w
'8*
v„^

THE GREEN BOOK
OF MATHEMATICAL PROBLEMS
Kenneth Hardy and Kenneth S. Williams
Carleton University, Ottawa
DOVER PUBLICATIONS, INC.
Mineola, New York

Copyright
Copyright © 1985 by Kenneth Hardy and Kenneth S. Williams.
All rights reserved under Pan American and International Copyright
Conventions.
Published in Canada by General Publishing Company, Ltd., 30 Les-
mill Road, Don Mills, Toronto, Ontario.
Published in the United Kingdom by Constable and Company, Ltd.,
3 The Lanchesters, 162-164 Fiilham Palace Road, London W6 9ER.
Bibliographical Note
This Dover edition, first published in 1997, is an unabridged and
slightly corrected republication of the work first published by Integer
Press, Ottawa, Ontario, Canada in 1985, under the title The Green Book:
100 Practice Problems for Undergraduate Mathematics Competitions.
Library of Congress Cataloging-in-Publication Data
Hardy, Kenneth.
[Green book]
The green book of mathematical problems / Kenneth Hardy and
Kenneth S. Williams,
p. cm.
Originally published: The green book. Ottawa, Ont., Canada :
Integer Press, 1985.
Includes bibliographical references.
ISBN 0-486-69573-5 (pbk.)
1. Mathematics—Problems, exercises, etc. I. Williams,
Kenneths. II. Title.
QA43.H268 1997
510'.76—dc21 96-47817
CIP
Manufactured in the United States of America
Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501

PREFACE
There is a famous set of fairy tale books, each volume of which is designated by the colour
of its cover: The Red Book, The Blue Book, The Yellow Book, etc. We are not presenting you
with The Green Book of fairy stories, but rather a book of mathematical problems. However,
the conceptual idea of all fairy stories, that of mystery, search, and discovery is also found in
our Green Book. It got its title simply because in its infancy it was contained and grew between
two ordinary green file covers.
The book contains 100 problems for undergraduate students training for mathematics
competitions, particularly the William Lowell Putnam Mathematical Competition. Along with
the problems come useful hints, and in the end Oust like in the fairy tales) the solutions to the
problems. Although the book is written especially for students training for competitions, it
will also be useful to anyone interested in the posing and solving of challenging mathematical
problems at the undergraduate level.
Many of the problems were suggested by ideas originating in articles and problems
in mathematical journals such as Crux Mathematicorum, Mathematics Magazine, and the
American Mathematical Monthly, as well as problems from the Putnam competition itself.
Where possible, acknowledgement to known sources is given at the end of the book.
We would, of course, be interested in your reaction to The Green Book, and invite
comments, alternate solutions, and even corrections. We make no claims that our solutions are
the "best possible" solutions, but we trust you will find them elegant enough, and that The Green
Book will be a practical tool in the training of young competitors.
We wish to thank our publisher, Integer Press; our literary adviser; and our typist,
David Conibear, for their invaluable assistance in this project.
Kenneth Hardy and Kenneth S. Williams
Ottawa, Canada
May, 1985
(iii)

Dedicated to the contestants of the
William Lowell Putnam Mathematical Competition

To Carole with love
KSW
(vi)

CONTENTS
Page
Notation ix
The Problems 1
The Hints 25
The Solutions 41
Abbreviations 169
References 171
(vii)

NOTATION
[x] denotes the greatest integer < x, where
x is a real number.
{x} denotes the fractional part of the real
number x, that is, {x} - x - [x].
In x denotes the natural logarithm of x.
exp x denotes the exponential function of x.
*>(n) denotes Euler's totient function defined
for any natural number n.
GCD(a,b) denotes the greatest conunon divisor of
the integers a and b.
n\ denotes the binomial coefficient n!/k!(n-k)!,
kf where n and k are non-negative integers
(the symbol having value zero when n < k ).
(a. .) denotes a matrix with a. . as the (i,j)th
3 entry.
det A denotes the determinant of a square matrix A.
fix)

THE PROBLEMS
Problems, problems,
problems all day long.
Will my problems work out right or wrong?
The Everly Brothers
1, If {b : n = 0,1,2,... } is a sequence of non-negative real
n
numbers, prove that the series
ao b
(1.0) y H_
I 3/2
n=0 (a+b0+b1+...+bnr''t
converges for every positive real number a.
2, Let a,b,c,d be positive real numbers, and let
. , , ,s a(a+b)(a+2b)...(a+(n-l)b)
Va'b'C'd) = c(c+dHc+2d)...(c+(n-l)d)
Evaluate the limit L = lim Q (a,b,c,d).
n-*-"> n
3. Prove the following inequality:
/■> n\ £n x . 1 (x+1) . n ,. .
(3.0) -=— < T—5 , x > 0, x * 1.
x -1 3 U+x)
1

2
PROBLEMS (4-12)
4, Do there exist non-constant polynomials p(z) in the complex
variable z such that |p(z)| < R on |z| = R , where R > 0 and
p(z) is monic and of degree n ?
5, Let f(x) be a continuous function on [0,a], where a > 0,
such that f(x) + f(a-x) does not vanish on [0,a] . Evaluate the
integral
'a f(x)
f(x) + f(a-x) aX •
6, For £ > 0 evaluate the limit
fX+1
lim x
2
sin(t )dt .
7, Prove that the equation
4 4 4 22 22 22
(7.0) x + y + z - 2y z - 2z x - 2x y =24
has no solution in integers x,y,z.
2
8, Let a and k be positive numbers such that a > 2k.
Set xn = a and define x recursively by
(8.0) xn = Vl + x^~ ' n = 1'2'3"-' *
n-1
Prove that
lim
n-i-oo
exists and determine its value.
x
n

PROBLEMS (4-12)
3
9, Let x. denote a fixed non-negative number, and let a and
b be positive numbers satisfying
v^b < a < 2v^b .
Define x recursively by
n
ax _. + b
(9.0) x 2_!__ n - 1,2,3,... .
n x , + a ' '
n-1
Prove that lim x exists and determine its value.
n-i-a>
10, Let a,b,c be real numbers satisfying
Evaluate
?
a>0,c>0,b >ac.
2 2
max (ax + 2bxy + cy ) .
x,y £R
x2+y2=l
11, Evaluate the s
um
[n/2]
n(n-l)...(n-(2r-D) 0n-2r
r^O (r!)2
(11.0) S - I "v" -'•••"y- *<> 2
for n a positive integer.
12, Prove that for m = 0,1,2,...
(12.0) S (n) =l2mfl+22mfi+... +n2mfl
m
is a polynomial in n(n+l).

4
PROBLEMS (13-21)
13, Let a,b,c be positive integers such that
GCD(a.b) = GCD(b.c) = GCD(c.a) = 1 .
Show that t = 2abc - (bc+ca+ab) is the largest integer such that
be x + ca y + ab z = £
is insolvable in non-negative integers x,y,z.
14, Determine a function f(n) such that the n term of the
sequence
(H.O) 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, ...
is given by [f(n)].
15, Let a,, a„, ..., a be given real numbers, which are not all
l I n
zero. Determine the least value of
2 j. j. 2
x, + ... + x ,
1 n '
where x., ..., x are real numbers satisfying
a.x, + .. . + a x » 1 .
11 n n
16, Evaluate the infinite series
23 33 43
17, F(x) is a differentiable function such that F'(a-x) = F'(x)
for all x satisfying 0 £ x £ a. Evaluate / F(x)dx and give an
example of such a function F(x).

PROBLEMS (13-21)
5
18. (a) Let r,s,t,u be the roots of the quartic equation
x4 + Ax3 + Bx2 + Cx + D = 0 .
2 2
Prove that if rs - tu then A D = C .
(b) Let a,b,c,d be the roots of the quartic equation
4 2
y +Py + qy + r ■= 0 .
Use (a) to determine the cubic equation (in terms of p,q,r) whose
roots are
ab - cd ac - bd ad - be
a + b-c-d' a + c-b-d' a + d-b-c'
19, Let p(x) be a monic polynomial of degree m £ 1 , and set
fn(x) - ep(xV(e-P(x)) ,
where n is a non-negative integer and D = — denotes
differentiation with respect to x.
Prove that f (x) is a polynomial in x of degree (inn - n).
Determine the ratio of the coefficient of x in f (x) to the
n
constant term in f (x) .
n
20, Determine the real function of x whose power series is
3 9 15
x_ x_ x
3! "*" 9! + 15! + ""• -
21, Determine the value of the integral
(21.0) I
n
0 , em x
sin nx , ,
dx ,
for all positive integral values of n .

6
PROBLEMS (22-31)
22. During the year 1985, a convenience store, which was open
7 days a week, sold at least one book each day, and a total of 600
books over the entire year. Must there have been a period of
consecutive days when exactly 129 books were sold?
23. Find a polynomial f(x,y) with rational coefficients such
that as m and n run through all positive integral values, f(m,n)
takes on all positive integral values once and once only.
24. Let m be a positive squarefree integer. Let R,S be
positive integers. Give a condition involving R,S,m which
guarantees that there do not exist rational numbers x,y,z and w such that
(24.0) R + 2Si/m = (x + y^m) + (z + w/m) .
25. Let k and h be integers with 1 S k < h . Evaluate the
limit
hn
(25.0) L = lim n ll - £.
n •+ oo r=kn+l
26. Let f(x) be a continuous function on [0,a] such that
f(x)f(a-x) = 1 , where a > 0 . Prove that there exist infinitely
many such functions f(x) , and evaluate
J0 1 + *<*>
27. The positive numbers a.,a.,a,,... satisfy
n , (• n -.2
(27.0) I V I a , n = 1,2,3,...
r-l r W rJ
Is it true that a = r for r = 1,2,3,... ?

PROBLEMS (22-31)
7
28, Let p > 0 be a real number and let n be a non-negative
integer. Evaluate
(28.0) u (p) = e pXsinnxdx .
J0
29, Evaluate
n-2
(29.0) I 2rtan *
r=0 2n"r '
for integers n > 2 .
30, Let n i 2 be an integer. A selection {s = a : i=l,2,...,k}
of k (2 < k £ n) elements from the set N = {1,2,3,...,n} such that
a. < a, < ••• < a. is called a k-selection. For any k-selection S ,
define
W(S) = min {a ^-a : i = l,2,...,k-l} .
If a k-selection S is chosen at random from N , what is the
probability that
W(S) = r ,
where r is a natural number?
31, Let k > 2 be a fixed integer. For n = 1,2,3,... define
_ fl , if n is not a multi
n \-(k-l) , if n is a mult
iple of k
iple of k
r n
Evaluate the series ) — .
i n
n=l
n
n

8
PROBLEMS (32-40)
32, Prove that
f:
, x e sin x dx = , . ~^ ,„ sin (m+l)ir/4
J0 2(m+2)/2
for m= 0,1,2,...
33, For a real number u set
(33.0) I(u) - I ln(l - 2ucosx + u2) dx .
00 - (<
Prove that
I(u) = I(-u) = -i-Ku2) ,
and hence evaluate I(u) for all values of u .
34, For each natural number k i 2 the set of natural numbers
is partitioned into a sequence of sets {A (k): n » 1,2,3,... } as
follows: Aj^(k) consists of the first k natural numbers, A„(k)
consists of the next k+1 natural numbers, A,(k) consists of the
next k+2 natural numbers, etc. The sum of the natural numbers in
A (k) is denoted by s (k) . Determine the least value of n - n(k)
n 3 n 2
such that s (k) > 3k - 5k , for k = 2,3,... .
n
35i Let {p : n = 1,2,3,... } be a sequence of real numbers
such that pn a 1 for n = 1,2,3,... . Does the series
(35"0) Ji <[p1]+l)<[p2]+l)...([pn]+l)
converge?
35, Let f(x) , g(x) be polynomials with real coefficients of
degrees n+1 , n respectively, where n 2: 0 , and with positive

PROBLEMS (32-40)
9
leading coefficients A , B respectively. Evaluate
L = lim g(x)
X-i-oo
fX f(t)-f(x).f.
e dt ,
in terms of A, B and n .
37, The lengths of two altitudes of a triangle are h and k ,
where h * k . Determine upper and lower bounds for the length
of the third altitude in terms of h and k .
38, Prove that
,, n+1... n+2. ,, n+r.
P . P (x) = Ik* )(!-* )...(l-x )
n'r n'r (l-x)(l-x2)...(l-xr)
is a polynomial in x of degree nr , where n and r are non-
negative integers. (When r = 0 the empty product is understood
to be 1 and we have P - = 1 for all n 2 0 .)
n,U
39, Let A, B, C, D, E be integers such that B * 0 and
F = AD2 - BCD + B2E * 0 .
Prove that the number N of pairs of integers (x,y) such that
(39.0) Ax2 + Bxy + Cx + Dy + E = 0 ,
satisfies
N £ 2d(|F|) ,
where, for integers n > 1 , d(n) denotes the number of positive
divisors of n .
n k
tf). Evaluate Y ~L 9 •
k=l k4 + k^ + 1

10
PROBLEMS (41-50)
ill Let P = P (n) denote the sum of all possible products of
~-L" mm
m different integers chosen from the set {l,2,...,n} . Find
formulae for P2(n) and P,(n) .
42, For a > b > 0 , evaluate the integral
(42.0)
ax bx
e - e
x(eaX+l)(ebX+l)
dx .
43, For integers nil, determine the sum of n terms of the
series
(43.0)
2n 2n(2n-2) 2n(2n-2)(2n-4)
2n-l + (2n-l)(2n-3) (2n-l)(2n-3)(2n-5)
44, Let m be a fixed positive integer and let z.,z , ...,z
be k (>1) complex numbers such that
(44.0)
zl + z\ + ••• + \ " ° »
for all s = m,m+l,m+2,...,m+k-l. Must z, = 0 for i = 1,2,...,k?
45. Let A = (a..) be the n*n matrix where
' n ij
x , if i = j ,
atj = 1 1 , if | i-j | = 1 ,
0 , otherwise,
where x > 2 . Evaluate D = det A
n n
45, Determine a necessary and sufficient condition for the
equations

PROBLEMS (41-50)
11
x + y + z = A ,
(46.0) { x2 + y2 + z2 = B ,
x3 + y3 + z3 = C ,
to have a solution with at least one of x,y,z equal to zero.
i|7, Let S be a set of k distinct integers chosen from
1,2,3,...,10 -1 , where n is a positive integer. Prove that if
(47.0) n<£n(i^li + i^li)/£nl0 ,
it is possible to find 2 disjoint subsets of S whose members
have the same sum.
£(8, Let n be a positive integer. Is it possible for 6n
distinct straight lines in the Euclidean plane to be situated so as
2
to have at least 6n -3n points where exactly three of these lines
intersect and at least 6n+l points where exactly two of these
lines intersect?
i|9, Let S be a set with n (>1) elements. Determine an
explicit formula for the number A(n) of subsets of S whose
cardinality is a multiple of 3 .
50, For each integer n > 1 , prove that there is a polynomial
p (x) with integral coefficients such that
x4n(l-x)4n = (l+x2)p (x) + (-l)n4n .
n
Define the rational number a by
n '
(50.0) a - -L-fi— P (x) dx , n=l,2,... .
4 J0

12
PROBLEMS (51-57)
Prove that a satisfies the inequality
n
4
51. In last year's boxing contest, each of the 23 boxers from
the blue team fought exactly one of the 23 boxers from the green
team, in accordance with the contest regulation that opponents may
only fight if the absolute difference of their weights is less than
one kilogram.
Assuming that this year the members of both teams remain the same
as last year and that their weights are unchanged, show that the
contest regulation is satisfied if the lightest member of the blue
team fights the lightest member of the green team, the next lightest
member of the blue team fights the next lightest member of the green
team, and so on.
52. Let S be the set of all composite positive odd integers
less than 79 .
(a) Show that S may be written as the union of three (not
necessarily disjoint) arithmetic progressions.
(b) Show that S cannot be written as the union of two
arithmetic progressions.
53. For b > 0 , prove that
f sinx , irl 1
by first showing that
(b .
sin x ,
a
e sin x dx
du

PROBLEMS (51-57)
13
54, Let a ,a ,...,a, be 44 natural numbers such that
0 < ax < a2 < ... < a44 £ 125 .
Prove that at least one of the 43 differences d. = a.,,-a. occurs
J J+l J
at least 10 times.
55, Show that for every natural number n there exists a prime
2 2
p such that p = a + b , where a and b are natural numbers
both greater than n . (You may appeal to the following two theorems:
(A) If p is a prime of the form 4t+l then there exist integers
2 2
a and b such that p = a + b
(B) If r and s are natural numbers such that GCD(r.s) = 1 ,
there exist infinitely many primes of the form rk+s , where k is a
natural number.)
55. Let a ,a ,...,a be n (si) integers such that
(1) 0 < a, < a. < . . . < a ,
l l xi
(2) all the differences a. - a. (1 £ j < i £ n) are distinct,
(3) a = a (mod b) (1 £ i £ n) , where a and b are positive
integers such that 1 £ a £ b-1 .
Prove that
n b 3
) a i Tt + (a - t) n
L, r 6 v 6
r=l
57, Let A = (a..) be the nxn matrix where
aij
2 cos t , if i = j ,
1 , if | i - j I = 1 ,
0 , otherwise ,
where -tt < t < ir . Evaluate D = det A .
n n

14
PROBLEMS (58-66)
58, Let a and b be fixed positive integers. Find the general
solution of the recurrence relation
+ a + Vb2 +
(58.0) x _,_. = x + a + Vb + 4ax , n = 0,1,2,... ,
n"rl n n
where x. " 0 .
59, Let a be a fixed real number satisfying 0 < a < tt , and set
,er> „\ t f 1 - r cos u ,
<59-°> Zv m 1 - 2r cos u + r2 du •
-a
Prove that
I, , lim I , lim I
1 r+l+ r-*l-
all exist and are all distinct.
60, Let I denote the class of all isosceles triangles. For
ie I , let h, denote the length of each of the two equal altitudes
of A and k. the length of the third altitude. Prove that there
does not exist a function f of h. such that
kA s f(hA) ,
for all A 6 I
61, Find the minimum value of the expression
(61.0) (x2+^rj - 2K1+C08 t)x + k<1+sint)l + (3 + 2 COs t + 2 sin t) ,
3
for x > 0 and 0 £ t S 2tt , where k > j + /I is a fixed real
number.

