =  ( log  ) ' 2 -1-  ( log  ) 2
,,.: d,t l d= l tI:
< z (log X)2 + 4 x (log log V)2 = 0 (,(.) + (J (x log log x.!) = 0 (&/' log :r).
Hence
 x  u (d) ( X ) 2
S=;'(tl)l+o(.dog.r.)=JJllog-, +o(xlog.c)
rl:a ( (l;;e l l
and, by Lell1ma 9,
S = 2 . log ; + 0 (x log &}').
rrhis proves Theorem 127.
74. An elementary proof of the prinlc number theorem. - Eve!.y-
thing is now prepared for finishing the proof of the prime
11 umber theorem. "-IVe have seen in Section 17 that this theoreUl
is equivalent to the proposition
(1)
I - ,0 (,/,) 1
I III ===.
.(-,)000 x

THE PRIl\IE UMBER THEORE}I
287
It follows from Theoreul 30 (Section 17) that for increasing x
the quotient
{) (J:)
:r
lIas a positive lower lilnit a and an upper liIrlit ..:1: then () < {{ < ..1.
Thus, to prove (1) \ve have to sho,,' that
(2)
a==_l=l,
The following' proof is Inaiuly based on Se!bc1.g's forulula which
was proved in the preceding section:
(3)
D (x)   f} ( :1.' \ I ) ' 1 )
- + -- 7. og JI - :. = 0 ( .
x x Iocr &'{'  1} I
t:) jJ ==- !I
the sun1 extending over all prhues p < !I = J/ :1', "or e alo need
the folIo wing forln ula, \'" hie 11 \VUS proved i11 Section ] 7 :
, log' }J .
L.l = log .(.' T U ( 1 ),
jJJ; p
(4)
the SUln extending over all prinles 1)  ,.I:,
"1 e first pro,e
L I · I ' f} {x} 1 , . · I " . f 1? (x) I
enUlla 1.:J. J lln sup --:- =..: 111;( 'I 1111]11 - = fl, ire. 10ft'
.r-CI'J ,,(., · .(-. x X
(5)
a + _1 =. :! .
Pruo): I t is possible to let £ teI)(} to infiuity in such a 'iay
that {) (x) tends to ...1. If F is a given poiti\"( number, we have
x
{) ( 'I.' ) . ,1'
- > (a - eJ
]J })
for every J' sufficiently large ant} for every Itrirue }J < !I = V x, and
therefore
  {) ( ,: ) 1 > 2 (a - B)  lo }I .
/ . 00'_
if: 10 0 ' ; .. } ) ".P - to')' x  P
I t:'t" jJ!I e Pll

288
CHAPTER YIII
] t follows from (-1) that tIle right-hand side of this inequality
tends to II - e when ,I" -+ 00. Thus, applJing formula (3) we get
 --...1 :.."> a-E. Since this holds for every positive e, we con-
clude that
(6)
.4. + a < 2.
On the other hand, it is possible to let L{: tend to infinity in
such a way that () (.r) tends to a. If E is a given positive num-
,)'
ber. ,ve have
() (  ) < (A + e) ,:t'
1) jJ
for every ,I.: sufficiently large and for ever}" prime 1) < 11 = J I x, and
therefore
2 2:. {) ' ) log P < 2 (.J + t')  log P .
I log x py \P log x p u 1)
It follows from (-1) that the right-hand side of this inequality
tends to .Ll + e ,vhen a -> 00. Thus appl)7ing fOrDlula (3) we get
 - a < 4 + e. Since this holds for every positive £, we con-
clude that
4. > It)
.."1 -r a = _.
This inequality together ,,,,ith (6) leads to (f»).
In the follolCi ng If(1 al,fa/ls let the l'll J'ialJle x tend to i nft 1zity
i J2 sueh a l1zanner that  (x) te1uls to ...-t.
x
Lellvna 14. If)" is a giz'eu llul1zber > ll, aud Ij' the SUl1Z
S(x) = }" lo P
."'-J 1)
r:dends Ol'el' all primeR p < .r.' ami sllth that {} () > x , the-II
S(x)
tltr quotient 1 t(ncls to zero Jar . -+ 00.
og' J:

THE PRIl\IE 1\ rl\IBER THEOREl\1
289
Proa .! : \\"7' e have b y } )uttinn' "/ =.:::. h
,..  . ,
"' ., ( . ) ",,' x'
 0 log }) -=  log 1) L.J log fJ
JJ.f. P 1'J'}> 'i::-
 ( ;). )  ( J. ) ( ) 2
=  1) ;, log jJ +  1) . log q -  log P
]>!I 1 fJU fJ. p:ay
= 2  U ( .1' ) log}) - (D(Vx)):!.
P1J ]J
Since, bJ Theorem 30, the last term has the ordel- 0 (.1'), we see
that Selberg's formula (3) ma)" also be ,vritten
(7)
D (i.C) 1, ( oT )
-+ 1 /,f) - logp-2=o(1).
x .t; og X J:r 1) .
Let e be a posjtjve number. For every x exceeding a certain
p
( X ) X
llun1ber it which depends on e we have {} - > (a - e) . rrhere
1) )J
exists a l)ositiye number b depending on tl and so 011 £ such that
(8)
.{} (  ) > (a - e)  - b
P 'p
'1"
for all the primes 1) such that' < u. Thus the latter inequality
p
holds for e'\er,y p < x.
If the sums ' extend over nIl primes p < x such that
{} ( ' ) > i.. x ,,"e have
} J 1J'
(9)
' Q (  ) 1 >., ,log"' p ( '3 _ ) ,logl)
 'U og }) = /I, x  > II. tl X 
l'  }> p
' 100" p
+ (o - e).1' L.J  .
P
If the sums " extenu over all pril11es p -:< x such that
{} ( X ) ' ).. X ( )
- < -) ,,,,,e have by 8
P 1)
19 - 5166;0 TT!lgve Nagell