PROBLEMS (58-66)
15
62. Let £ > 0 . Around every point in the xy-plane with integral
co-ordinates draw a circle of radius £ . Prove that every straight
line through the origin must intersect an infinity of these circles.
63. Let n be a positive integer. For k = 0,1,2,...,2n-2
define
(63.0)
\ =
2n, n,,
x +x +1
dx .
Prove that I. * I , , k - 0,1,2,...,2n-2 .
k n-1 ....
64. Let D be the region in Euclidean n-space consisting of all
n-tuples (x1,x„,...,x ) satisfying
x. a 0 , x. a 0 , ... , x a 0 , x, + x„ + ... + x si.
1 I n 1 2 n
Evaluate the multiple integral
tt t k. k. k k .
(64.0) JJ...J xl x2 ...xnn (l-x1-x2-...-xn) n dx1...dxn ,
where ]£_...,k . are positive integers.
65i Evaluate the limit
1 n
L = lim - Y
n ^
n-i-o> k=l
2/n
A
66. Let p and q be distinct primes. Let S be the sequence
consisting of the members of the set
r m n « , „ t
{p q : m,n = 0,1,2,... }
arranged in increasing order. For any pair (a,b) of non-negative

16 PROBLEMS (67-74)
integers, give an explicit expression involving a, b, p and q for
the position of p q in the sequence S .
57, Let p denote an odd prime and let Z denote the finite
field consisting of the p elements 0,1,2,...,p-l . For a an
element of Z , determine the number N(a) of 2x2 matrices X ,
P
with entries from Z , such that
P
2
(67.0) X = A , where A
a 0
0 a
58, Let n be a non-negative integer and let f(x) be the
unique differentiable function defined for all real x by
(68.0) (f(x))2n+1 + f(x) - x = 0 .
Evaluate the integral
fX
f(t) dt ,
J0
for x a 0
59, Let f(n) denote the number of zeros in the usual decimal
representation of the positive integer n , so that for example,
f(1009) = 2 . For a > 0 and N a positive integer, evaluate the
limit
t = 11m I" S(N)
where
N
S(N) = I a
k-1
f(k)

PROBLEMS (67-74)
17
70, Let n a 2 be an integer and let k be an integer with
2 £ k £ n . Evaluate
M » max min (a .-a ) ,
S Hsisk-l J
where S runs over all selections S - {a.,a , ...,a,}
{l,2,...,n} such that a. < a. < ... <
from
\
2
71, Let az + bz + c be a polynomial with complex coefficients
such that a and b are nonzero. Prove that the zeros of this
polynomial lie in the region
(71.0) |z| s J + £
72, Determine a monic polynomial f(x) with integral coefficients
such that f(x) = 0 (mod p) is solvable for every prime p but
f(x) =0 is not solvable with x an integer.
73, Let n be a fixed positive integer. Determine
M - max £ lx1~xi I "
0^xkSl l*i<jsn J
k-l,2,...,n
74, Let {x±: i - l,2,...,n} and {y±: i - l,2,...,n} be two
sequences of real numbers with
X. 2 X. 2 ... 2 x .
12 n
How must y.,...,y be rearranged so that the sum
n
(74.0) I (x - y )'
i»l
is as small as possible?

18
PROBLEMS (75-84)
75 Let P be an odd prime and let Z denote the finite field
/ —'i * p
consisting of 0,1,2,...,p-l . Let g be a given function on Z
with values in Z . Determine all functions f on Z with values
P P
in Z , which satisfy the functional equation
P
(75.0) f(x) + f(xfl) = g(x)
for all x in Z .
7K Evaluate the double integral
,1 el
(76.0)
- ff dxdy
77 Let a and b be integers and m an integer > 1
Evaluate
a+b
2a+b
(m-l)a4b
73 Let a.,...,a be n (>1) distinct real numbers. Set
2 2 2
S = a, + ... + a , M= min (a. - a.)
1 n lsi<jsn i j
Prove that
1 > n(n-l)(n+l)
M " 12
79, Let x.,...,x be n real numbers such that
k=l K k=l k
Prove that
(79.0)
i. i k
k=l
2 2n

PROBLEMS (75-84)
19
30, Prove that the sum of two consecutive odd primes is the
product of at least three (possibly repeated) prime factors.
31, Let f(x) be an integrable function on the closed interval
[ir/2,ir] and suppose that
f f(x)sinkxdx = (° > 1 SkSn-1 ,
Prove that |f(x)| a —;—r on a set of positive measure.
1 ' tt Zn 2
32, For n = 0,1,2,... , let
(82.0) s = 7a. + Va , + Va , + . .. + Va„
n n n-1 n-2 0
where a = —-7- . Show that lim s exists and determine its
n n+1 _ n
n->o>
value.
33, Let f(x) be a non-negative strictly increasing function on
the interval [a,b], where a < b . Let A(x) denote the area below
the curve y = f(x) and above the interval [a,x], where a < x <, b,
so that A(a) = 0 .
Let F(x) be a function such that F(a) - 0 and
(83.0) (x1 - x)f(x) < F(x') - F(x) < (x'-x)f(x')
for all a s x < x' < b . Prove that A(x) = F(x) for a <, x <, b .
34, Let a and b be two given positive numbers with a < b .
How should the number r be chosen in the interval [a,b] in order
to minimize
r - x
(84.0) M(r) = max
a<x<b

20
PROBLEMS (85-94)
§5, Let {a : n = 1,2,... } be a sequence of positive real
numbers with lim a = 0 and satisfying the condition
n + » n
a -a , > a ,-a , _ > 0 . For any e > 0 , let N be a positive
n n+1 n+1 n+2
integer such that a s 2e . Prove that L - £ (-1) a, satisfies
k=l
the inequality
N .
(85.0) |L - I (-D a. | < e .
k=l K
35, Determine all positive continuous functions f(x) defined
on the interval [0,ir] for which
J0
(86.0) | f(x)cosnxdx = (-l)n(2n+l) , n» 0,1,2,3,4
37, Let P and P' be points on opposite sides of a non-
circular ellipse E such that the tangents to E through P and
P' respectively are parallel and such that the tangents and normals
to E at P and P' determine a rectangle R of maximum area.
Determine the equation of E with respect to a rectangular
coordinate system, with origin at the centre of E and whose y-axis is
parallel to the longer side of R .
88. If fo"r distinct points lie in the plane such that any three
of them can be covered by a disk of unit radius, prove that all four
points may be covered by a disk of unit radius.
89, Evaluate the sum
00 00
m=l n=l m - n
m*n

PROBLEMS (85-94)
21
QQ If n is a positive integer which can be expressed in the
2 2 2
form n = a + b + c , where a,b,c are positive integers, prove
2k
that, for each positive integer k , n can be expressed in the
2 2 2
form A + B + C , where A,B,C are positive integers.
91, Let G be the group generated by a and b subject to the
3 5
relations aba = b and b = 1 . Prove that G is abelian.
92, Let {a : n = 1,2,3,... } be a sequence of real numbers
00
satisfying 0 < a < 1 for all n and such that £ a diverges
00 2 n n=1
while 1 a converges. Let f(x) be a function defined on [0,1]
n»l =>
such that f"(x) exists and is bounded on [0,1]. If Y f(a )
~ , n
II
converges, prove that £ If(a )l also converges.
n=l n
93, Let a,b,c be real numbers such that the roots of the cubic
equation
3 2
(93.0) x +ax + bx + c « 0
are all real. Prove that these roots are bounded above by
(2/a2-3b - a)/3 .
9U, Let Z - {0,1,2,3,4} denote the finite field with 5
elements. Let a,b,c,d be elements of Z, with a * 0 . Prove that
the number N of distinct solutions in Z, of the cubic equation
2 3
f(x) » a + bx + ex + dx =0
is given by N = 4 - R , where R denotes the rank of the matrix

22
PROBLEMS (95-100)
A =
abed
b c d a
c d a b
d a b c
95, Prove that
(95.0)
is a rational number.
m,n=l (mn)
(m,n)=l
96, Prove that there does not exist a rational function f(x)
with real coefficients such that
(96.0) f(^) - p(x) ,
where p(x) is a non-constant polynomial with real coefficients.
97, For n a positive integer, set
n ,
S(n) = I -±-
"0
Prove that
j.i n+1 ok
S(n) mJS±L I 1-
2n+1 k-1 k
98. Let "(x) fae a non-trivial solution of the differential
equation
u!' + pu = 0 ,
defined on the interval I = [!,">) , where p = p(x) is continuous

PROBLEMS (95-100)
23
on I . Prove that u has only finitely many zeros in any interval
[a,b], 1 £ a < b .
(A zero of u(x) is a point z , 1 £ z < ■» , with u(z) = 0).
99, Let P (j = 0,1,2,...,n-l) be n (£2) equally spaced
points on a circle of unit radius. Evaluate the sum
S(n) = I \K?\2 ,
0Sj<k<n-l J K
where |PQ| denotes the distance between the points P and Q .
100. Let M be a 3x3 matrix with entries chosen at random from
the finite field Z = {0,1} . What is the probability that M is
invertible?

THE HINTS
1, Define
The little fishes of the sea,
They sent an answer back to me.
The little fishes' answer was
"We cannot do it, Sir, because —
Lewis Carroll
a+b„+b, +...+b , n > 0,
0 1 n
and prove an inequality of the type
a - a ,
n n-1 ,
372— * c
a
n
1 1
1/2 1/2
a , a
n-1 n
, n > 1 ,
where c is a constant.
2. Consider five cases according as
(a) b > d ,
(b) b = d and a > c ,
(c) b < d ,
(d) b = d and a < c ,
(e) b = d and a ■= c .
In case (a) show that L = + ■» by bounding Q (a,b,c,d) from below
H n
by a multiple of (—) . In case (b) show that L = + ■» by
estimating 0 (a,b,c,d) from below in terms of the harmonic series.
25

26
HINTS (3-14)
Cases (c) and (d) are easily treated by considering -pr-7—r tt-
(^(.a.b.c.d;
The final case (e) is trivial.
3, A straightforward approach to this problem is to show that the
function
F(X) » (x3-l)(*+l) _ 3lnx 5 x > 0 5
(xJ+x)
suggested by the inequality (3.0), is increasing.
l\, Apply Rouche's theorem to the polynomials f(z) = -z and
g(z) = p(z) . Rouche's theorem states that if f(z) and g(z) are
analytic within and on a simple closed contour C and satisfy
\g(z)\ < \f(z)\ on C , where f(z) does not vanish, then f(z)
and f(z) + g(z) have the same number of zeros inside C .
5, Apply the change of variable x - a - t to
I =
6. Integrate
£(x) .„
f(x) + f(a-x) aX
-__ /^ dt
z m [™ 2t sin (t^)
Jx 2t
by parts and obtain an upper bound for |l| .
7. Consider (7.0) modulo 16.
8. By squaring (8.0), obtain the lower bound /2kn+a for x .
n
Using this bound in (8.0) obtain an upper bound for x
n

HINTS (3-14)
27
9, Assume that the required limit exists, and use (9.0) to
determine its value L . Again use (9.0)to estimate |x - L| .
10, Either set x = cos 9 , y = sin 9 and maximize the resulting
2 2
function of 9 , or express ax + 2bxy + cy in the form
2 2 2
A(x + y ) - (Bx + Cy) for appropriate constants A,B,C .
11, Consider the coefficient of x in both sides of the identity
(1 + x)2n = ((1 + 2x) + x2)n .
12, Express (IU + 1) )k - (« - l)£)k , (k = 1,2,3,. . . ) as a
polynomial-in t , then sum over £ ■= 1, 2, ..., n to obtain
k
(n(n + 1)) as a linear combination of
S[k/2](n), .... Sk_l(n) .
Complete the argument using induction.
13, Prove that the equation
be x + cay + ab z = 2abc - (be + ca + ab) + k
is solvable in non-negative integers x,y,z for every integer k i 1.
Then show that the equation with k = 0 is insolvable in non-negative
integers x,y,z .
14, Let u be the n term of the sequence (14.0) and show
that u = k for n = (k~1)k + 1 + I , I = 0, 1, 2, ..., k-1 ,
n i ^
and deduce that k £•=-(! + /8n-7) < k + 1 .

28
HINTS (15-27)
15. Use Cauchy's inequality to prove that
-1
n 2 > f n 2]
I Xi * [J aij
i«l i«l
and choose the x. so that equality holds.
3
16. Express (n + 1) in the form
An(n - l)(n - 2) + Bn(n - 1) + Cn + D
for suitable constants A,B,C,D .
17. Integrate F*(a-x) - F'(x) twice.
18. For part (b), find the quartic equation whose roots are
a-z,b-z,c-z,d-z, and use part (a) to ensure that the
product of two of these roots is equal to the product of the other
two.
19. Differentiate f (x) to obtain the difference-differential
equation
fn+1(x) - fn(x)-p'(x)fn(x) .
2 1 *—
20. Consider sinh x + sinhwx + sinh w x, where w ■ y(_l + S-3)
21. Show that
1 " x i
n n-1
Sin(2n ~ l) * dx , n S 2 ,
0 SlnX
and then use a similar idea to evaluate the integral on the right side.

HINTS (15-27)
29
221 Let a. (i - 1, 2, ..., 365) denote the number of books sold
i j.t.
during the period from the first day to the i day inclusive.
Apply Dirichlet's box principle to
at, a2, ..., a365, a^+129, a2+129, ..., a^+129 .
231 Show that a polynomial of the required type is
f(x,y) . <«+7- lHx + y,-2)+x t
by showing that f(x,y) - k , where k is a positive integer, has a
unique solution in positive integers x and y which may be expressed
in terms of the integers r and m defined by
(r - l)(r - 2) , ,. ^ r(r - 1) (r - 1) (r - 2)
2" < K S ^ > m « k 5 .
24. Consider the complex conjugate of (24.0).
251 Consider
* ^ f1 ~ &}
r-kn+1"- n >
and use the expansion
- In (1 - x) - J ^ , |x| < 1 .
k«l
26. For the evaluation, set x - a - y in the integral.
27, Use mathematical induction to prove that a - r for all
positive integers r .