290 CHAPTER VIII
" ( X ) ,,100' P
(10) LJ () p log}J > (a - e) x 2.J ; - bfJ (x).
Fron1 (9) and (10) we deduce that
 D (  ) log p = ' {) (  ) log p + ".f) ( 'I! ) log 1)
}I'/' 1 P P
 100' }1  ' lo 'P
> (a - e) x   + (A. - a) x .LJ -1J{) (x).
p.l P P
Substituting from this result into formula (7), we obtain for x
8 (x)
tel1ding to infinity in such a manner that 7 tends to ...J 
' !og J

..-1 + a-e + (A---a) Lirn sup ]J < .
;r log x
Hence, l"ecalling that a + A = 2,
V" log 1)
'" =
lim sup ...... p
.r log .f'
< e
=A-a
Since A. - a > 0, and since e can be chosen al"bitrarily small this
gives the desired lemma.
L1'11lnZa 1:). f Il is a gi fen positire Jlflt11beJ" < ..1, a11d '1,1 the siun
R(x) =' CO;l J) (-iq)
extentls ot.eJ. all the ]Jritnes p and q It:hick satis./y the ,f'ollonojug
conditions:
p < t x, q < 1/  ' () ( :q ) s ;: '
then the quotient
R(x)
(log .l.)
feud, to zero for .r;  00.

THE PBI1trIE NU1trIBER THEOBEl\1
291
x
Proof Replacing x by - in Selberg's formula we obtain, bJ
p
tt - V 'lO
pu lng or = p ,
( x ) ) x ( x ) 9 ( X )
{} =  + 0 - -   {} - - log q.
P P 1) logq-e pq
p
Introducing this expression for {} () into the same formula. we
get, by putting :lJ = JI x,
2 x  lOll P 4 J'"
-0 (x) = 2 x + 0 (x) - ---- - (2 + 0 (1 .) + - -- ,
log x p log x
p:!. ,II
where
v =  {} ( . ) (lo p) (g' q) ,
p, q P '1 100' -
h jJ
the sum extending over all primes p and q such that p < V x 
< li re
fj = -. Since, by formula, (4),
1 J
lop _ 1 1 0(1)
 -;i og x + ,
P1I P
it follows that
41'7"
{) ( J o ) = 1 - + 0 (x).
. og
... -
In every ternl of the sum l'" we have p < V x and q < V  , and
. P
therefore lJIJ. = p (pq!)t < x J . Thus, if  is any positive number,
we have
{} ( - ) <(.A +) x
pq pq
for x sufficiently large.
Let us write
7 ' V"
J=+,

292
CHAPT:ER VIII
where the first sum extends oyer all primes 1) nl1d q such that
p < J/;(;, q < 1 / x and {} ( x )  !L , and the second sum over all
. p ptj l)q
prinles lJ and q such that }) < V;, q < 1 / and {} ( x ) > /L x .
l' P'l pq
TheIl we have
J r < ' 1 log p log q ( 1  ) " 1 log p log q
= p, ,v  · · +  + u .}.  - -. · ,
l xP q I xP q
o- o
p p
where the SUlns are taken as above. 'Ve further get
, 1
JP < (A + t5) ;)" JfT - (.4 +  - p,) X 7,
...... 3.'0
100'
p
lo P log- q
. _e_.
}) q
where
JJ' =  1 . lo!? P . lo£!' q =  _ 1_ . log- P  I?£!, If ,
I x P ( 1  .1.' J J .... (I
p, q 00" - lJ1I lo q 7: :L
 ]J b 1)
the sums extending' over all primes jJ < ]1 = Jlx and over all
primes q <  = V -:x: . Applying formula (-t) we obtain
]J
 lo p
W =  - (+ 0 (1)) = llog x + 0 (log ,1:).
p:iy P
Hence
-t. ,,' 1 log'}) 100' q
D(r) < (...4. + t5)x - (...-1 +  - /L) x  -. · - + l}J.'O
log x  x p q ,
10<7 -
 1)
where '1J tends to zero for .1' -.. 00. From this we deduce tluLt
4 Q (.A + IS - fL) L, dog )! . log- q < A 4.  _ D (.r) -j- rJ.
(log x)"" }J q .f;'
where the sum is the sanle as in Lemn1a 15. Hence, for.t:
tending to infinity in such a manner that {} x) tends to A,