30
HINTS (28^2)
28i Use integration by parts to establish the recurrence relation
n(n-l) „ > o
" = —rr~7" u -> ) n i 2 .
n n2+p2 n-2 '
29• The series may be summed by using the identity
tan A « cot A - 2 cot 2A .
30. Prove that the number of k-selections S from N such that
W(S) > r , r = 1,2,3,..., is
n - (k-D(r-l)
k
31i For each n i 1 define integers q and r uniquely by
n=kq +r , 0 £ r <k. Express the nth partial sum s of
Hn n ' n r v n
the series in terms of n and q , and determine lim s by appeal-
n ^m n
n "*■ °°
ing to the result
lim (1 +•=•+... + In m) = c ,
where c denotes Euler's constant.
32. Recognize the given integral as the imaginary part of the
integral I x e dx . Evaluate the latter integral using
integration by parts.
1 2
33. For the evaluation, iterate I(u) = yl(u ) to obtain
1 ?n
I(u) = — 10/ ) (n = 1,2,3,...)
2
and then let n ■*■ + <*> in the case 0 < u < 1 .

HINTS (2&42)
31
34. Determine an exact expression for S-Ck) and then compare
the values of 8^(10 and sk(k) with 3k3 - 5k2 .
<*> p -1
35. Show that (35.0) converges by comparison with Y .
n-1 Pl'--pn
36. Apply l'Hopital's rule.
37. Relate h,k,£ to the lengths of the sides of the triangle,
and thea use the triangle inequality.
38. Obtain the recurrence relation P , , "» P . , ,+xP ,
n+l,r n+l,r-l n,r
and apply the principle of mathematical induction.
39. Show that Bx + D is a divisor of F .
40. The first few terms of the series are
3 = l[l ~ ij '
-i. id - i)
21 2{3 7J '
.1 Ifl J_
91 2 7 13
41, Use the identity
(l-x)(l-2x)...(l-nx) = 1 - P.x + P,x2 - P-x3 + ... + (-l)nP xn
1 l o n
x
42, F°r a suitable constant C , set f(x) = — + C , and
eX + 1
show that for t > 0

32
HINTS (43-56)
f , ax bXv
! (e - e ) ,
i j—- dx =
J0*'eaX+l)(ebX+l)
t
f(ax)
x
0
-dx -
f(fax)dx
X
0
43. Let S denote the sum of n terms of (43.0). Calculate
the first fev values of S , conjecture the value of S in general,
and prove it by mathematical induction.
44i Consider the polynomial whose roots are z.,z ,...,z , and
use (44.0) to show that its constant term is zero.
45 ■ Obtain a recurrence relation for D by expanding D by
its first rov.
£j6i Express xyz in terms of A, B and C .
47. Consider the sums of the integers in subsets of S and apply
Dirichlet's box principle.
48. Count pairs of lines in the proposed configuration.
(a) ' I [I
k-0 <• •
49i Show that A(n) - £ I, and evaluate this sum by considering
k-0 W
k=0 (mod 3)
(1+1) + (1+w) + (1+w ) , where w is a complex cube root of unity.
50, To prove the required inequality, replace p (x) by
-—-— 1 + v» *n (50.0), and then use the inequalities
4n,, _.N4n ,_,\ti,n
in (50.0), and thei
,1
< 1 and x(l-x) < 7- to estimate
1+X2 - - «"" "^ "' - 4
x4n(l-x)4n,
TT*— dx-

HINTS (43-56)
33
51, Let B. ,B_,...,B_3 (resp. G.,G_ G_,) be the members of
the blue (resp. green) team, ordered with respect to increasing weight.
For each r (1 £ r £ 23) consider last year's opponents of
B ,,...,B„, or G .......G_, according as B is heavier or
r+1 23 r+1 '23 r
lighter than G .
r
52■ Each member of S can be written in the form (2r+l)(2r+2s+l),
for suitable integers r £ 1 and s £ 0 . Use this fact to construct
the three arithmetic progressions.
53. For y > 0 prove that
-ux . , i ,
e sin x dx du =
(1-e ') dx
and then show that
lim
,, -xy-.si.nx ,
(1-e ') dx =
x
b .
sinx
dx
43
54i Consider £ d.
j-1 J
55, F°r any natural number n , construct a prime p of the form
n 2 2
p - 4k II (r +qr - q ,
r-1
where k is a natural number and q > n is a prime of the form
2 2
4t + 3 , so that p=a +b ,0<a<b. Then, assuming a S n ,
2 2
obtain a contradiction by considering the factor a +q of b
56. For 2 £ r £ n obtain a lower found for a -a, in terms of
r 1
b and r by considering the differences a.-a. , 1 S j < i £ r .

34
HINTS (57-68)
57. Evaluate D ,D ,D and conjecture the value of D for all
) 1 2. j n
n . Prove your conjecture by using the recurrence relation which may
be obtained by expanding D by its first row.
58. Prove that
x = x , + a - /bs + 4ax , ,
n n+1 n+1
and use this to obtain the recurrence relation
x ,, - 2x + x , = 2a .
n+1 n n-1
59. For r > 0 and r * 1 show that
^-2>r -
J~ PL
I = a +
r 2K '\ 1 - 2rcosu + r2 '
and evaluate the integral using the transformation t = tan u/2 .
60. Construct a class of isosceles triangles whose members have
two equal altitudes of fixed length h , while their third altitudes
are arbitrarily long.
61. Recognize the expression in (61.0) as the square of the
distance between a point on a certain circle and a point on another plane
curve.
62. When the line L through the origin has irrational slope,
use Hurwitz's theorem to obtain an infinity of lattice points whose
distances from L are suitably small.
In 1881 Hurwitz proved the following basic result: If b is
an irrational number then there exist infinitely many pairs of integers
(m,n) with n + 0 and GCd(m3n) = 1 such that

HINTS (57-68)
35
b -
/5n2
This inequality is best possible in the sense that the result
becomes false if /F is replaced by any larger constant.
65. Show that I, = I- , , and use the arithmetic-geometric
KJJ' k 2n-k-2
mean inequality.
64, Express the multiple integral (64.0) as a repeated integral
and use the value of L x (a-x) dx , where r and s are positive
integers and a is a positive real number, successively in the
repeated integral.
65, Show that for a suitable integer f(n)
4n
m -
f(n)
= I
s=l
(2s+l)2
4n
(2s+2)2
and thus compute L in terms of well-known series.
66i P 4 is the n term of the sequence S , where n is the
number of pairs of integers (r,s) such that pq Spq , r £ 0 ,
s > 0 .
67i * straightforward approach is to determine explicitly all
2
matrices X such that X = A . The form of X depends on whether
or not a is a square in Z
68. Recall that if y = g(x) is differentiable with positive
derivative for x i 0 and g(0) - 0 , then
rx rg(x)
g(t) dt +
g_1(t) dt = xg(x) , x > 0

36
HINTS (69-81)
59t Evaluate S(10 -1) exactly and use it to estimate S(N) .
70. Show that M - £r| •
2
71, Express the roots of az + bz + c in terms of a, b and c
and estimate the moduli of these roots.
2
72, Choose integers a, b and c such that x + a = 0 (mod p)
2
is solvable for primes p = 1 (mod 4) and p«2;x + b = 0 (mod p)
2
is solvable for p = 3 (mod 8) ; x + c = 0 (mod p) is solvable for
p = 7 (mod 8) ; and set
f(x) - (x2+a)(x2+b)(x2+c) .
73. Assume without loss of generality that 0£x.£x_£...£x £1
and show that
n
1 |x. " x | - I x (2k-n-l)
l£i<j£n x J k-1 K
Consider those terms in the sum for which k £ ^-(n+1) and deduce
that M - [n2/4] .
74, Show that the smallest sum (74.0) is obtained when the y.
are arranged in decreasing order.
75, Replace x by x+k (k « 0,1,2,...,p-l) in (75.0) and form
the alternating sum
p;1 k
I (-1) g(x+k) .
k-0

HINTS (69-81)
37
76. Express the improper double integral I as a limit of proper
double integrals over appropriate subregions of the unit square and
2
use standard methods to show that I « ir /6 .
77, Use the identity
BH-1*"-
where k is any positive integer and e is any real number.
2 2
78. Reorder the a's in ascending order and define min a ■ a ,
t c- j v • * • l<i<n J
for a fixed subscript j .
Set b « a + /S(i - j) (i = 1,2,...,n) and prove that a * b .
n 2
Deduce the required inequality from S > £ b. .
i-1 X
79. Establish and use the inequality
k n
£ 1 - - , 1 £ k £ n .
n
80. Denote the n prime by p , and show that if
k I n
p + p , = 2 p , for some odd prime p , then k + t 2 3 .
rn rn+l r
81. Estimate the integral
In/2
|£(x)|
y sinkx
k-1
dx
from above under the assumption that |f(x)| < . . on [-t-.itj
except for a set of measure 0. Use (81.0) to obtain a lower bound
and derive a contradiction.

38
HINTS (82-96)
92. Show that s is non-decreasing and bounded above.
83, Assume that A(x) and F(x) differ at some point c in (a,b]
and obtain a contradiction by partitioning [a,c] and using (83.0) on
each subinterval.
84. A direct approach recognizes M(r) as max(— - 1,1 -T-) and
then minimizes M(r) with an appropriate choice of r .
n , .
85. Let S = I (-1) a, and show that |S -l| < Is ,-l| and
n , . k n n-1
k=l
a - |S -Ll + |S ,-L| .
n ' n ' ' n-1 '
2 2
86. Express (cos x + cosx +1) as a linear combination of
cos nx (n = 0,1,2,3,4) and consider
f1T 2 2
f(x)(cos x + cos x +1) dx .
87. Begin by determining R when the ellipse is in standard
position and then rotate the axes through an appropriate angle.
88. Recall Belly's theorem: Given n(z4) convex regions in the
plane suoh that any three have non-empty intersection, then all n
regions have non-empty intersection.
89. Use partial fractions and the result
f N 1
Um I - - In N
N-i-<x>l.k=l
= c ,

HINTS (82-96)
39
where c is Euler's constant, to evaluate
N 1
lim T 2 2
u m —n
n=l
n*m
, m -n
90. Use the identity
2222 722? ? ?
(x +y +z ) = (xV-z ) + (2xzT + (2yz) .
gj_, Prove that a and b commute by using the relation aba = b
in the form b ab = b a to deduce ab = b a .
92, Start by applying the extended mean value theorem to f on
[0,an].
93, Let p be the largest root of (93.0). Consider the discrim-
3 2
inant of (x +ax +bx+c)/(x-p) .
Cji)i Let B be the Vandermonde matrix given by B =
and consider the rank of BA .
1 1 l2 l3'
1 2 22 23
1 3 32 33
1 4 42 43
95, Collect together terms having the same value for GCD(r,s)
°° 1
in L<rs)2 •
r,s=l
95, Suppose that such a rational function f(x) exists and use
the decomposition of its numerator and denominator into linear factors
to obtain a contradiction.

40
HINTS (97-100)
97. Sum the identity
2 1
(n+1)!
(n+1)
for k - 0,1,2,...,n-l
r??i
= k!(n-k)! - (k+1) ! (n-k-1)! ,
98. Assume that the set of zeros of u(x) on [a,b], 1 s a < b ,
is infinite. Deduce the existence of an accumulation point c in
[a,b] with u(c) » u'(c) = 0 , and then show that u(x) = 0 on [a,b].
99. Take P (j = 0,1,2,
the unit circle I z
,n-l) to be the point exp(2irji/n)
1 in the complex plane, and express
|P.P. | in term>* of exp(2ir(k-j)i/n) .
IDO. Let M = (a ) (1 s 1,] i 3) and with the usual notation
let det M = a..A.. + a..A . + a.-A , . Begin by counting the number
of triples (a..,a..,a.-) for which det M = 0 , distinguishing two
cases according as CA »Aio'A13^ = (°»°>°) or not-

THE SOLUTIONS
It seemed that the next minute
they would discover a solution.
Yet it was clear to both of them
that the end was still far, far off,
and that the hardest and most complicated
part was only just beginning.
Anton Chekhov
1, If {b : n = 0,1,2,...} is a sequence of non-negative real
numbers, prove that the series
oo b
(1-0) Jo (a+bn+b,+n..+b)3/2
U 1 n
converges for every positive real number a .
Solution: For -a > 0 we set
a = a + b„ + b, +...+b , n a 0 ,
n 0 1 n
so that
a - a , = b , njl
n n-1 n
As b i 0 we have, for n > 1 ,
n
a„ > a > 0 and a > a , > 0
0 n n-1
41

42
SOLUTIONS (1-2)
Now, for n i 1 , we may deduce that
(a+br,+b.+. ..+b )
U 1 n
371
a - a
n r
37?"
a
n
1
77?
n-l
"371
Hence, for m i 1 , we have
b
= I
"n-1
Tft
-l
n
n
n-l
i
a i
n-l
1
1
"77?
17?
l . a.
n-l n
1/2
n-l
J72
a
1 +
172
n-l
m " (a-An+b.+. ..+b )
n=0 0 1 n
37T
T7I
372
+ 2
n-l
77?
171
2 I
n-l
1
n-l
172
"T7I "T72
a , a
n-l n
"77?
17?
o
X + 2
^7? "77?
ao ao
137?
T7?
As the partial sums s of (1.0) are bounded, the infinite series
m
(1.0) converges for every a > 0.

SOLUTIONS (1-2)
43
2, Let a,b,c,d be positive real numbers, and let
n , u ,,"> - a(a+b)(a+2b)...(a+(n-l)b)
Qn(a,b,c,d) - c(c+d)(c+2d)...(c+(n-i)d)
Evaluate the limit L = lim 0 (a,b,c,d).
Solution: Considering the five cases specified in THE HINTS, we show
that
1 + °°, in cases (a),(b) ,
0 , in cases (c),(d) ,
1 , in case (e) .
We set k - |<j| + 1 , so that c < kd .
[f]
(a) When b > d , as c + jd < (j + k)d , we have for n - k
„ , , ,s > a(n-l)!b
Qn(a,b,c,d) - k(k+1)...(k+(n-i))dn
_ aCk-Dib11"1
(k+ti-l)(k+ti-2)...ndn
(k+ti-l)k d 1 d
which tends to infinity as n tends to infinity, showing that L= + =>.
(b) When b = d and a > c we have
n-1
Q (a,b,c,d) = n
n 1-0
a + lb
c + lb
n-1 ,
n i+ a -c
1=0 V c + lb
n-1 ,- i n-1 ,
> n [i+^ > ^ l^r
t;0 I1 - RWJ
i-0

44
SOLUTIONS (3^t)
which tends to infinity as n tends to infinity since the harmonic
series diverges, showing that L = + => .
(c)(d) By considering the reciprocal of Q (a,b,c,d), we obtain from
n
the above results L - 0 , if b<d,orif b=d and a < c .
(e) Clearly Q(a,b,c,d) - 1 in this case, so that L = 1.
3. Prove the following inequality:
rn n\ £n X . 1 (x+1) . n _ .
(3.0) —z < -T -^-3 » X > 0, X * 1.
x -1 (x +x)
Solution: For x > 0 we define
F(X) . (x3-l)(x+l) _ 3£nx f
(xJ+x)
so that
F(x) - x , x l - 3 £n x .
(xJ+x)
Differentiating F(x) with respect to x , we obtain
F'(x) - (^x3+3x2-l) (x3+x) - (x4+x3-x-l)(3x2+l) _ 3_
, 3. -.2 x
(x +x)
, x6+3x4+4x3+3x2+l _ 3
, 3, -.2 x
(x +x)
that is
(3.1) F'(x) = x ~3x "*"3x _2x +3x _3x+1
(x3+x)2
The polynomial p(x) in the numerator on the right in (3.1) has the
property that p(x) = x p(—) ,
cubic polynomial in x + 1/x .
fi 1 —^
property that p(x) = x p(—) , and so x p(x) can be written as a

SOLUTIONS (3-4)
45
We have
x |, .1,3 ,,..,1,2
F'(x) - * , (x-f-r - 3(x+^)' + 4
(x3+x)Z<- X X
As X3-3X+4 = (X*l)(X-2)2 we obtain
F.(x).^W(x+i)-2)2
(x +l; *■ '
((x4)2+|)(x-l)2
x (x +1)
so that F'(x) > 0 for x > 0 , while F'(l) =0. Thus F(x) is a
strictly increasing function of x for all x > 0 . Hence in
particular we have
F(x) > F(l) , for x > 1 ,
and so
ii ii In x 1 (x+1) . ,
(3.2) —=— < -r- —5 , for x > 1 .
x -1 i (xJ+x)
Replacing x by — in (3.2), we obtain
(i i\ In x 1 (x+1) . n ,
(3.3) —5— < -s- —5 , for 0 < x < 1 .
x -1 (x +x)
Inequalities (3.2) and (3.3) give the required inequality.
£|, Do there exist non-constant polynomials p(z) in the complex
variable z such that |p(z)| < R on |z| • R , where R > 0 and
p(z) is monic and of degree n ?
Solution: We show that no such polynomial p(z) exists, for suppose
there exists a non-constant polynomial
p(z) ■= z + a ,z + ... + a.z + an
such that |p(z)| < Rn on |z| = R .

46
SOLUTIONS (5-7)
Then we have
|zn + (a _ z11- + ... + a^ + a ) | < |-zn| on |z| = R ,
and so, by Rouche's theorem,
n, n-1 ,»nn
z +(a ,z + ... + a.z + an; - z and -z
n-1 1 U
have the same number of zeros counted with respect to multiplicity
inside |z| = R , that is, a _.z + ... + a.z + a- has n zeros,
which is clearly a contradiction. Hence no such polynomial p(z)
exists.
5, Let f(x) be a continuous function on [0,a] , where a > 0 ,
such that f(x) + f(a-x) does not vanish on [0,a] . Evaluate the
integral
fa f(x) .
JQf(x) + f(a-x) dX •
Solution: Set
fa ,, % ra
I
Clearly we have
fa f(*) A y f f(a-x) ,
= jQf(x) + f(a-x) dx ' J = Jjf(x) + f(a-x) dx
I + J - I 1 dx = a .
Jo
On the other hand, changing the variable from x to a-x in I , we
obtain
I = J .
Hence we have
I = J = j a .

SOLUTIONS (5-7)
47
6. For e > 0 evaluate the limit
fX+1
lim x
x-»-=>
1-e
sin (t ) dt
Solution: Integrating by parts we obtain
2. , , ,.2N
I-
in(t2) dt = "C°
s(Q 1 fcos(0 ..
2t 2 \— dt
so that for x > 0
fX+1
giving
so that
l-X+l
Lsi
x+l
in(t2) dt = ""g'^P + C0.S *' - lJ*+lc0Sjt2)
sin(t ) dt
2 (x+l) "•" 2x 2^ .2
, 1 .11 fx+1dt _ 1
S 2(x+l) +!Z + l\x ^2 " x
dt
1-e
rx+l
sin(t ) dt
Since — •+ 0 , as x -»■ + =>, we deduce that
x
1-e fx+1 2
lim x1 *■ sin(0 dt = 0 .
x-»-o> Jjj
7. Prove that the equation
(7.0)
has no solutions in integers x,y,z
444 ,22 ,22 ,22
x +y + z -2yz -2zx -2xy =24

48
SOLUTIONS (7-8)
Solution: Suppose that (7.0) is solvable in integers x, y and z .
4 4 4
Clearly x +y +z must be even. However x,y,z cannot
all be even, as 24 is not divisible by 1-6 . Hence exactly one of
x,y,z is even", and, without loss of generality, we may suppose that
x = 0 (mod 2) , y = z = 1 (mod 2) .
Thus we have
x = 0 (mod 16) , y = z = 1 (mod 16) ,
-2y z = -2 (mod 16) , -2z x = -2x y = -2x (mod 16),
and so (7.0) gives
-4x2 = 8 (mod 16) ,
that is
2
x = 2 (mod 4) ,
which is impossible.
Second solution: We begin by expressing the left side of (7.0) as
the product of four linear factors. It is easy to
check that
A2 + B2 + C2 - 2BC - 2CA - 2AB = (A+B-C)2 - 4AB
= ((A+B-C) - 2/Sf) ((A+B-C) + 2/AB) .
2 2 2
Replacing A,B,C by x ,y ,z respectively, we obtain
4^4^4 ,22 ,22 ,22
x +y +z -2yz -2zx -2xy
= (x + y - z - 2xy) (x + y - z + 2xy)
- ((x-y)2 - z2)((x+y)2 - z2)
= (x - y - z)(x - y + z)(x + y - z)(x + y + z) ,
so that (7.0) becomes
(x-y - z)(x - y + z)(x + y - z)(x + y + z) = 24 .

SOLUTIONS (7-8)
49
In view of the form of the left side of (7.0), we may assume
without loss of generality that any solution (x,y,z) satisfies
x £ y £ z £ 1 , so that
x-y-z£x-y + z£x + y-z£x + y + z .
3
Moreover x - y - z and x - y + z cannot both be 1 . As 24 = 2 *3,
we have
(x-y-z, x-y+z, x+y-z, x+y+z)
= (1,2,2,6),(1,2,3,4) or (2,2,2,3).
However none of the resulting linear systems is solvable in positive
integers x,y,z .
2
8, Let a and k be positive numbers such that a > 2k.
Set x. = a and define x recursively by
0 n
(8.0) x =x +-^-, n = 1,2,3
n n_l x -
n-1
Prove that
x
, . n
lim -=
M
n-*-=> *
exists and determine its value.
Solution: We will show that
x
lim -§ = /2k .
n + » <F
Clearly x > 0 for all n > 0. Since x = x , + for
n n n-1 x ,
n = 1,2, . . . , we have
2 2 , „ , k2
(8.1) x » x . + 2k + -£
n n-1 2

50
SOLUTIONS (8-9)
and so
2 2 2 2 2 2
x > x .+2k > x „+4k > x ,+6k > ... > x.+2kn = a +2kn ,
n n-1 n-2 n-3 0
that is
(8.2)
x Z /2kn + a2 , n= 0,1,2,...
n
On the other hand, we have, using (8.1) and (8.2),
, 2
2 2
x £ x . + 2k +
n n-1
2k(n-l)+a
, n - 1,2,... ,
and thus
n-1
7 7 7 1
x. < x. + 2kn + k V ,, . *
n 0 .Ln 2ki +
+ 2kn + k
i=0
2 dx
2kx + a2
2kn + a2 + | In
2k(n-l) + a
a2-2k
giving
(8.3) x
:n<,/2kn + a2+^£nf2krl+(a2-2^
2«n a2-2k L) • n = 0,1,2,...
Hence, from (8.2) and (8.3), we obtain
1 £
i+a2 "V 2 2kn + a2
/2kn+,
and thus
lim
/2kn + a2
Since lim /- = /2k , we obtain lim -£ = fK .
/n /n
n -»■ ■» n -»■ =>

SOLUTIONS (8-9)
51
9, Let x. denote a fixed non-negative number, and let a and
b be positive numbers satisfying
/b < a < 2/b .
Define x recursively by
n
(9.0)
ax , + b
n-1
n x , + a
n-1
— , n - 1,2,3,...
Prove that lim x exists and determine its value.
n-*-<»
Solution: As xf)£0,a>0,b>0, the recurrence relation shows
that x > 0 for n = 1,2,... . If lim x exists, say
n n ' '
equal to L , then from (9.0) we obtain
aL + b
so that L = b , L = + ,/b .
Next we have
fi\
L + a '
ax . + b
n-1
x , + a
n-1
/b
(a
(a -
n-i
x , + a
n-1
A>ivi-
- A)
^1
x , + a
n-1
. (a - A) i r-\
* —i—ivi" ^i
"n-1 " *l
so that
|xn - A| <
1*0 ' M
Letting n tend to infinity, we obtain
lim x " A> .
n
n-*=>

_52 SOLUTIONS (10)
10, Let a,b,c be real numbers satisfying
2
a > 0, c > 0, b > ac .
Evaluate , ,
max (ax + 2bxy + cy ) .
x,y eR
x2+y»-l
2 2
Solution: All pairs (x,y) e RxR satisfying x + y = 1 are given
by x ■ cos 8,y"sin8,0£8£2Tr. Hence we have
2 2
max (ax + 2bxy + cy ) = max F(8) ,
(x,y)eR2 0S8S2TT
x2+y2=l
where
2 2
F(8) = a cos 8 + 2b cos 8 sin 8 + c sin 8
- ^(1 + cos 26 ) + b sin 28 + ^-(1 - cos 28 )
= -^-(a+c) + b sin 28 + -^(a-c) cos 28
= -^(a+c) + •^'(a-cJ'+Ab2 sin (28 + a) ,
where
tana
2b
Clearly max F(8) is attained when sin (28 + a) = 1 , and the
0£8£2tt
required maximum is
•^•(a+c) + -.-/(a-c)2 + 4b2
Second solution: We seek real numbers A,B,C such that
(10.1) ax2 + 2bxy + cy2 = A(x2+y2) - (Bx+Cy)2 .
Equating coefficients we obtain
(10.2) A - B2 - a ,
(10.3) -2BC -2b ,
(10.4) A - C2 = c .

SOLUTIONS (10)
53
Subtracting (10.2) from (10.4) we obtain
(10.5)
Then,from (10.3) and (10.5), we have
2 2
B - C - c - a .
(B2 + C2)2 - (B2 - C2)2 + (2BC)2
- (c - a)2 + 4b2 ,
so that
(10.6) B2 + C2 = +/(a-c)2 + 4b2 .
Adding and subtracting (10.5) and (10.6), and taking square roots, we
get
(10.7) B
//(a-c)2 +
4b2 - (a-c)
, C
v(a-c)2 + 4b2 + (a-c)
Then, from (10.2) and (10.7), we have
A = -|(/(a-c)2 + 4b2 + (a+c)) .
2 2
Finally, from (10.1), we see that the largest value of ax +2bxy+cy
2 2-
on the circle x +y =1 occurs when Bx + Cy » 0 , that is, at the
points
(x.y)
±C
?B
2 2 ' 2 2
B + CT B + C
and we have
max (ax +2bxy+cy ) = A - •5-(v/(a-c)2+4b2 + (a+c))
x2+y2-l

54
SOLUTIONS (11-12)
11, Evaluate the sum
(U.O)
„ [n£2]n(n-l)...(n-(2r-l ,n-2r
S = I 1 2
r=0 (r!)
for n a positive integer.
Solution: We have
[n(2] n!
S _ vlQ (r!)»(n-2r)! 2
n-2r
[n/2]
I
r=0
nl
n
n-r
n-r 1 _n-2r
n-2r
s ] ,2s-n
2s-n
n s
I I
s=0 t=0
2s-t=n
:»'
which is the coefficient of x in
n s
F(x) = I I
s=0 t=0
s] ,t 2n-2s+t
t 2 X
Now
F(x)
mm2V }<»2'-
i
s=0
° (1 + 2x)S(x2)n_s
((1 + 2x) + x2)°
(1 + x)
2n
As the coefficient of x in (1+x) is ( ) , we have S » 7—ri~-
n ^n!)2

SOLUTIONS (11-12)
22, Prove that for m = 0,1,2,.
(12.0)
is a polynomial in n(n+l)
. . ,2nrt-l „2nrt-l 2nrt-l
S (n) = 1 +2 +...+n
m
Solution: We prove that
2 Q SQ(n) = n(n+l)
Jjs^n) - (n(n+l))2 ,
2^01) + 2^]s2(n) = (n(n+l))3 ,
1,S2(n) + 2
3Js3(n) - (n(n+l))4
and generally for k » 1,2,3,...
k
(12.1)
2J0[r]SlHWn) " (n(n+1))k
r+k odd 2
An easy induction argument then shows that S (n) (m - 0,1
m
a polynomial in n(n+l).
We now prove (12.1). We have
k
2K
r-0
r|Sr+k-l(n)
2 I k I lI+k
r+k odd
r+k odd
n k
l'l r-0
r+k odd
kj .r+k
r

56
SOLUTIONS (13)
'kj^r^k r _ (_£)k r}
I I
l'l r=0
= i {a2+i)k - a2-i)k)
- I (C£te+i))k - (oe-iu)k)
= (n(n+l))k
as required.
13, Let a,b,c be positive integers such that
GCD(a.b) = GCD(b.c) - GCD(c.a) - 1 .
Show that t - 2abc - (bc+ca+ab) is the largest integer such that
be x + ca y + ab z » Z
is insolvable in non-negative integers x,y,z .
Solution: We begin by proving the following simple fact which will
be needed below:
Let AjBjCj be real numbers smsh that
A + B + C < -2 .
Then there exist integers t,u,v satisfying
t - u > A ,
u - v > B ,
v - t > C .

SOLUTIONS (13)
57
To see this, choose
t - [A] + 1 ,
u •= 0 ,
v - -[B] - 1 ,
so that t,u,v are Integers with
t - u = [A] + 1 > A ,
u - v - [B] + 1 > B ,
v - t = -[A] -[B]-2>-A-B-2>C.
The required result will follow from the two results below:
(a) If k is an integer £ 1 , then
box + oay + ab z = 2abo - (bo + oa + ab) + k
is always solvable in non-negative integers x,y,z .
(b) The equation box + oay + ab z = 2abo - (bo + oa +■ ab) is
insolvable in non-negative integers x,y,z.
Proof of (a): As GCD(ab,bc,ca) ■ 1 , there exist integers ■x..,y.,z[
such that
be xn + ca yn + ab z_ » k .
1 ,
1 ,
x,
y0
0

58
SOLUTIONS (13-14)
so that .x v z
f 0 y0 0
[a b c
= -4--2
abc
< -2 .
Hence,by our initial simple fact, there are integers t,u,v such
that
t-u>-^-l ,
a
s ^ 1
u - v > -_- 1 ,
v-t>-!° .
Thus we have
a + x. + at - au > 0 ,
b + yQ + bu - bv > 0 ,
z. + cv - ct > 0 .
Set
x - a-1 + x. + at - au ,
y = b-1 + y_ + bu - bv ,
z - -1 + z. + cv - ct ,
so that x,y,z are non-negative integers.
Moreover
be x + cay + ab z = 2abc - (ab + be + ca) + k
as required.
Proof of (b): Suppose the equation is solvable, then
2abc = bc(x+l) + ca(y+l) + ab(z+l) ,
where x+l,y+l,z+l are positive integers.

59
Clearly, as GCD(a.b) - GCD(b.c) - GCD(c.a) = 1 , we have that a
divides x+1 , b divides y+1 , and c divides z+1 . Thus there
are positive integers r,s,t such that
x+1 = ar , y+1 - bs , z+1 = ct .
Hence we have
2abc = abc(r + s + t) ,
that is
2 = r + s + t£3,
which is impossible.
This completes the solution.
1M, Determine a function f(n) such that the n term of the
sequence
(14.0) 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, ...
is given by [f (n)] .
Solution: Let u be the n term of the sequence (14.0) . The
n
integer k first occurs in the sequence when
n=l + 2+3+... + (k-1) + 1 - (k"21)k + 1 .
Hence u = k for
n
(14.1) n = (k"21)k + 1 + I , 1 = 0,1,2,. ...k-1 .
From (14.1) we obtain
0 £ n - (k"21)k - 1 < k-1
and so

60
SOLUTIONS (15-16)
Multiplying (14.2) by 8 and completing the square, we have
(2k-l)2 + 7 £ 8n £ (2k+l)2 - 1 ,
(2k-l)2 £ 8n - 7 £ (2k+l)2 - 8 < (2k+l)2 ,
2k-l £ /Sii-y < 2k+l ,
k£iW§sEZ<k + 1 .
that is
un = k = [(1 + /8ii::7)/2]
15, Let ai»a7 an be given real numbers, which are not all
zero. Determine the least value of
2 _. j. 2
x, + ... + x ,
1 n
where x.,...,x are real numbers satisfying
a.x, + ... + a x - 1 .
11 n n
Solution: We have, using Cauchy's inequality,
1 - a, x,+. . .+a x £
1 1 1 n n1
n ^\l2f n 2-|l/2
x. ,
t n ,il/2,- n
U=l xi U«l
so that
? 2 > 1
) x. £ .
• i J- n ->
i=l v 2
.' ai
i=l x
If we choose

SOLUTIONS (15-16)
61
we have
n
V a.x. - 1
and
n
2 1
.L, l n
i-1 X
n 2 n 1
so the minimum value of \ x. subject to \ a.x. "1 is
1=1 1=1 y a2
i-1x
16, Evaluate the infinite series
23 33 43
s " 1 " IT + 2T IT + •••
Solution: We have
so that
(n+1)3 = n(n-l)(n-2) + 6n(n-l) + 7n + 1
, ? (-l)n(n+l)3
" K Si
n=0
r (-l)°(n(n-l)(n-2) + 6n(n-l) + 7n + 1)
- y <zl£ + 6 y Jzl)l + , y ("Dn + y ("Dn
- ij^TT+ 6 ij^zTT+ 7 i.-sr- + t-rrr
n=3 n«2 n=l n=0
= _yizi^+6yill^.7yilir+ y (-Dm
m=0 m»0 m=0 m=0

62
SOLUTIONS (17-18)
17, F(x) is a differentiable function such that F'(a-x) = F'(x)
for all x satisfying 0 £ x £ a . Evaluate J F(x)dx and give an
example of such a function F(x) .
Solution: As
F'(a-x) = F'(x) , 0 £ x < a ,
we have by integrating
-F(a-x) = F(x) + C ,
where C is a constant. Taking x = 0 we obtain C = -F(0) - F(a) ,
so that
F(x) + F(a-x) - F(0) + F(a) .
Integrating again we get
r
a ra
F(x) dx + [ F(a-x) dx = a(F(0) + F(a)) .
As
a ra
F(a-x) dx = I F(x) dx ,
r
the desired integral has the value -y(F(0) + F(a)) .
Two examples of such functions are
kcos— and k(2x3 - 3ax2) ,
a
where k is an arbitrary constant.
18. (a) Let r,s,t,u be the roots of the quartic equation
4 3 2
x + Ax + Bx + Cx + D = 0 .
2 2
Prove that if rs = tu then A D = C .