TDE PRIl\IE NUMBER THEOREl\I
293
1 ' loo- p 100' q 6
lin1 sup oJ / I  · '""' <
J:->OO (log :1')..  1) q - --l: (-1 - !l)
Since ...1-!L > 0, and since b is an arbitrarJ positive llUlIl bel',
this proves the lernma.
B" 111eans of Len1Jnata 13, 1-1- and 1 f) it is now easy to prove
the relat.ion
(11 )
tl = ...4. = 1.
SUpl)Ose that t ).'. a. Let a be a posit.ive number > 1 such that
aa < :1, and let  be a positive nunlber so small that
(1 )
1 - II a ::::  a -j- 2.
Furtller. denote by.'" a natural number.
We con sidet" tIle SUU1
.s =-  logp . og q  log' ,," "
2 P q 3 j"
where )'. extends oyer all the priules }J and q such that
2
" V '- x .". ( X ) :£
1) < "3 q < -'}Jq > 1\. -0 -- > (1-)-
. P 1) q 1) !J.
and where  extends over all the primes 1. such that
...- :.:
J)q < r < al J q.
a
If there are no primes 1., the sum 3 is = O.
For every term in the sum  we have
. 3
r <. ap q -= api (1) q2)! < ax1- xi = ax! < {J'
when :1.: is 8ufficiently large. For the same terms we shall prove
the inequality
{ 13)
( x ) x
'0 - > (a + )-
r ,.
wlln x is sufficientl)"p large. This inequality is true for all
r < }) fJ, since

294
CHAPTER '\"111
() (  ) > {j ( _X ) > (.A _) "7 x > (A _) x > (a + ) x
r pq pq ar ,.
in virtue of (12).
Consider now the terms with ". > p q. If we put ;r = II and
I-
x
-- = Z', we get 'It < l.' < au_ If in Selberg's formula
pq
(log x) {j (.r) + 2  D (  ) log]J == 2 x log x + 0 (x log x),
p:i!' jJ
where y = JI x, we first replace x by v and then x by tl, we
obtain on subtraction
(log r) {) (l:) - (log .u) '9 (14) < 2 l-log t' - 2 "log" + 0 (ulog u),
or
D (II) > 1 1 - o l' () (t:) - 2 (I' - u) - 2t1 Io  - log If + 0 (u).
og 11 og 'I'
Thus
{) (ll) > (A - ) 1: - 2 (r -- u) + 0 (It) = 2u - (2 - .A + ) l.' + 0 (ll)
aild. since a + A = 2 and 4.4 - (l a > 6 a + 2 6,
{j (u) > (a + 26) 'If + (J (-u),
If x is sufficiently large, we therefore have
( x ) x
'9 - = D (u) > (a + d) II = (a + 6) .
r r
This proves inequality (13) for aliI"
Consequently
s <  lo r  Io p . log q ,
r r  p '1
where the first sum extends over all the primes r < x and such
that {j ( X ) > (a + ) , and where the SUD) , extends over all
r,. "
- V X 1-
primesp and fJ such thatl) <. J'x, q < , - < }Jq<ar.
jJ (J

THE PRIME NUMBER THEOREM
295
Thus we have, by putting y = JI x and f = a r 
]J
 loa' P . lo q < a  100' '"' loa'
' P q - r  e P  e q
py qt
=   (log 1)) {} ( a r ) < C 1  log P < c log x,
1 1JY P p$.y lJ
where C 1 and C 2 are positive consta.nta: Consequently
.. _  o r
S < r. lo!? X/I --- ,
 IO.J ....- I-
T
where the sunl extends over all the primes 1- < :1.' and such that
( x ) ( X
o - > a + 6) -. B y Lemma 14 we then obtain
l' 1-
(14)
s = 1Jl (log l')!,
where '111 tends to zero for x -+ 00.
N o,v consider the sum
1" -=  109" P , lo q ,
 1) q
extending over all pritnes 1) and q such that
1) < h, q < l , p q > S.
1)
,
Putting y = J'X, z = 11 x and 12 = V N , we then have
T ;?:. (  lo 71 ) ( 2 log '1 ) .
]I -n 1 q:.:n f}
Hence by formula (4)
15)
T > l'3 (log x):! 
C;J being a positive constant.
Let us put
T = , 102' P , log q + V,' lo ,'P _ to!!" fJ ,
""'2 P q  1> q

296
CHAPTER VIII
,vhere the latter sum extends o,er all the prImes 1) und q satis-
fying the conditions
p < f:r. I}  I
r
X
f} ( Ii: ) < (_1 _ c5) r .
})(I ]Jl}
1)
This latter SUDl. in virtue of Lelnn18 15. is equal to }2 (log J'):!.
where 1} tends to zero for jo -7 00. Hence we have
, log}J . log q = T --. 1 ] ,' (100' :r)
......, 2 jJ ([ ...  ,
and. in virtue of inequalitJ' (15),
",' log p log' q 1 ( ' 1 ) 2
" : . > oj C s og :c
...... 2 }) q ... ...
(16)
for :J: sufficient.l" large.
In the sun1 ', we now consider for a fixed value of x the
...... 2
primes }J and fJ which haye the pr0l'erty that the sum :J 10; I'
takes its ll1illinl11111 v'alue p,; /.l depends 011 x only.
Then we ha¥e. by (16),
 >  log J) , log q ........ 1 . (1 ' ' 
J. = II. .", -., II ( .. O!! ,r ) .
r- :.! p fJ _.,,---
If we compare this result ,vith inequality (14), we obtain for
..I.: --> 00 that
 100' 1"
- 
P, - -- -  o.
3 ,.
Consequently, to every posit.i,e number e and to every l1atnral
number N there cOl.responds a uOD1ber t = P q > N such that
r!1t 100- r
",' roo <"
, ... E,
...:...J r
rll::-t
the sum extending over all prinles r which are
Hence
.L
I
> aljc1 < (J t.
(J
r rif
 log' r ....:: E (J t,
ro t