SOLUTIONS (17-18)
63
(b) Let a,b,c,d be the roots of the quartic equation
4 2
y +py +qy+r=0.
Use (a) to determine the cubic equation (in terms of p,q,r) whose
roots are
ab - cd ac - bd ad - be
a + b-c-d' a + c-b-d' a + d-b-c'
Solution: (a) As r,s,t,u are the roots of the quartic equation
4 3 2
x + Ax + Bx + Cx + D = 0 , we have
r + s + t+u = -A ,
rst + rsu + rtu + stu - -C ,
rstu = D .
Since rs » tu we have
2 2 2 2
AD= (r + s + t+u) rs
2 2 2
= (r s + rs + rst + rsu)
2
» (rtu + stu + rst + rsu)
= C2 .
(b) As the roots of the equation
4 2
y +py + qy + r = 0
are a,b,c,d , we have
(18.1) a+b+c+d-0 ,
(18.2) ab + ac + ad + be + bd + cd - p ,
(18.3) abc + abd + acd + bed = -q ,
(18.4) abed = r
Let z be a real or complex number. We begin by finding the quartic
equation whose roots are a-z,b-z,c-z,d-z .

64
SOLUTIONS (18-19)
From (18.1) , we obtain
(18.5) (a-z) + (b-z) + (c-z) + (d-z) = -4z .
Similarly, from (18.1) and (18.2), we obtain
(a-z)(b-z) + (a-z)(c-z) + (a-z)(d-z) + (b-z) (c-z) + (b-z)(d-z) +(c-z)(d-z)
= (ab + ac + ad + be + bd + cd) - 3(a + b + c + d)z + 6z ,
that is
(18.6) (a-z)(b-z) + (a-z)(c-z) + ... + (c-z)(d-z) = p + 6z2 .
Next, from (18.1), (18.2) and (18.3), we have
(a-z)(b-z)(c-z)+(a-z)(b-z)(d-z)+(a-z)(c-z)(d-z)+(b-z)(c-z)(d-z)
2 3
= (abe+abd+acd+bed) - 2(ab+ac+ad+bc+bd+cd)z + 3(a+b+c+d)z - 4z ,
so that
(a-z)(b-z)(c-z) + ... + (b-z)(c-z)(d-z)
(18.7) 3
= -q - 2pz - 4z
Also, from (18.1), (18.2), (18.3), (18.4), we have
(a-z)(b-z) (c-z) (d-z)
- abed - (abc+...+bcd)z + (ab+...+cd)z - (a+b+c+d)z + z ,
so that
(18.8) (a-z)(b-z)(c-z)(d-z) -= r + qz + pz2 + z4 .
Hence the desired quartic equation, whose roots are
a-z , b-z , c-z , d-z ,
is
I "X 0 0 "i 0 I
y + 4zy + (p+6z )y + (q+2pz+4z )y + (r+qz+pz +z ) = 0 .
To finish the problem we take z, = —-r t , so that
r la + b-c-d'
(a - z.)(b - z^ = (c - z^(d - z^ ,

SOLUTIONS (18-19)
65
and thus by (a) we have
Kiz^v + qz1 + pz1 + z^) " (q + 2pz1 + 4z^ ,
so that z. is a root of
(18.9) 8qz3 + 4(4r - p2)z2 - 4pqz - q2 = 0 .
„. ., ac - bd , ad - be .
Similarly z, = ; r ■=■ and z, = —= r are also
2 a + c-b-d 3 a + d-b-c
roots of (18.9), which is the required cubic equation.
19, Let p(x) be a monic polynomial of degree mil, and set
f (x) - ep(x)Dn(e_p(x)) ,
n
where n is a non-negative integer and D = -=— denotes
differentiation with respect to x.
Prove that f (x) is a polynomial in x of degree (mn - n)
Determine the ratio of the coefficient of x in f (x) to the
n
constant term in f (x) .
n
Solution: Differentiating f (x) by the product rule, we obtain
fn(x) - ep(x)Dn+1(e_p(x)) +p'(x)ep(x)Dn(e-p(x)) ,
so that
f'(x) = f ..(x) +p'(x)f (x) ,
n n*ri n
and so
(19.1) fn+l(x) * fn(x) " P'Wyx) •
Clearly fQ(x) = 1 is a polynomial of degree 0, f.(x) = -p'(x) is
a polynomial of degree m-1 , and f.(x) » -p"(x) + p'(x) is a
polynomial of degree 2m-2 . With the inductive hypothesis that f (x)
n
is a polynomial of degree mn-n , we easily deduce from (19.1) that

66 SOLUTIONS (19-20)
f , (x) is a polynomial of degree m(n+l) - (n+1) , and hence the
n+1
principle of mathematical induction implies that f (x) is a
polynomial of degree (mn-n) for all n .
Setting
and
/ \ m m— 1
p(x) - x + P^^ + ••• + P0
, , v mn-n mn-n-1
f (x; « a x +a ,x + . .. + a. ,
n mn-n mn-n-1 0
we obtain, from (19.1),
mn+m-n-1
+ ... +aQ
, ,, mn-n-1 , , , % mn-n- 2 ,
» (mn-n)a x + (mn-n-Ua ,x + ... + a,
mn-n mn-n-1 1
/ m—1 , / n * m-2 , v
- (mx + (m-l)p x + ... +p )
m—i i
, mn-n , mn-n-1 , .
(a x + a ,x + ... + a„) .
mn-n mn-n-I 0
- ^. ££. . c mn+m—n—1 .^ .
Equating coefficients of x , we obtain
a , , ™ -ma
mn+m-n-1 mn-n
Solving this recurrence relation, we obtain
a™,_„ = (-m) an '
mn n U
that is
a
mn-n , -, n
—-— - (-m)
a0
20i Determine the real function of x whose power series is
x3 x9 x15
fr + fr + T5T+ ••• •

SOLUTIONS (19-20)
67
Solution: We make use of the complex cube of unity
-1 + tt
so that
(20.1)
Now, for all real x , we have
3 2
w ■ 1 , w + w + 1 = 0 .
3 5 7 9
sinh x = x + *y + |j+ JJ + JT +
3 2 5 7 9
., , X , W x . WX . X ,
sinh wx = wx + -jy+ yj- + 7T + 'qT +
, , 3 5 2 7 9
• u ,,2 „2 , x , w x , w x , x
sinh wx»wx+yt+ Ti" Ti" + '5T +
Adding these equations and using (20.1), we obtain
sinhx + sinhwx + sinhw x = 3Ury + •tj
Now
sinh wx = sinh
■ -sinh
and similarly
sinhw*x = -sinh
and so
¥)■
sinh
-(¥)
+ cosh
sinh
■i^
•r-lcosl—^r1! + icosh
w°-m
icosh
giving
3 9 2
f)sin(^] ,
fx] . (xvT|
2jsinm '
2sinh -_- cos —!
:osh
f) - cos
n-
■$

68
SOLUTIONS (21-22)
21, Determine the value of the integral
|2
n Jq [ sin x J
(21.0) I_ - | 1.71 „ 1 <*x ,
for all positive integral values of n
Solution: We will show that I = rnr , n = 1,2,3,... .
~^__— n
From (21.0), we have for n > 2
D - I - I
n n
-r-
Jo
(sin nx - sin (n-l)x)
dx
(sin nx - sin (n-l)x)(sin nx + sin (n-l)x)
dx
2 sin-r- cos(nx--r-) 2 sin(nx-•=•) cos •=■
sin2x
Ldx
that is
(21.1)
where
sin x • sin (2n-l)x
dx
sin (2n-l)x
dx ,
D - J, n > 2 ,
n 2n-l '
sinmx
dx , m - 0,1,2,.
Now, for m 2 2 , we have
J - J ,
F
inmx - sin (m-2)x)
dx

SOLUTIONS (21-22)
69
I
2 sinx cos (m-l)x ,
IQ sinx
2I cos(m-l)x dx
Esin(m-l)xj
. m-! Jo
so that
2
0 ,
j - j - j - . - r° " ' if m even'
m m-2 m-4 '*' \J, = tt , if m odd.
/Jo ' 0 ,
Hence, from (21.1), we obtain D ■ i , n 2 2 , so that I ■ rnr ,
n n
nil, as I, ■ tt .
22i During the year 1985, a convenience store, which was open 7
days a week, sold at least one book each day, and a total of 600
books over the entire year. Must there have been a period of
consecutive days when exactly 129 books were sold?
Solution: Let a. , i - 1,2,3, .. .,365, denote the number of books
sold by the store during the period from the first day to
the i day inclusive, so that
and thus
1 £ a1 < a < ... < a365 » 600 ,
130 £ a,+129 < a,+129 < ... < a,,c+129 = 729
1 2 365
Hence a.,...,a,,,,a.+129,...,a,,,+129 are 730 positive integers
between 1 and 729 inclusive. Thus, by Dirichlet's box principle,
two of these numbers must be the same. As &1 ,...,&,,, are all
distinct and a,+129,...,a,,,+129 are all distinct, one of the a.
1 365 i

70
SOLUTIONS (23-25)
must be the same as one of the a.+129 , say,
a, = a, + 129 , 1 £ I < k £ 365 .
k -c
Hence a, - a» ■ 129 and so 129 books were sold between the (£+1)
th
day and the k day inclusive.
23. Find a polynomial f(x,y) with rational coefficients such
that as m and n run through all positive integral values, f(m,n)
takes on all positive integral values once and once only.
Solution: For any positive integer k we can define a unique pair
of integers (r,m) by
(r^l)(r-2) < fc ^ r(r-l)
2 " " a 2
(r-l)(r-2)
Clearly we have
m = k
n <■ „, < r(r-1) (r-l)(r-2) _
U < m £ x x ■ r-1 ,
that is
1 £ m < r ,
so that r and m are positive integers. Moreover, we can define
a positive integer n uniquely by r * m + n , which gives
. (m+n-l)(m+n-2) ,
K = r + m ,
and a polynomial of the required type is therefore
f(x,y) = (x+y-DU+v-2) + x _

SOLUTIONS (23-25)
71
24, Let m be a positive squarefree integer. Let R,S be
positive integers. Give a condition involving R,S,m which
guarantees that there do not exist rational numbers x,y,z and w such that
(24.0) R + 2S»G - (x + y^ii)2 + (z + w^)2 .
Solution: If there exist rational numbers x,y,z and w such that
(24.0) holds then
7 7
R - 2S^m = (x-yi^m) + (z-w^m) X 0 ,
and so a condition that will guarantee the non-solvability of (24.0)
is R - 2S^m < 0 , that is
2S < ^ *
25, Let k and h be integers with 1 £ k < h . Evaluate the
limit
hn (• -I
L - lim n 1 -£, •
n ■* <*> r=kn+l "• '
Solution: For |x| < 1 we have
£n(l - x)
and so
x + in (1 - x)
giving
|x + in (1 - x)|
V X
s=l
oo g
•f x
s-2'
oo g
It
s=2s
* t
00 i i s
x
s=2

72
SOLUTIONS (25-26)
L s-2
ITF
xTT
Taking x ■ — 2 (kn < r £ hn) we obtain
h + ^^-f.)
2»*<l-£>
2n\l - J)
Thus we obtain
| hn
i
Ir-kn+l
in i \ , hn
cn+l v ' ?n ( 1 ——) t-=1cn-t
2n (1—) r=kn+l
n
hn „
J y
2n (1—) r=l
n
h n
2»*<l-g)
showing that
(25.1)
Next we have
hn
lim I
n -*■ oo r«kn+l
2nd " *>
0 as n -*
f«+&(1-»)
hf £ . I Jhn(hnfl) _ kn(kn+l)\ (h2-k2),(h-k)
r-kn+l"2 nH 2 " 2 / " 2 + 2n •

SOLUTIONS (25-26)
73
so that
(25.2)
.. f r (h2-k2)
llm <• n* 2
n ■* <*> r=kn+l
Thus from (25.1) and (25.2) we obtain
hn
and so
lim In n (1 - --) - -i(h2-k2)
n -»■ ■» r=kn+l
hn -?(h -k )
lim n (1 - £2) - e
n-».o> r=kn+l
26, Let f(x) be a continuous function on [0,a] such that
f(x)f(a-x) ■ 1 , where a > 0 . Prove that there exist infinitely
many such functions f(x) , and evaluate
dx
1 + f(x)
Solution: The function f(x) = e is continuous for all x and
satisfies f(x)f(a-x) = 1 , so that f(x)n (n = 0,1,2,..
gives an infinite family of functions of the required type.
Setting x = a - y we obtain
x. r «* . f° -ir
JQ1 + f(x) Jal + f(a-y)
= fa dy . fa £(y)dy
JqI + f(a-y) JjfCy) + 1
.^f-rrW}^

74
SOLUTIONS (27-28)
fa dy
a - i ■'-
Jo1 +
f(y)
a
- a - I , so that I « -r .
27i The positive numbers a.,a,,a,,... satisfy
n , ,-n -,2
(27.0) laj- Jal , n - 1,2,3
r»l '■r"l '
Is it true that a » r for r - 1,2,3,... ?
r
Solution: The answer is yes. We proceed by mathematical induction,
making use of the identity
l3 + 23 + ... + n3 - (1 + 2 + ... + n)2 .
3 2
Taking n « 1 in (27.0) gives a. - a. , which means that a. - 1
because a1 > 0 .
Next, assume that a, = k for k = l,2,...,n-l . The equation
(27.0) gives
l3 + 23 + ... + (n-1)3 + a3 - (1 + 2 + ... + (n-1) + a )2
n n
so that
(1+2+. ..+(n-D) 2 + a3 = (l+2+...+(n-l))2 + 2( 1+2+.. .+(n-l))a + a2 ,
n n n
that is,
(27.1) a3 = (n-l)na + a2 .
n n n
As a > 0 , we see that (27.1) gives a = n , thus completing the
n n
inductive step.