THE PRIME NU)IBER TI-IEORE1\l
297
ana
(] 7)
. t )
/7 \0' t) -. 0 (0' < EO't.
If )."'r, and therefore also t, are sufficient.ly large, ,ve ha ,e
tJ (at) > (a -- E) at
anti
o U) < (..t + E) ; ·
Hence it follows froln (17) that
t -+-
I )  I 
tel - l;' a - < 8 a.
a
This inequality holds for eve:.y positive nU1l1her E. HelIce ,ve
obtaiu a a - 4 <. O. On the other hand, ,ve have a a <: 1 and
!l :...,.. ().
Tlms, every number 0' < A has the property that Go,! < .\. If
a .. it
. ] t - -t b . tl t ( ...4. ) 2 < .A .A. -- 1 . f · < A
a tenc so, we 0 taln la --- = or - "'-. . IDce a =
a (l a a
and a + ....4 =, we clearl)' have a = A = 1. Q. E. D.
The basic new thing- in the abo\e proof is the asymptotic for-
lllula (3) lTheorem 17). Fl.Otn this forulula there are several
'ays to deduc(.A the prin1e llulllber theorem. The first proof of
Selberg dates froni 1 U-I-8. The ])roof developed in this ectjon is
related to it it fol1ow8 an expositioll g'iveu by van Jer Corput
and based on notes of SOIne lectures held by ErJ..ts in 1 94
.'
(DenlonstratifJ}l ehnze12t{(£1.e (he theo.e}ne SU1' la dis'lrib?llioll des norn-
1)1.(....; jJrellli(lr......, criptum 1, Centre I\fathematique, Amsterdam 19-19).
In a pu per published ill 1 B-t.9 (.J n f'lernentar.ll proo./' of the lJti1ne
12u1nlJ(lr theore11, Anuals of Matheluatics, VoL 50) Selberg uses
another method for ucc1ucillg the prill1e llulnber theoreUl from
his lJasic fOrll1ula (13).
It is eaiIy seen that the proofs g'iven in this cha.pter can be
1110dified so as to a,.oid cOlllpletely the application of concepts
and results frolll analysis.
A 11 e \v epoch has bee]l ina ug'Ul'a ted in th e theor. of prinl e
11Ullluers by the itleas of Selberg.

298
CI1APTER YIII
Exercises
172. Prove the approximate formula
lp (n) 6
 - = ..x + o (log .l:},
 11- n'"
1J.r
the sum extending over all positive integ'ers '12 < ;1.'.
173. Show that there exists a positive constant k such that, for
all natural numbers 12,
m(n » k n .
T Ion" loa- n
z:, e
174. Prove the approximate formula
 ! = log log x + C + 0(1),
p.t P
the sum extending over all primes p < x. 0 is an absolute
constant.
175. Prove the approximate formula
( 1 ) e-' ( 1 )
II 1- = +0 ·
p:t }) log .£ log x
the product extending over all primes 1) < x. j1 is Euler's
COllstal1t.
176. Denote by (1,) (n) the nl1nlber of different prime factors of n.
Prove the approximate formula
!,ro{n) =xloglogx + G{£ + 0(.1:),
n;Sz
1 ,..,..
t L
,vhere C is an absolute constant, the same as in Exercise 17-1-.
The sum extends over all positive integers 12 < JJ.
Denote by Pn the nth prime. BJ' applying tIle prime number
theorenl prove that
1 . P1 1
lnl = .
n-oo 11 log 11
lonversely, deduce the prilne n u1l1her theorem fronl this
rela tion.

THE PRIME NUMBER THEOREM
299
178. Show that the infinite series
00 1
 P (log log p)h '
P
extending over all primes 1J, is convergent or divergent
according as /1, is > 1 or < 1.
179. By applying the prime number theorem prove that
 p.(tz) = o (x),
n=-z
the sum extending over all positive integers 12 < x.
180. By applying the prime llumber theorem prove that
 t_(n) = o.
 n
n=l

300
Table
indicating the least positive primitive root 9 of the first 150 odd
primes p. An asterislr attached to the priD1e Ineans t.hat it, has
the primitive root. 10.
p
I ,II :1 }J
.Q II
p
.ll Ir p
!/ II
p
I ,II I
II I , II
.
1:1* 'I II Ii
3 .) .) :! n:3 .) II 479 1:3 (j-'" ;)
I Ii I ,,;,
'I
f) .) II 137 ') 307 .3 487* :3 U77 0)
_ t.J -
... :;: 3 139  i :U 1 17 !U L ::. .) I o8: .)
, I, ,I _
11 .) tl J.Jn* 2 31: .: 10 499::: ,.. ti J 1 :{
, I
13 .) II 151 6 II : J 7 ) .)fI: ::; :) 7n 1.:: .)
_ ,
I I
17 :: : II lij7 !j '1 331 :J II 509* .) r 7u9* .,
I
19* .) I 163  337 .t- It) " lJ21 .) 71U 1 J
- I II .) I,
23::: 5 I 167: 5 :347' .J !) 2:3 .) - .) - ,-:
I I I 1- , ..,
II I
:!u* .) 173 2 I 349 0) .341* .) I 7 :3: c;
, ...
31 3 179-:: 2 I' 35:{ 3 I ;")""7 .) II 73! ')
- oj
18]* I I -; 4:*
37 ..) .) :J:>9 7 I 557 .) I, .j
.. -
41 (j 191 19 367 0 :: 6 I ijli: :) 7;')1 3
_ II
43 3 193* 5 I 373 ) 5GH " 7r..7 2
I I ... ..., I
47* I :}79 : I :)7(:: II
I oJ lU7 .) II .) I :J 7tjJ u
.- .-
53 2 199 3 38 :r i : 5 I ---.:. 5 7u9 II
I ;) , I I
I I
5tf ) I 211 2 I 38U* 2 , :>87 .) I: i73 .)
.-
61";: I '):)3 ojO 3 I :3£17 II .)n: :::
.) ;) a 787 2
II ..-
I il
07 .) II 227 2 .JOt :3 .390 1 I 797 .)
I, I Ii
71 7 ,.}() .!\ 6 409 I II uo! , ()9 3
I __0
73  233* 3 II 4] fJ .::  I ufJ7 a 8Jl* 3
;j
79  239 ,.. 421 .) !I tj13 0 g:!J* , .)
 :1 _ -
b3 .) 241 - 431 7 017 :J :1 823* 3
,
89 3 251 6 \: 4- .').:;: 5 \I GI9: .) 8:)"" .')
'I I _ IJ ..I I ..
97::: [, I 257* 3 439 L3 I o:1 : II M:!B '>
Ii , I ..
101 .) I 263* 5 443 .) 041 ., II :39 I 11
, .- / .
II I
10:3 5 I 269* ,) :1 440 3 I 643 II S:;:J .)
.-
fj -J. 7 ::: I 8:».*
107  I 271 0 I 457 1:1 5 Ii 3
109::: U I 277 ;j I 461::: 2 i U :):  859 .,
II I' \)
113 ::: II 281 3 46: G::;fJ ::: .
{ : 2 Ii S fj :r- ;j
1 ')- 0' 283 3 ,I 467 ., I (jOI .) h;; 2
- .) I -
I
II

301
The fundanlental solutions of the equations x 2 - D )0 2 = + 1
for D < 103.
The table indicates the fundamental solution of equation .l:2 - fJ .'/2
= - 1 or of .r 2 - D .7/2 =  1 according as an asterisk is attached
to the number D or not.
J) I ,I. I .II I  [J I .,' I // il LJ I ,l' I !I
:)* I II :38 :1
1 1 \1 :ri 6 il 71 :{ 4sn 413
:3 .) 1 II 3B :!5 4 7j 17 .) I
-$ 40 19 3 i3* ] OUS 1 :!.;
:) .) 1
(i 5 .) 41* :2 f)  I 74* 4: 5
-
7 8 : 42 1 : 2 I 75 :!u .)

I il
8 3 1 4:J 3 48 ;;31 76 ;'); 7tH) t) n:JO
10* :3 1 44- JU9 3U II 77 :.;:;1 40
11 10 : 4;; UH 24 II i8 .13 ij
I:! 7 2 I 40 24 3:Jfi :} 588 II 7U sn 9
I
]3* I
18 5 47 48 7 I 80 U 1 I
I I
I
J4: 15 4 . 48 7 1 8$ {) 1
r r:'* I
t5 4 1 I ,,0 7 1 II 83 8 {)
17* 4 1 I :>1 :;0  84: .,.J U
.
I 2 S5:i' 378 41
18 17 -I: , 6-tU 90 I
I >-
10 170 30 II 53'" IS::! :! 86 It) O;) 1 t:!
:!O U  I 54 48,') 60 I 87 8 3
:?1 55 12 ! 5.') 89 12 88 197 21
.} .) 197 42 56 lfi 2 80* fion 53
_oJ II
3 24 5 II 57 151 o 90 10 2
::!4 5 1 \, 58* 99 1 : Ul 1 574- 1(;5
26* 5 I 1 59 ;):o HU 9 I 151 I=!0
27 26 5 I 60 31 4 93 12 151 1260
I
28 1.1"- 24 61* 29 718 3 80.3 94- 2 !j-l: n.) 221 064
_I I
29* 70 13 I u2 63 8 95 :J9 4 I
I
I I 49 5
30 11 2 63 8 1 d !)O
31 I 5O 273 I 0"'* 8 1 U7* 5 f)(.4 5fiB
:)
:i2 17 :J 66 65 S 98 99 ]0
33 23 4 \ l17 48 8* (; 967 99 IH I
:i4 35 6 \ (;8 33 4- ]01* 10 1
35 6 1 t 69 ,.. ,..  .. 036 102 ]01 10
illi)
37* 6 1 I 70 2,;1 :JO 103 :!7 5:!S :! 419
r