SOLUTIONS (27-28)
75
28i Let p > 0 be a real number and let n be a non-negative
integer. Evaluate
(28.0)
u (p) - I e sin x dx
Jo
Solution: For n £ 2 and p > 0 , integrating u = u (p) by
parts we obtain
1 -px . n
u = e r sin x
n p
0 Jo
-px
+ n sin x cos x
dx
n [ . n-1 -px ,
« — s in x cos x e dx
p Jo
Integrating by parts again, we get
n f 1 -px , n-1
u " — i — e sin xcosx
n p \. p
+ Jo ((n-1)
n-2 2 J n. e
sin x cos x - sin x
-px
dx
p2 -O
(n-l)sin x(l-sin x) - sin x e p dx
n(n-l)
P
-px j n-2 . n | -px . n ,
e r sin x dx r I e sin x dx ,
that is
Thus we have
n(n-l) n ,
u = -i *■ u »u , na2
n p n-2 2 n
n(n-l) ,
un = 272 V2 ' n 2 2 •
n +p

76
SOLUTIONS (29)
giving
' n(n-l) # (n-2)(n-3) 2'1
2 2 2 2
n V (n-2)%z
nCn-1) . (n-2)(n-3)
i fi2+p2 (n-2)2+p2
u_ , if n even,
2 V °
3*2
—5—r u. , if n odd
3 +p *
1
One easily sees that u « — and u. « r- ,
P 1 + P
so that
"" 22
P n ((2i)z +PZ)
, if n even,
i=l
n!
(n-l)/2
H ((2i+l) + p )
, if n odd.
»■ i-0
29^ Evaluate
(29.0)
for integers n k 2
n-2
) 2 tan ,
r-0
Solution: We use the identity
tan A - cot A - 2 cot 2A ,
which is easily verified as
2 2
, ,. ,. _ 2 cos 2A _ cos A - sin A
2 cot 2A - _.,_ „, - —T7—TZ—.— - cot A - tan A
sin 2A sin A cos A
Then we have
"Vtan -^ ~ "Vfcot -^ - 2 cot-^1
r-0 2 r=0 *■ 2 r 2 -1

SOLUTIONS (29)
77
E 2 cot Tn^r" - Ln2 cot 3Tr^T
r-0 2 r-0 2
n-2
n-1
I 2rcot ^ _ I 2r,
r-0
,n-r .
2 r«l
cot
.0 . it _n-l ir
2 cot 2 cot ;;—rr
2n 2n-(n-l)
cot 2 cot -z
2n 2
cot
Second solution (due to L. Smith):
For 0 < B < 2ir we consider the integral
KB)
fa ra— i.
l\l
•b "-r-0
-2 rr
l2l
-o •to
n-2 fB
2rtan -±-
2n-:
A
dA
tan
dA
r—
n-2
r -r -.n-r . A
Y 2 -2 In cos
** _ n—
r-0
2° Uncos-f-
r-0 2
n n_2 H
2nln n cos -r-
r-0
2n£n
. B
sin-r
2n

78
SOLUTIONS (30-31)
- 2n In sin | - 2n In sin \ - 2° In 2n_1
Z 2
Differentiating we obtain
I'(B) - cot-*-- 2n_1cotf .
2n 2
Taking B - it we get
nf2rtan -f- - cot ^ .
r=0 2 2
30. Let n a 2 be an integer. A selection {s - a : i»l,2 k)
of k (2 <. k £ n) elements from the set N =» {1,2,3 n} such that
a. < a. < .. . < a, is called a k-selection. For any k-selection S ,
define
W(S) - min {ai+1-a±: i - 1,2 k-l} .
If a k-selection S is chosen at random from N , what is the
probability that
W(S) = r ,
where r is a natural number?
Solution: Let f (n,k) , 2 s k s n , denote the number of
k-selections S from N such that W(S) i r ,
r - 1,2,3 We will show that
n-(k-l)(r-l)l
(30.1) fr(n,k)
, r - 1,2,3
where
™ " 0 for any integer m < k . When r - 1 , f (n,k)
enumerates all k-selections S from N , so that f (n,k) » " ,
and hence (30.1) holds in this case. Now suppose r > 2 . Let
S - {a^a, a^] be a k-selection from N with W(S) 2 r ,

SOLUTIONS (30-31)
79
so that a. + r < a. . for i = l,2,...,k-l . The mapping F
defined by
(30.2) F(S) - {a^ a2 - (r-1), a3 - 2(r-l), .... ak - (k-l)(r-l)}
associates with a k-selection S from N having W(S) i r , a k-
selection from the set
M - {1, 2, ..., n - (k-D(r-l)} .
Clearly F is one-to-one and onto, so that f (n,k) is just the
number of k-selections from M , which is given by the right side of
(30.1).
Thus the required probability is
fr(n,k) - fr+1(n,k) fn-Oc-lHr-nl _ fn-(k-l)r)
f^n.k)
(")
31, Let k £ 2 be a fixed integer. For n » 1,2,3,... define
1 , if n is not a multiple of k ,
-(k-1) , if n is a multiple of k .
Evaluate the series
t -
n=l n
Solution: Let s be the sum of the first n terms of the given
n
series. For each n i 1 we have uniquely
and since
n-kq+r, 0 < r < k ,
ti n n
-(k-l)/tk » 1/tk - 1/t
for t=l,2,...,q ,we have

SOLUTIONS (31-32)
s
n
(i+U...+!)- (i + i + ...+-L) .
Now, n - q (k + r /q ) so
n n n
£n n « fa q + fa (k + r /q ) ,
n n n
and hence
s - (1 +T+ ... +--fa n) - (1+T+ ... +— - fa q ) + fa (k + r /q )
n i n l q n n n
■ u - v + w .
n n n
The sequence {u : n - 1,2,3,... } converges to Euler's constant c;
the sequence {v : n « 1,2,3,... } also converges to c as n ■* •*>
implies q ■*■">; and the sequence {w : n - 1,2,3,... } converges to
n n <» a
fa k as r is bounded. Thus, we have £, —a - fa k .
n n»l n
32. Prove that
x e sin x dx - /J^W? s*n (mfl)'n'/4
for m « 0,1,2
Solution: We set for m - 0,1,2,...
, m -x .
S - | x e s in x dx ,
^ i m -x
C ■ | x e cos x dx ,
_ , m (i-l)x ,
E - x ev dx .
m '

SOLUTIONS (31-32)
81
As
we have
cos x + 1 sin x ,
E - C + i S
mm m
Integrating E (mil) by parts, we obtain
(i-lb
E = x ■ .
m l-l
m-1 e
(i-l)x
i-1
dx
Hence we have
■r—r E . , m i 1 .
i-l _m-l
(-l)mml „ . .
E - -— EA , m i 0
m ,. , sm 0
(i-1)"
Clearly E„ = - t—r , so that for m £ 0
0 l-l
(-1)"*1 ml ml (i+l)"*1
(l-l)
-nri-1
Finally, we obtain for m £ 0
S = •£-(£ - E )
m 2i m m
2^2i
r(
(l+i)
nrt-1
-d-i)^1! ,
and so, by Demoivre's theorem, we have
ml
m _nrt-2.
2 i
nri-1 mfl
2 2 (cosdiH-Dy+i sin(mfl)^) - 2 2 (cos(nrt-l)£- i sin(nrt-l)£)
4 4 4 4
, fort)/2 «»<»«>$ • ">^0-

82
SOLUTIONS (33)
33i For a real number u set
(33.0)
Prove that
(u) - f -
J0
I(u) - ln{\ - 2u cosx + u ) dx .
I(u) - l(-u) - }l(u2) ,
and hence evaluate I(u) for all values of u .
Solution: We will show that
(33.1)
Ku)
0 , if |u| £ 1 ,
2ir^Jt|u| , if lul > 1 .
First, we prove that
(33.2) I(u) - I(-u) .
Setting x «tt- y in (33.0), we obtain
Ku)
£lt(l + 2ucosy + u ) dy
0
K-u) .
I(u) + I(-u) - I(u ) .
Next we show that
(33.3)
We have
(1 - 2ucosx + u )(1 + 2ucosx + u ) - (1 + u ) - (2ucosx)2
L 7 ?
- 1 + u + 2u (1 - 2cos x)
" 1 - 2u cos2x + u ,
so that
£li(l - 2ucosx+u ) + £w(l + 2u cos x + u ) =* Zyi(l — 2u cos 2x + u ) ,
and thus
Ku) + I(
-u) - f &t(l - 2u2cos 2x + u4) dx ,

SOLUTIONS (33)
83
and setting y » 2x , we obtain
f2ir
I(u) + I(-u) - j
JO
2 A
ln(.l - 2u cosy + u ) dy
1 2 1 f ? A
I I(u ) + | £n(l - 2u cos y + u ) dy .
Setting y ■= 2tt - z in the last integral, we obtain
2A ?
•£n(l - 2u cos y + u H) dy - l(u ) ,
proving (33.3) as required.
From (33.2) and (33.3), we deduce
(33.4)
I(u) - I(-u) - -|l(u2)
For |u| =1 , that is u = ±1, we have
1(1) - I(-l) - j 1(1)
1(1) - I(-l) = 0
For u < 1 we have
I(u) - i I(u2) = X I(u4) - X I(u8)
2 2J
1 2
i I(u } '
for all positive integers n . Letting n •+• +■» , we have u •+• 0
as |u| < 1 , and I(u) being continuous gives
lim I(u ) = 1(0) .
n-*-<»
1
[In fact it follows from (33.4) that 1(0) = 0 .] Hence, as — •+• 0 ,
2n
we obtain
,n
lim -^ I(u ) = 0 ,
n-»-"> 2

84
SOLUTIONS (34-35)
giving
I(u) = 0 , for |u| < 1 .
Next, for |u| > 1 , we set u - - , so that 0 < |v| < 1 . Then,
for u > 0 , we have
<„, - £
I(u) - I ln(l - |cosx + -j) dx
{ln{\ - 2vcosx + v ) - 2-dnv} dx
0
I(v) - 2lnv I dx
"fd
= 0 - 2-nlnv
= 2-aln.n .
Finally, if u<0 , we have
I(u) - I(-u) - 2itln(-u) ,
so that for all u with |u| > 1 we have
I(u) - 2-nln. lul .
34i For each natural number k > 2 the set of natural numbers
is partitioned into a sequence of sets {A (k): n - 1,2,3,... } as
n
follows: A.(k) consists of the first k natural numbers, A.(k)
consists of the next k+1 natural numbers, A,(k) consists of the
next k+2 natural numbers, etc. The sum of the natural numbers in
A (k) is denoted by s (k) . Determine the least value of n - n(k)
n n
such that s (k) > 3k3 - 5k2 , for k - 2,3
n

SOLUTIONS (34-35)
85
Solution: The last element in A (k) is the number
n
k + (k+1) + (k+2) + ... + (k+n-1) - nk + (1+2+...+(n-l))
,nk+iH^)n.
Since there are k+n-1 numbers in A (k) , the first element in
n
A (k) is
n
[nk+(n"21)n] - (k+n-1) + 1 - (n-l)k + |(n2-3n + 4) .
Hence we have
(34.1) sp(k) = (k + ^-!> ((2n-l)k+(n2-2n + 2)) .
Taking k = 2,3,4,5 in (34.1) suggests that n = k may be the
required value of n . To prove this conjecture we calculate s, _. (k)
and s, (k) . We have
sk-l(k) = (k~ 1X3k2-7k + 5) " 3k3-10k2 + 12k-5
and
Sk(k) = (2k-l)(3k2-3k + 2)=3k3_9k2+7k_1 #
One easily checks that for k - 2,3,...
sk(k) > 3k3 - 5k2 > s^flc) ,
so that n(k) = k is the required minimal value.
35i Let {p : n » 1,2,3,... } be a sequence of real numbers such
n
that p SI for n = 1,2,3 Does the series
[pn]-l
(35,0) Ji ([P1]+i)([P2]+i)...([Pn]+D
converge?

86
SOLUTIONS (35-36)
Solution: The answer is yes.
p -1
Let a = — , n = 1,2,...
n Pr..Pn
Then we have for m 2 1 ,
m
S
n=l n n=l^r--Vl VPJ
1 l
S 1
V"pm
1
Pr-p»fi
= SnH-l '
so that {S : m = 1,2,3,... } is an increasing sequence which is
m o
bounded above by 1 . Hence lim S exists, showing that E.a
m n=l n
converges. Finally, as
^a_, n=l,2,3
([P1]+l)([p2]+l)-..([pn]+l) n
the series given by (35.0) converges by the comparison test.
36i Let f(x) , g(x) be polynomials with real coefficients of
degrees n+1 , n respectively, where n 2 0 , and with positive
leading coefficients A , B respectively. Evaluate
L - lim g(x)
x-»-">
*.'«>-'<*>* ,
in terms of A , B and n .
Solution: As A > 0 and B > 0 , we have

SOLUTIONS (35-36)
87
and
Moreover
lim
X-i-" JO
lim
ef<«dt-+.
ef(x)
« -t-oo
lii
lim
lim
dx[
e dt
■b
dl
dxj
f4^1
1 gto J
g(x)2
f'(x)g(x) - g'(x)
B2x2n+...
2n
x-*-» AB(n+l)x"" + ...
B
(n+l)A
so, by L'H6pital's rule, we have
L » lim -> k
x + » e W/g(x)
_d_
dx
lim
(C-",,«)
_d_
dx
f(x). . .
e /g(x)
B
(n+l)A

88
SOLUTIONS (37-38)
37■ The lengths of two altitudes of a triangle are h and k ,
where h * k . Determine upper and lower bounds for the length of
the third altitude in terms of h and k .
Solution: We show that
(37.1)
h+k < *■ < Wk]
Let the points P,Q,R be chosen on the sides BC.CA.AB (possibly
extended) respectively of the triangle ABC so that AP,BQ,CR are
the altitudes of the triangle. Set a - |BC|, b = |CA| , c - |AB|,
h - |AP| , k = |BQ| , I - |CR| . Clearly
ah = bk = d ,
so that
c h ' c k •
Without loss of generality we may suppose that h < k . From the
inequality
a < b + c ,
we obtain
so that
that is
c c h k
£<£ +
1 ,
I - I < 1
h k '
I <
hk
k-h
Also from the inequality
we obtain
c < a + b ,
c c h k

SOLUTIONS (37-38)
89
that is
*■ h+k
This completes the proof of (37.1).
38. Prove that
,, n+K., n+2,. ,, n-rN
p » p (x) - (1-x )(l-x )...(l-x )
n'r n'r (l-x)(l-x2)...(l-xr)
is a polynomial in x of degree nr , where n and r are non-
negative integers. (When r = 0 the empty product is understood
to be 1 and we have P _ - 1 for all n £ 0 .)
n,U
Solution: For n jO and r jl we have
P - xrP = (l-xn+2)...(l-xn+r+1) _ „r (l^-d-*"")
n+l,r n,r
(1-x).
(1-x ).
,, n+2,,
(1-x ).
(1-x).
P
n+l,r
..(l-xr) (l-x)...(l-xr)
,. n+rs <•,. n+r+U r,. n+K-.
..(1-x )|(l-x ) - x (1-x ) 1
(l-x)...(l-xr)
,, n+r,,
..(1-x ) T
..(1-x )
xrP - P ^ , •
n,r n+l,r-l
so that
(38.1)
We now make the inductive hypothesis that P is a polynomial
n, r
of degree nr for all pairs (n,r) of non-negative integers
satisfying n + r £ k , where k is a non-negative integer. This is
clearly true when k » 0 , as, in this case, we must have n = r = 0,
and P_ _ - 1 . Now let (n,r) be a pair of non-negative integers
such that n + r = k + 1 . For n % 1 and r £ 1 , by (38.1), we have

90
SOLUTIONS (39-40)
(38.2) P = xrP . + P . .
n,r n-l,r n,r-l
As (n-1) + r = n + (r-1) = k , by the inductive hypothesis, P _.
is a polynomial of degree (n-l)r and P _. is a polynomial of
degree n(r-l) . Hence, by (38.2), P is a polynomial of degree
n,r
max(r + (n-l)r , n(r-l)) = nr .
P is clearly a polynomial of degree 0 in the remaining cases
n = 0 and r = 0 . The result now follows by the principle of
mathematical induction.
39. Let A, B, C, D, E be integers such that B * 0 and
F = AD2 - BCD + B E * 0 .
Prove that the number N of pairs of integers (x,y) such that
(39.0) Ax + Bxy + Cx + Dy + E = 0 ,
satisfies
N £ 2d(|F|) ,
where, for integers nil, d(n) denotes the number of positive
divisors of n .
Solution: Let (x,y) be a solution in integers of (39.0), so that
(39.1) -(Bx + D)y = (Ax2 + Cx + E) .
We define an integer z by
(39.2) z = - (Bx + D) , so that x = -|-(z + D) .
Clearly z * 0 , for otherwise x = -D/B and from (39.1) we would
2
have —•* s- + E = 0 , contradicting F * 0 . From (39.1) and
B2 B