NAME INDEX
BACHET DE MZllUAC) C. G., 191, 22;,
262
B..\UER, )1., 168
BERl.'ELEX, N. P., 64
}JJt;RTUAXO, J. J... F., 67, 125
BIJ..LIXG, G., 232, 269
BORU, H., 67
HRl'"S-, V'., 66, 67
CATAJ...-\x, E. CH., 124
CA eCBY, .A. L., 144
CU.\.TEJ..ET, P., 259
CORPUT, V...\:S DER, J. G., 297
DIOPHAXTOS, 6, 32, 227, 252
DIHICHJ..ET, P. G. L., 38, 66, 197, 252
EISEXTEIX, F. G. 1\1., 141, 241
}1BATOSTIIEXES, 61
ERD08, P., 297
Et.:CLJD, 0, 14. 15, 22, 34, 44, 226
El-LEIt, 1..., 6, 23, 38, 44, 71, 133, 144,
199, 246, 261
I,;'AREY, J., 46
I-"'ERrAT, P., 6, 44, 71, 197, 227, 229,
246, 262, 261
GAlS, c. F., 6, 64, 68, 113, 139, 144,
177, 241
GOI..DDACH, CH., 66
HADA)JARD, J. S., 64
HARDY, G. H., 67
IIERl\lrrE, CII., 40, 126
rGU_\)J, A. E., 67
.TA('OBI, K. G. J., 113, 144, 146, 241
.T AC()UTnAL, E., 182
I{UOYECKBR, L., 144
Kl.)J)IER, E. E., 252, 253
LAGRANGE, J. L., 6, 99, 192,197,236
l...\)JUERT, J. H., 40
LAXDAU, E. G. H., 57
J..EBESOl'"E, V. A., 194, 261
LEGENDUE, A. 1I., 6, 64;- 138, 219, 246,
252
I.EII::\JER, D. N., 61
I..Isn, C. E., 232
I..IsDE::\L\SN, C. L. F., 40
LJOUYII..J..E, J., 35
I.ITTLE\\-OOD, J. E., 57
I...JUSGGREX, 'V., 266
Lt.:c AS, E., 52, 128
JtIAILJ..,ET, E. TH., 261, 264
}IAXGOI..DT, H. C. F., 67
}IEISSEJ.., D. '. E., 64
JtIERSEXE, 11., 44
lERTENSJ F. C. J., 127
)IILI..S, 'V. H., 65
:p.IIXKO\\-KI, H., 34
)lORDELl., L. J., 235, 263, 259, 265
IloBIUS, A. F., 27
XIVEX, I., 40
OR}, 0., 90
PJ.:J..L, J., 197
}'IPI)ISG, N. J., 66
JJYTIIA(;ORAS, 5, 34
RIE){'ASS-, G. F. B., 66, 66
ScnOI..z, A., 122
SEI.nERG, .A., 6., 66, 286, 297
SIEGEL, C. I..., 260, 263, 264

SKOLE3f, Tn., 266
STOR3IER, F. C. 1\1., 267
SU TSE, 78
SYI.. VESTElt, J. J., 60
TCBEDYCBEJ.', P. L., 65, 60
THUE, A., 122, 260, 261, 262
TITCH3fARSH, E. C., 67
NAl\lE INDEX
303
V ALI..EE POLSSIX, DE LA, CD. J., 66
'VINOGRADO\'"', I. )1., 66
''''ElL, A., 259, 264
'VII..SOX, J., 99
WOI..STEBOL3IE, J., 128
ZASSESJIAUS, H., 66

SUBJECT INDEX
A
Algebraic congruences .:1, 79, 83, 8=>,
DO
Algebraic <-urves, 254, 260
" nUmhf'rR, 35
Algorithm, Euelidenn, 21, 22
AlDbiguous c.'lasses, 205
ArithlDetical functions, 26
progression s, 66
Associated numbers, 237
); solutions, 204
B
Bachet's theorelD, 191
Basis of tl moc.lul, 1Q
). of rat.ional points, 59
Bauer's theorelD, 168
Bernoulli numbers, 253
Bertrand's conjecture, 67
BinolDial coefficients, 49
('unJl:ruenl'e, 110
Binary quadratic forms, 215
Biquadratic congruences, 73
residues, 115, 13Q
Birational transforlDation, 21)';
Box principle of Dirit'h1et, 38, ] 23
Brun's sieve lDethod, 60
theorelD on prime twins, 67
c
Canon arithm.eticus, 113
Classes of residues, 69
).
in a <lundrntic
...
field, 2... 0
Classes of residues, prin1e to n, 71
Classes of solutions, 204-
). J', ambiguous, 205
,,» ,conjugate, 205
COIDination uf Ii thins. 49
COlDplete residue systelD, .. 0
COlDposite l111nlbers, l:i
Congruences, 68
aJp;ehrai<-, ,S
" , lJiuolUial, 115
e:}Jonential, 114
functicJnu), 7 
in a ql1ac.lratic lield, 24(
, linear, 7H
q uac.lrati('. 132
, sinnlltaneons, i7
Congruent n unl1Jcrs, US
polynomials, i 4
Conjecture, Bertrancrs, 6i
, Goldbac.'h's. 66
Conjugate clnsses, 205
llnnll>ers, 2:35
Cubic "011gruencefoC, 73
Cubic curves, 254
of genus one, 254
of genus zero, 254
, unicursal, 254
Diophantine e(lun.tiol1, 241,
240, 25-1, 250
Cubic resic1\1s, 115
Curves, algelJraic, 254
, :\Iaillet's, 264
, nnirursal, 254
CyclotolDic polynolDial, index of. 158
,) \', irreducihil-
ity of, 160
CyclotolDic polynolDial, 158
).
»
, prin1e di\.j.
sors of, 164
D
DecolDposition into prime factors, 15
Degree of an algebraic- congruence, 73
of an nlgebrt\ic Dl1ml)er, 262