SOLUTIONS (39-40)
91
(39.2) we have
that is
B2zy - A(z+D)2 - BC(z+D) + B2E ,
z(B2y - Az - (2AD-BC)) - F ,
so that z is a divisor of F .
Thus the total number of possibilities for z is £ 2d(|F|) .
For each such z there is at most one possibility for x ,
namely, x = --=-(z + D) if this is an integer. As (39.1) implies
that each x determines at most one y , the total number of pairs
(x,y) is < 2d(|F|) .
i|0. Evaluate £
4 2
k=l k + k + 1
Solution: We have for k 2 1
k4 + k2 + 1 = (k4 + 2k2 +1) - k2 = (k2 - k + 1)(k2 + k + 1)
and
2k 1
4 2 2 2
k+k+1 k-k + 1 k+k + 1
f(k-l) - f(k) ,
where
so that
f(x) - 1
2
x + x + 1
k=l k +k +1 k=l
| (f(0) - f(n))

92
SOLUTIONS (4M2)
n + n + 1
i. n + n
2 2
n + n + 1
41i Let P = P (n) denote the sum of all positive products of
m m
m different integers chosen from the set {1,2,...,n} . Find
formulae for P.(n) and P~(n) .
Solution: We will show that
P2 = ± n(n+l)(n-1)(3n+2) , ?3 = ± n2(n+l)2(n-l)(n-2)
We begin by considering
(l-x)(l-2x)...(l-nx) = 1 - P x + P2x2 - P3x3 +
+ (-1)"?/ ,
so that, with P- = 1 and x sufficiently small,
id l (-orp xr] - I ma - icx)
lr-0 > k»l
? r kV
k»l £=1 *•
Hence we obtain
I (-l)rP xr = exp
r=0 r
™ £ r n „
-It l^
d»lc lk»l
" ^r n /
£=1 *• ^k-1

SOLUTIONS (41-42)
93
that is
(41.1)
n °<> .
I«-»V ■ X FT
00 ^r n c
" I T I k
£=1 *■ lk«l
r=0 h=0
and so, for r - 0,1,2,...,n ,
P = (-1) coefficient of x on the right of (41.1)
Thus we have
and
coeff. of x in £
2 ... I ("Dh
1 U2+tt\ U
k=l
h=0
n -,2
h!
<*> I r n .
i t [lk
£=!*- lk=l
k«l
ln(n+l)(2n+l) ^ lfn(iri-l)]2
2l 2 J
24
n(irt-l)(n-l)(3n+2) ,
P, = coeff. of x in \K '
'3
h=0
h!
<*> If n „■
£-1 *• lk-i
yj n2(n+l)2 - -L n2(n+l)2(2n+l) + -^ n3(n+l)3
^ n2(n+l)2(n-l)(n-2)
i|2. For a > b > 0 , evaluate the integral
(42.0) = ==- dx
, ax , -, , bx ,.
x(e +l)(e +1)

94
SOLUTIONS (42^3)
Solution: For any constant C , the function
f(x)
eX+l
+ c , x a o ,
is such that
f(ax) - f(bx)
ax bx
e - e
(eaX+l)(ebX+l)
For t > 0 we have
(t
J(t) =
ax bx
e - e
x(e +l)(e +1)
dx
f(ax) - f(bx)
dx
f(ax)
dx
f(bx)
dx ,
provided both the latter integrals exist.
f (v)
This is guaranteed if lun J exists, which holds if and only
y + 0 y '
if C = -j . With the choice C
, we have
J(t)
at
f(y)
dy
bt
f(y)
dy
at
Jbt
f(y)
dy .
Now lim f (x) =-y so that given any e > 0 there exists a positive
real number x. = xn(e) such that
■j - e < f (x) < j + z .
If t > xn/b , then at > bt > xn , and so we have

SOLUTIONS (42-43)
95
<I " e> H
at4 - t)
dy
Jbt
< J(t)
(y + e)
dy
%t
(i + e) fcif
Since £ is arbitrary we obtain
lim J(t) - i£n| ,
that is
ax bx
e - e , 1 0 a
dx = — In-r
ax ,,,, bx , ,, 2 &
x(e +l)(e" +1)
i|3i For integers nil, determine the sum of n terms of the
series
(L-x m 2n . 2n(2n-2) , 2n(2n-2) (2n-4)
(43-0) "2n^f + (2n-l)(2n-3) + (2n-l)(2n-3)(2n-5) + •" *
Solution: Let S denote the sum of n terms of the given series
(43.0). We have
Sl 1 Z '
<? - 4 I 4'2 - 4 I 8 -r 12 -r A
2 3 3-1 3 3 3
e 6 ^ 6-4 ^ 6-4-2 18+24+48 90
3 5 5-3 5-3-1 15 15

96
SOLUTIONS (43-44)
These values suggest the conjecture S = 2n for all positive
integers n . First, as S. = 2 , the conjecture holds for n = 1 .
Assume that S = 2n holds for n - m
n
Then we have
2m+2 , 2m+2 f 2m 2m(2m-2) , , „ ,.
Sm+1 * "2m+T + 2m+T 2m^T + <2m-l)(2m-3) + • • • (m terms)
2m+2 . 2m+2 e
2m+l
2m+2
2m+l
2m+2
2m+l
2m+2
' 2m+l "m
2m+2
+ 2m+l
(1 + 2m)
9
2m
showing that S = 2n is true for n = m + 1 . Hence, by the
principle of mathematical induction, S = 2n is true for all positive
integers n .
£j£|, Let m be a fixed positive integer and let z ,z_,...,z
be k (SI) complex numbers such that
(44.0) z® + zS2 + ... + z® - 0 ,
for all s = m,m+l,m+2,...,m+k-l. Must z. - 0 for i - 1,2,...,k?
Solution: The answer is yes. To see this, let z.,z7,...,z be the
roots of the equation
k _,_ k-1 , k-2
z + a. , z + a, ,z + .. . + a, z + a„ = 0 .
k-1 k-2 1 0
We will show that a.^ = 0 . Suppose an * 0 . Clearly z.,z.,...,z
are also roots of
m+k-1 m+k-2 m m-1
z + a._.z + ... + a. z + a z = 0 .

SOLUTIONS (43-44)
97
Hence for i = 1,2 k we have
... ,. m+k-1 m+k-2 , m m-1 n
(44.1) z. + Vl«i + ... +alZ. + aQz. - 0 .
Adding the equations in (44.1) and appealing to (44.0) we obtain
k
ur
, m-l
a0 .i zi
i-l
As an * 0 we have
X *m_1 - o .
i-1 x
Clearly zifz7 z, are roots of
,,, ,.. m+k-2 m+k-3 , , m-l m-2
(44.2) z + a,.z + ... + a.z + a.z » 0 .
Taking z = z. , i = 1,2,...,k , in (44.2) and adding the equations
we obtain as before
* m-2
I zT< - o .
i-1x
Repeating the argument we eventually obtain
k
> z. - 0 ,
i-1 X
and one more application then gives a k » 0 , which is impossible.
Hence we must have a_ » 0 , that is
(-l)kZl...zk - 0 ,
and so at least one of the z. is 0 , say z = 0 . The argument
can then be applied to z.,...,z . to prove that at least one of
these is 0 , say z, _. » 0 . Continuing in this way we obtain
« - z2 - ... - zk= 0 .

98
SOLUTIONS (45-46)
lfi> Let A = (a,.) be the nxn matrix where
n>l n ij
a. .
x , if i * j ,
1 , if |i-j| = 1 ,
0 , otherwise,
where x > 2 . Evaluate D - det A
n n
Solution; Expanding D by the first row, we obtain
so that
(45.1)
D = xD . - D , ,
n n-1 n-2
D -xD ,+D ,=0.
n n-1 n-2
The auxiliary equation for this difference equation is
2
t - xt + 1 = 0 ,
which has the distinct real solutions
x ± /x2-4
as x > 2 . Thus the solution of the difference equation (45.1) is
given by
Dn - A(|(x+v^r4))n + B^x-^ir?))11 f
for some constants A and B .
We now set for convenience
a - i(x + ȣ^4) ,
that, as |(x + /x2-4) • ±(x - ^^4) = 1 ,

SOLUTIONS (45-46)
99
which gives
Now
D - Aa11 + Ba~n , n - 1,2,3
n
1 » B
D, »x»a+ — » Aa + —
1 a a
and
D2 = x2 _ ! = a + I
i 2 1 . . 2 B
1 » a + —^ + 1 = Aa + —
a a
so that
(45.2)
2 2
aA + B = a+l ,
4 4 2
aA + B-a +a +1.
Solving (45.2) for A and B yields
2
a2-l
, B
1
a2-l
so that
that is
an 11
n 2 , 2 , n
a -1 a -1 a
2n+2_.
\--T7T-' n- 1,2,3,...
a"(a -1)
i|5, Determine a necessary and sufficient condition for the
equations
(46.0)
x + y + z = A ,
2 ^ 2 2
x + y + z - B ,
3 3 3
x + y + z = C ,
to have a solution with at least one of x,y,z equal to zero.

100
SOLUTIONS (47-49)
Solution: Let x,y,z be a solution of (46.0). Then, from the
identity
2 2 2 2
(x+y+z) = x + y + z + 2(xy+yz+zx) ,
and (46.0), we deduce that
(46.1) xy + yz + zx - -|(A2 - B) .
Next, from the identity
3 3 3 2 2 2
x + y + z - 3xyz ■= (x + y + z) (x +y +z - (xy+yz+zx)) ,
we obtain using (46.1)
C - 3xyz = A(B - |(A2-B)) = -|ab - |a3 ,
so that
(46.2) 3xyz = yA3 - |aB + C .
Hence a solution (x,y,z) of (46.0) has at least one of x,y,z
zero if and only if xyz = 0 , that is,by (46.2), if and only if the
3
condition A - 3AB + 2C = 0 holds.
1)7, Let S be a set of k distinct integers chosen from
1,2,3,...,10 -1 , where n is a positive integer. Prove that if
(47.0) n < In
<4*l+fc"l]/4»10.
it is possible to find 2 disjoint subsets of S whose members
have the same sum.
Solution: The integers in S are all £ 10 -1 . Hence the sum of
the integers in any subset of S is
£ (10n-k) + ... + (10n-2) + (10n - 1) » k-10n - |k(k+l) .

SOLUTIONS (47-49)
101
k
The number of non-empty subsets of S is 2 -1 . From (47.0) we have
2k - 1 > k-10n - |k(k+l) ,
and so, by Dirichlet's box principle, there must exist at least two
different subsets of S , say S. and S, , which have the same sum.
If S. and S, are disjoint the problem is solved.
If not, removal of the common elements from S. and S. yields
two new subsets S' and Si with the required property.
£j8i Let n be a positive integer. Is it possible for 6n
distinct straight lines in the Euclidean plane to be situated so as
2
to have at least 6n -3n points where exactly three of these lines
intersect and at least 6n+l points where exactly two of these
lines intersect?
Solution: Any two distinct lines in the plane meet in at most one
point. There are altogether . ■ 3n(6n-l) pairs of
lines. A triple intersection accounts for 3 of these pairs of
lines, and a simple intersection accounts for one pair. As
(6n2-3n)3 + (6n+l)l
- 18n2 - 3n + 1
> 3n(6n - 1)
the configuration is impossible.
i|9. Let S be a set with n (£1) elements. Determine an
explicit formula for the number A(n) of subsets of S whose
cardinality is a multiple of 3 .

102
SOLUTIONS (49-50)
Solution: The number of subsets of S containing 3£ elements is
, I = 0,1,2,...,[n/3] . Thus, we have
(49.1)
[n/3]r , , ,
AW - I £ " I LI •
1=0 *■-"-•' k=CP
k=0 (mod 3)
Let w = y(-l + i/3) so that
w2 = |(-1 - i/3) , w3 = 1 ,
and, for r = 0,1,2 , define
k-r (mod 3)
Then, by the binomial theorem, we have
n
(49.2)
(49.3)
(49.4)
k=0
n
n-(i+i)n-,u3-W2.
k=0
3
(1+W)n= LSW =S0+wSl+wS2'
(1 + w)n = i0ww =so + A+wS2-
Adding (49.2), (49.3), (49.4), we obtain, as 1 +w + w = 0 ,
so that
Hence we have
A(n)
2n + (l+w)n + (1+w2)11 - 3SQ = 3A(n) ,
A(n) = |(2n+ (-w2)n+ (-w)n) .
■j(2n + 2(-l)n) , if n E 0 (mod 3) ,
|(2n - (-l)n) , if n t 0 (mod 3) .

SOLUTIONS (49-50)
103
50i For each integer nil, prove that there is a polynomial
p (x) with integral coefficients such that
4n,, ,4n ,, , 2% / \ , / i\n/n
x (1-x) - (1+x )p (x) + (-1) 4 .
n
Define the rational number a by
n '
(50.0)
(-1)
n-1
n ,n-l
4
p (x) dx , n - 1,2, .. .
Prove that a satisfies the inequality
ir " a I <
n1 ,5n-l
4
, n = 1,2,... .
Solution (due to L. Smith): Let Z denote the domain of rational
integers and Z[i] » {a+bi: a,b £ Z}
the domain of gaussian integers. For n - 1,2,3,... set
(50.1)
, ,. 4n,, ,4n
q (x) = x (1-x)
(-l)n4n ,
so q (x) e Z[x] . As q (±i) = 0 , q (x) is divisible by x+i and
x-i in Z[i][x] , and so p (x) = q (x)/(x +1) eZ[i][x] . However
p (x)eR[x] , and as R[x]nZ[i][x] - Z[x] , we have p (x)eZ[x] .
This proves the first part of the question.
For the second part, we note that
4n,, ,4n , ,,n,n
x (1-x) _ „ / \ . (-D 4
g = pn(x) + y- ,
1+X
1+x
so that
4n,. ,4n
x (1-x)
1+x2
dx
p (x) dx + (-l)n4n
n
dx
1+x2
Now, as
dx
j^j—j- = j , we have using (50.0)

104
SOLUTIONS (51-52)
n1 ,n-l
4
4n,, ,4n
x (1^> dx .
0 1+x
Now
*4nQ-*>4ri < X4n(i-X)4n £ _L
_ x u x; a . ,
1+x 4
as x(l-x) £ 7- , and thus we have
hr - a <
'n' ,5n-l '
4
completing the second part of the question.
51■ In last year's boxing contest, each of the 23 boxers from
the blue team fought exactly one of the 23 boxers from the green
team, in accordance with the contest regulation that opponents may
only fight if the absolute difference of their weights is less than
one kilogram.
Assuming that this year the members of both teams remain the same
as last year and that their weights are unchanged, show that the
contest regulation is satisfied if the lightest member of the blue
team fights the lightest member of the green team, the next lightest
member of the blue team fights the next lightest member of the green
team, and so on.
Solution: More generally we consider teams, each containing n
members, such that the absolute difference of weights of
opponents last year was less than d kilograms.
Let B,,B„,...,B denote the members of the blue team with weights
1 Z n "
bl* b2* ■■■* bn'

SOLUTIONS (51-52)
105
and let G1,G,,...,G denote the members of the green team with
weights
8l £ g2 £ ... £ gn .
For each r with 1 £ r £ n , we consider this year's opponents B
and G . We show that lb - g I £ d . We treat only the case
r r r'
b 2 g as the case b £ g is similar. If there exists s with
r < s £ n and t with 1 £ t £ r such that B fought G last
year, then b -g £ b -g£d. If not, then every boxer B
with r < s £ n was paired with an opponent G with r < t £ n
last year, and thus B must have been paired with some G with
r r u
1 £ u £ r last year. Thus we have
b -g £ b -g £ d .
r er r u
This completes the proof.
52i Let S be the set of all composite positive odd integers
less than 79 .
(a) Show that S may be written as the union of three (not
necessarily disjoint) arithmetic progressions.
(b) Show that S cannot be written as the union of two
arithmetic progressions.
Solution: (a) Each member of S can be written in the form
(2r + l)(2r + 2s + 1) ,
for suitable integers r > 1 and s £ 0 , and so belongs to the
2
arithmetic progression with first term (2r+l) and common
difference 2(2r+l) . Taking r - 1,2,3 we define arithmetic
progressions A ,A ,A as follows:

106
SOLUTIONS (52-53)
A1 = {9,15,21,27,33,39,45,51,57,63,69,75} ,
A2 - {25,35,45,55,65,75} ,
A3 - {49,63,77} .
It is easily checked that
(b) Suppose that
where
and
A u A u A .
S = A u B ,
{a , a+b , ... , a+(m-l)b}
B - {c , c+d , ... , c+(n-l)d} .
Without loss of generality we may take a » 9 . Then we have either
a + b - 15 or c » 15 . In the former case A » A. and so c - 25 ,
c + d = 35 giving B - A. . This is impossible as 49 is neither
in A nor B . In the latter case either a + b ■= 21 or c + d = 21.
If a + b - 21 we have A - {9,21,33,45,57,69} and so c+d - 27
giving B - {15,27,39,51,63,75} . This is impossible as 49 belongs
neither to A nor B . If c + d » 21 we have B » A,
1
{9} so
a + b = 25 giving A - {9,25,41,...} which is impossible as 41 is
prime.
53. For b > 0 , prove that
fb
s in x , tt
<t.
by first showing that
b .
smx
dx
•to
-ux . ,
e smx dx
du .