SUBJECT INDEX
305
Diophantine equations, 32
)" ), of the ti rtlt
degree, 29
Diophantine equations or t be sc<."ond
df'gree, 188
Diophantine equations of tht- third
l1el'E'e, 241, 24G, 254, 265
Diophantine equations of the fourth
llegrtAe, 227, 232, 235, 267
Diophantine equations of higlJer de-
gree, 227
Diophantine problems, 32
Diophantos's ArithInetica, 5, 32,
26, 252
Dirichlet's box principle, 38, 123
Dirichlet's theorelD, 66
Disquisitiones arithJneticae, 0, 68
Divisibility by 9 .r) modulo n, i 6
)) of uumLel"s, 12
)f of polynomials, 93, 160
Divisor, 11, 12
» , greatest con1mon of two or
nlore numbers, 17
Divisor, reatest common of two poly-
nomials Inodulo p, D6
Divisors, fixed, of polynoluials, 122
). of I(x) modulo tt, 76
» , proper, 12
) , trivial, 12
), , trivial of f(3.') modulo 1), 94
E
Equations, Diophantine, 32
 , indeterluinate, 32
Eratosthenes's sieve lDethod, 61, 62
Euclid's algorithm, 21, 22
» Elementa., 5, 14, 15, 34, 44,
226
Euler's constant, 120, 276, 298
;) criterion, 1 SS
» -function, 23
» identity, 66, 192
» theorem, 72
Even numbers, 12
Exceptional point, 259
20 - filBd70 'r!lgve Nagel!
Exponent of an integer modulo )', 102
Exponential congruences, 114
F
Farey series, 46
FerIDat's last theorem 251
,
» primes, 44
» theorem, 71
Field, 21
>I , Gaussian, 241
» , imaginary quadratic, 21, 235
}} , q uadra tic, 21
» , rational, 21
» . real, 21
» ,real quadratie, 21
Fixed divisors or polynomials, 122
FOrIDula, Hermite's, 126
» , Selberg's, 283, 286
Function &(x), 60
 C ('), 66
» Li (x), 66
» p. (n), 27
» n (), 54
» T (n), 26
» tp(n), 23
Functional congruences, i 4
Functions, arithmetical, 26
Fundamental solution of a class, 205
FundaIDental solution of an equa-
tion, 17, 201
FundalDental theorelD of number
theory, 15
G
Gaussian field, 241
» sums, 177
Gauss's lemma, 139
» polynumial identity, 174
Generating system, ] 9
Generator, 259
Genus of an algebraic curve, 264
Goldbach's conjectures, 86
Greatest cOlDIDon divisor, 1 i I 240
» »  of poly-
nODlia]s modulo p, 96
Greatest integer  c, 18

306
SUBJECT INDEX
H
Height of a solution, 244
Hermite's formula, 126
Homogeneous coordinates, 2J;
I
Ideals, 262
Identical congruence, 74
Identity, Euler's, 66
» , I...ebesgue's, 194
IIDaginary quadratic fields, 21, 236
Incongruent l1uwbers modulo tl, 68
» roots of a congruence, 73
» solutions of a congru-
 3
en(ae, .
Indeterminate equations, 32
Index: calculus, 111
Index: of a cyclotomic polynomial, 168
), of a natural number, 112
» of the equation a ;£'+b y'-t (' z'.!
= 0, 218
Induction, mathematical, 11
» , multiplicative, 16
Inequalities, Tchebychefs, 60, 62
Infinite descent, 229
Initial solution, 235
Integers, 11, 236
Integral logarithlD, 56
Integral polynolDials, 73
Inversion formula, .:\1 obi us, 28
I. r. (integer representing) polyno.
mials, 120
Irrationality of e and 1l, 38
Irrational nmnbers, 34
Irreducibility of eyclotowic polyno-
mials, 160
Irreducible polynolDial Dl0dulo p, 96
J
Jacobi's symbol, 145
Jacobsthal's theorelD, 182
K
Ku:m.merian prhnes, 253
K nTnTn er's theorem, 263
L
Lattice, 32
Lattice points in the plane, 32
» »on conics, 212
» » on plane algebraic
curves, 260
Least absolute remainder of a mod-
ulo b, 13
Least non-negative remainder, 12
Least positive cOlDlDon lDultiple, 16
Lebesgue's identity, 194
Legendre's symbol, 133
LelDlDa of Gauss, 139
Linear congruences, 78, 76
» Diophantine equations, 29
M
Maillet's curves, 2601
» theorelD, 264
MathelDatical induction, 11
Mersenne's primes, 44
Moduls, 19
Modulus of a ('ongrnenre, 68
Mordell's theorelD, 263
Multiple, least positive common of
t\\"o or more numbers, ] 6
Multiple of an integer, 12
» of I(x) modulo tl, 76
Multiple root of a congruence, 86
Multiplicative induction, 16
Mobius function, 27
» inversion fornlula, 28
N
Natural numbers, 11
Non-KUDlIDerian primes, 263
Non-residues, quadratic, 116
Norm of a number, 236
Normal polynolDial modulo p, 94
nth roots of unity, 166
Nmnber axis, 32
Numbers, algebraic, 35
» , associated, 237
)p , Bernoulli, 263
» , com posi te, 13

SUBJECT INDEX
307
Numbers, congruent, 68
» , conjugate, 236
 , even, 12
» , incongruent, 68
>, , irrational, 34
... , odd, 12
» , perfect, ...
» , prime to tJ, 23
» , relatively prime, 23
 , relati,"ely prime in pairs, 23
» sq nare-free, 27
» , transcendental, 36
o
Odd numbers, 12
Order of an inteoger modulo n. 102
Orders of lDagnitude o(x) and o (X),
r." ..
.a
p
Partitions of Dum lJers, 273
Pell's equation, 1 D7
Perfect nUlDbers, 44
Point lattice, 32
Points, exceptional, 269
1> , rational, 216, 2601
;" , singular, 254
,. , tangential, 260
PolynolDial c()efficient, 49
» identity of Gauss, 1 i 4
PolynolDials, e"clotomic, ] 68
» , divisibilit- of, 160
. integra], 73
) , i. r. '.integer rpreR('nt
ing\ 120
PolynolDials, irreuuribl(', 160
.) irreducil,le n}o<1ulo p, {t;)
nornlal l1\od ulo p, 9 t
primary luoduJo p, 94
prinlitivc, 73
rednrible, 160
reducihle Dlodulo }"', 5
Power residues, 1 J;:;
PriIDary polynonlialR nlodll1o p, 4
Prime divisors of an intgE'r, 13
PriIDe divisors of a t"'yclotomic poly-
nomial, ] 64
Prime divisors of pol"DomiaI8, 81
;,) ,> of (IUadratic polyno-
luials, ] 49
Prime factors of an integer, 16
PrilDe function Inodulo p, 95
PriIDe number theorelD, 55,275,286
PriIDes, 13
» , 'ermat, 44
» in arithmetical progressions,
66, 163, 168, 173
Primes, I{ummerian, 263
» , lersenne, 44
» , non-Kummerian, 263
of the form 11.'1 - 1, 173
Prime twins, 52
PriIDitive pol"nomials, i3
root.s of unity, 157
Primitive root of a congruenre, 104-
I" "of a nlodnlus, 107
Principal remainder, 12
Products, trigono metrical , 173
Proper dh-isors, 12
8olution8, 2] {)
Q
Quadratic congruenc'es.. 73, 132
), fields, 21, 235
non-r(\si(hlP, 132
,; residue::. 115, 132
Quadratic reciprocity law, ) '. 1
R
Rank of a cnhic ('urvE', 259
of a luodul, I 
Rational field X(1), 2J
Rational nUlDber in a 1ielct 25 &
Rational points in U 11eld, 21 t)
) jn thE' pJan f '. 21 fj
,. on eonics, 216
), 11 un plain aJgE'hrai c '
('urvc. 2;) t
Real field, 21
Reciprocity law, quadratic. 141

308
SUBJECT INDEX
Reduced residue system, 71
Relatively prime l1um bers, 23, 240
» »polynomials mod-
ulo p, 96
Remainder, least absolute of a mod-
ulo h, 13
RelDainder, least non-negat.ive of a
modulo h, 12
Remainder, principal of Q. modulo b,
12
Residue classe8, 69, 94
» >I in a quadratic fiE'ld,
240
Residue cla88e8, prime io 11, 71
Residues, biquadrati(', 1] 5
 , cubic, 116
» , IDodnlo tJ, 69
" , (luadrat.ic, 116
Residue system, eompleote, 70
;) N, redured, 71
Riemann's zeta function, 66
Rings, 20
Root, of an algebraic congruencE', 73
) primitive, 104, 107
Roots, inrongrnent, i3
Rule8 for congruen('es, 68, 69, 70
» for index calculus, 112
s
Scholz's theorelD, 123
Selberg's formula, 283, 286
Selberg's proof of the prime nUmhtaT
theorem, 286
Siegel's theorem on approximation
of algebraic numbers, 263
Siegel's theorem on lattice points on
algebraic ('urt'es, 264
Sieve method, Erat.osthenes's, 61, 62
» », Rrun's, 66
Simplo root of n cungruene(\, 85
Singular point, 264
Solution, height of a, 244
>, , initial, 236
» of an algebraic congru-
ence, 73
Solution, proper, 21 D
Solutions, associated, 20..
Solvability of (,ol1gruencE', i 3
Square-free Dum beor, 2 i
StOrmer's theorelD, 2Ai
Sums, Gaussian, Iii
» of integral sflunre, 188
Sylvester's theorelD, 50
Symbol [c], 13
Symbol, Jacobi's, 345
» , LE'gendre'8, 133
Symbols 0 (x) and 0 (x), 2;0
System of residues, rOD1p1ett', ;0
»  » , reduced, 71
T
Tangential point, 260
Tchebychef's inequalities, 60, 62
TheorelD, Bachet's, 191
)# , Hauer's, 168
" , Brun's, 67
#) , Dirichlet's, 66
> , Eisenstein's, 141
» , Euler'R, 72
p , Fermat's, 71
,) , Jacobsthal's, 182
" , Kummer's, 253
i. 1Iaillt's, 264
, J.lertens', 127
, Jordel1's, 269
,) , Scholz s, 123
, Siegel's, 263, 264
, Rtormer's, 267
, S).lvester'1 50
, Tc-bel.y('bet'Ei, 60, 62
, TIlue's, 263
, 'VilSOD'R, 99
, 'V nlstenholme's, 128
Theory of ideals, 252
Thue's relDainder theorem, 122
Thue's theorem on approximation of
algebraic nnmbers, 263
Thue's theorem on Diophantine equu.-
tions, 263
Totient of )l, 23

SUBJECT INDEX
309
TraD8cendeatal numbers, 216
Tl' au.,to l'lDation, biratioual, 267
Tr!gonometftcal pl'Oductll, 1 ';S
Trivial divi80r8 of f() modulo P, 94
» » of an integer, 12
u
UDicunal curves, 254
UDlque lactorizaUoD, 16, 239, 252
UDiw in quadratic fields, 286
UDltJ', Jztb roots of, 166
» ,primitive roots of, 166
w
Wil8on'. theorem, 99
» », geot'ralized, 100
z
Zeta IuDctton, Riemann's, 56
\ \