SOLUTIONS (52-53)
107
Solution: We begin by showing that for b > 0
fb
c
(53.1)
Vie have for y > 0
b .
sin x
dx » lim
y + oo JQ
,, -xy. smx
(1 - e J) —-— dx .
b . ,b
^2^dx - 1(1 - e_Xy) ^*dx
-0 x Jo
-xy sm x ,
e J dx
x
£ max
0<x£b
- M(b)
- M(b)
< M(b)
i
e_Xydx
-xy
(1 - e~by)
Letting y-t-°o we obtain (53.1).
Next we have
,. -xy. smx ,
(1 - e ') —— dx
ilc
Jo Jo
xUj
xe du
•dx
-ux .
e smxdx
du
Letting y-t-°° we obtain
) X JO I JO
-ux .
e sm x dx
du

108
SOLUTIONS (54-55)
-bu
(1 - e (u sin b + cosb))
du
1 + u
r-bu. . ,
e (u sm b +
l+u2
cos b)
du .
Hence we have
b .
ii^dx-I
0 X 2
-bu
e (u sin b + cos b)
l+u2
du
-bu /l+u2 ,
e -TH?du
£ I e"budu
as required.
5Ai Let ai»a7 a4i be 44 natural numbers such that
0 < a. < a, < ... < a., i 125 .
12 44
Prove that at least one of the 43 differences d. = a. , - a.
J J+l J
occurs at least 10 times.
Solution: We have
43
JlVj^J+rV =a44"al S 125-1 - 124

SOLUTIONS (54-55)
109
If each difference d. occurs at most 9 times then
43
1 d. > 9(1+2+3+4) + 7(5) = 125.
j-1 J
This is clearly a contradiction so at least one difference must occur
at least 10 times-
55. Show that for every natural number n there exists a prime
2 2
p such that p = a + b , where a and b are natural numbers
both greater than n . (You may appeal to the following two theorems:
(A) If p is a prime of the form 4t+l then there exist integers
2 2
a and b such that p = a + b
(B) If r and s are natural numbers such that GCD(r,s) = 1 ,
there exist infinitely many primes of the form rk+s , where k is
a natural number.)
Solution: Let n be a natural number. By (B) there exists a prime
q > n of the form 4t+3 . Set
m = 2(l2+q)(22+q)...(n2+q) .
Clearly we have
GCD(m.q) - 1 .
Hence, by (B), there exists a natural number k such that the number
2,
p = m k - q
is a prime. Clearly p is of the form 4u+l . Hence, by (A), there
exist natural numbers a and b such that
p ■ a + b
Without loss of generality we may assume that a < b . Suppose now
that a £ n . Then we have

SOLUTIONS (56-57)
2 2 2, 2
c - p - a = m k - q - a
4k n (r2+q)2 - (a2+q)
r=l
(a2+q)
2 2 2
4(a>q) n (r +q) - 1
r=l
r*a
where the factors on the right hand side of the equality are coprime.
Consequently they must be squares, but this is impossible as the
second factor is of the form 4v-l . Thus we must have b > a > n .
56, Let a ,a , ...,a be n (51) integers such that
(i) 0 < a1 < a2 < ... < a ,
(ii) all the differences a. - a. (1 S j < i S n) are distinct,
(iii) a. = a (mod b) (1 £ i £ n) , where a and b are positive
integers such that 1 £ a £ b-1 .
Prove that
fa i £n3 + (a - £)n .
L, r 6 6
r=l
Solution: Let r be an integer such that 2 £ r £ n . For
1 S j < i i r there are !j = -|r (r-1) distinct
differences a. - a. , and these are all divisible by b . Thus the
largest difference among these, namely a - a. , must be at least
h r i
jv (r-1) , that is
ar - ax * |r (r-1) , 2 i r £ n ,
and so
ar > a + |r(r-l) , 2 < r i n .

SOLUTIONS (56-57)
111
As a = a (mod b) there is an integer t such that a = a + bt .
If t £ -1 then a = a + bt £ a - b £ -1 , which is impossible as
a. > 0 . Hence we have t £ 0 and so a. 1 a , giving
(56.1) ar i a + |r (r-1) , 2 £ r £ n .
The inequality (56.1) clearly holds for r - 1 . Thus we have
n , n . 2 , v
V •» , b r , ,. b n(n -1)
I ar > an + j \ r(r-l) - an + ■=■ • -^-t—'- ,
r»l r=l
so that
l ar > |n3 + (a - |)n
r=l
57. Let A = (a..) be the nxn matrix given by
a. .
2 cos t , if i = j ,
1 , if |i - j| - 1 ,
0 , otherwise ,
where -ir < t < tt . Evaluate D = det A
n n
Solution: Expanding D = det A by the first row, we obtain the
recurrence relation
(57.1) D =2costD ,-D ,, n>2.
n n-1 n—2
We now consider two cases according as t * 0 or t = 0 . For
t * 0 the values of D1 and D, may be obtained by direct
calculation as follows:
,. , „ sin 2t
D. » 2 cos t » —:—— ,
1 smt
and

112
SOLUTIONS (57-58)
2
D_ = 4 cos t - 1
2
, 4 sin t cos t - sm t
sin t
2 2
2 sin t cos t + sin t (2 cos t - 1)
sin t
sin 2t cos t + sin t cos 2t
These values
(57.2)
sin t
sin (2t + t)
sin t
sin 3t
sin t
suggest that
D _ sin (n+l)t
n sin t
, n = 1,2,3, ... .
In order to prove (57.2), assume that (57.2) holds for n = l,2,...,k-l,
Then we have, by the recurrence relation (57.1),
\ m 2 cos C Dk-1 " \-2
- 2 cos t 12£i£ - »in<fc-Dt
sint sint
2 sinkt cos t - sin (k-l)t
sin t
m sin (k+l)t
sin t
Thus (57.2) holds for all n by the principle of mathematical
induction.
For t = 0 we have
D1 = 2 , D2 - 3 , D3 - 4 ,
which suggests that D =n+l, nSl. This can also be proved
by mathematical induction. We note that

SOLUTIONS (57-58)
113
lia, Sin <n+1)t -n + 1 .
t + o sint
Remark: The auxiliary equation for the difference equation (57.1) is
2
x - 2(cos t)x+l - 0 ,
which has the roots
t * 0 ,
t - 0 ,
giving
fexp(it) , exp(-it) ,
(_ 1 (repeated) ,
[A exp(nit) + B exp(-nit) , t * 0 ,
D = I L L
n 1 A2 + B2n , t - 0 ,
for complex constants A1 and B and real constants A. and B. ,
which can be determined from the initial values D. = 2 cos t ,
2
D. = 4cos t - 1 , using DeMoivre's theorem in the case t * 0 . One
finds
A - yd - i cot t) , Bj ' |(1 + i cot t) ,
A2 - B2 - 1 .
58i Let a and b be fixed positive integers. Find the general
solution of the recurrence relation
+ a + Vb +
(58.0) x ,, - x + a + Vb + 4ax , n •= 0,1.2, ... ,
n+1 n n ' ' '
where x„ = 0 .
Solution: From (58.0) we have
b + 4ax ,, ■ b + 4a <x_ + a + &
~ ■ ' -' ■ +a+vb
2 , D- , ,. 2.
+ 4aVb '
4a + 4aVb +4ax + (b +4ax )

114
SOLUTIONS (58-59)
= (2a + \4 +4ax ) ,
n
so that
(58.1) Aft) +4ax . = 2a +Vo +4ax
Hence, by (58.0) and (58.1), we have
+ a - \fi?-
(58.2) xn = xn+1 + a - Vb +4axn+1
Replacing n by n-1 in (58.2) , we get
n
(58.3) x . =x + a - Vt>2+4ax
n-1 n i
Adding (58.0) and (58.3) we obtain
x , + x , - 2x + 2a ,
n+1 n-1 n
and hence
(58.4) x _,_, - 2x + x . = 2a .
n+1 n n-1
Setting y : » x - x . in (58.4), we obtain
n-1 n n-1
(58.5) yn - yn_1 = 2a .
Adding (58.5) for n = 1,2,3,... yields (as y » x. - x = a + b)
y = 2an + (a + b) ,
2
and so x = an + bn
n
59i Let a be a fixed real number satisfying 0 < a < T , and set
fa
(59.0) Ir
1 - r cos u
1 - 2r cos u + r^ "
•a
Prove that
I, , lii I , lim I
1 r + 1+ r r + 1_ r

SOLUTIONS (58-59)
115
all exist and are distinct.
Solution: We begin by calculating I1 . We have
(59.1)
1 - cos u
2-2 cos u
du
1 ,
■j du = a ,
where we have taken the value of the integrand to be -^ when u = 0.
Now, for r > 0 and r * 1 , we have
1 - 2r cos u + r > 2r - 2r cos u = 2r(l - cos u) £ 0 ,
so that the integrand of the integral in (59.0) is continuous on [-a,a],
We have
I = | f1 | (1 - r2>
r J 1 2 2(1 - 2r cos u + r2)
-a ^
du
= a + i(l - r2)
du
1 - 2r cos u + v2 '
which gives
(59.2)
where
r 2 r
(59.3)
du
r J 1 - 2r cos u + v1
-a
Let t = tan •=■ with -a £ u £ a so that
1 - t _, 2
cos u = =■ , du = =- dt .
1 + t
1 + t
Using the above transformation in (59.3), we obtain

116
SOLUTIONS (59-60)
(59.4)
(1+r)
dt
+ fel
J-t,
2 *
1+r J
where
t. - tan a/2
Evaluating the standard integral in (59.4), we deduce that
1+r
(59.5)
r li 2
1-r
4 „ -1
tan
1-r
tana/2
From (59.2) and (59.5) we get
! . a + 2Iil£) tan"l
I1-r2I
1+r
1-r
tana/2
Hence for r > 1 we have
I - a - 2 tan
r
r+1
r-1
tana/2 ,
and for 0 < r < 1 we have
I » a + 2 tan
r
Taking limits we obtain
1+r
1-r
tana/2
lim i-a-2.i-a-Tr,
r +1+ r L
lim I ■=a+2«y=a + Tr
r -t-1~
Thus the quantities I. , lim I , lim I all exist and are all
_,. . r + 1+ r r + 1" r
distinct.

SOLUTIONS (59-60)
117
50i Let I denote the class of all isosceles triangles. For
A £ I , let h, denote the length of each of the two equal altitude
of A and k, the length of the third altitude. Prove that there
does not exist a function f of
for all A e I .
h, such that
kA £ f(V •
Solution: Let h be a fixed positive real number. For t > 1 let
(60.1)
h(t+i)2
!<»i> •
As b < 2a there exists an isosceles triangle A(t) with vertices
A,B,C such that |AB| = |AC| — a , |BC| = b . It will follow that
the choice (60.1) is such that
'A(t)
h , k
A(t)
h
2
''4
i
'"J
Let P , Q , R be the feet of the perpendiculars from A to BC ,
B to CA , C to AB respectively. Then we have
hA( ) = BQ2 = CR2 * a2sin2A " a2(l ~ cos2A) .
Applying the cosine law to A(t) we obtain
2a2 - b2
cos A
2a2
Hence it follows that
(60.2) h2(t)=a2|l
2a2 - b2]2
2a2
b2
4a2
(4a2 - b2)
Next, from (60.1), we see that
2a-b
h
(t+-)
«<-!>
2a+b
' h
t(t+i)
b2
-4.2
<t-i»2
<t+i»2 '

118
SOLUTIONS (60-62)
so that
l> .. 2 ,2. ,2
—j (4a -b ) = h
4a
and thus from (60.2) we have
hA(t) * h •
Applying Pythagoras' theorem in triangle ABP, we have
2 Ap2 2 b2 h2<t+7)2
kA(t)" ** *a x= 4 7i m '
so that
(t-i)'
'MO
t + —
t
«-*
Finally, suppose there exists a function f = f(h,) such that
kA S f(hA) ,
for all A e I . Then, in particular, one sees that
which implies
that is
(60.3)
kA(t) "- f<W ' C * X •
<**
f i
< f(h) , t > 1 ,
C + t „ 2 f(h)
1 " h
t > 1 .
C~ t
As the left side of (60.3) tends to infinity as t"•"+■» while the
right side is fixed, we have obtained a contradiction, and therefore
no such function f can exist.

SOLUTIONS (60-62)
119
61, Find the minimum value of the expression
(61.0) [x2+%] - 2[(l+cos t)x + k(1+^nt)1 + (3+2 cost+ 2 sin t) ,
for x > 0 and 0 £ t £ 2tt , where k > -r + /2 is a fixed real
number.
Solution: The expression given in (61.0) can be written in the form
(x - (1+cos t))2 + (| - (1+sin t))2 ,
which is the square of the distance between the point (x , —) (x > 0)
on the rectangular hyperbola xy - k in the first quadrant, and the
point (1+cos t , 1+sin t) (0 i t £ 2tt) on the circle centre (1,1)
with radius 1 . The condition k > -=• + /l ensures that the two
curves are non-intersecting. Clearly the minimum distance between
these two curves occurs for the point (A: , /k) on the hyperbola and
the point (1 + -w , 1 + "Tf) on the circle. Hence the required
minimum is
2(/k - U+7f))2 •
62, Let £ > 0 . Around every point in the xy-plane with integral
co-ordinates draw a circle of radius £ . Prove that every straight
line through the origin must intersect an infinity of these circles.
Solution: Let L be a line through the origin with slope b . If
b is rational, say b » -j , where k and t are integers
satisfying t 5 1 and (k,£) = 1 , then L passes through the
centres of the infinity of circles all of radius e , centred at
lattice points (It , kt) , where t is an integer.

120
SOLUTIONS (62)
If b is irrational, then by Hurwitz's theorem there are
infinitely many pairs of integers (m,n) with n * 0 and GCD(m.n) = 1
for which
£_b < 1 < 1 .
n c- 2 2
on n
Choosing only those pairs (m,n) for which
1
n ~> i~' ,
EYb +1
we see that there are infinitely many pairs (m,n) for which
that is,
(62.1)
for which
m
n
- b
<
m - bn
^
eyb +1
n
< e
Since the left side of (62.1) is the distance between the line L
and the point (m,n), L intersects the infinity of circles all with
radius £ centred at these lattice points.
Second solution (due to L. Smith): Let L be a line through the
origin with slope b . The case
when b is rational is treated as in the first solution.
When b is irrational, we construct an infinite sequence of
lattice points whose distances from L are less than any given
positive £ .
Pick any lattice point (x , y ) with d.
'1
bx
< 1
Clearly d. is positive as b is irrational. Set
KJ
* i ,
so that

SOLUTIONS (62)
121
(62.2)
1 j y 1
a.+l 1 a.
Let (x, , y.) be the lattice point given by
(x2 , y2) =
> bx
1 '
(alXl ' alyl " X) ' if yl
(alXl ' alyl + 1) ' if yl < bXl '
and set d, = |y_ - bx.| . Clearly, as d~ * 0 , we may set
(62.3)
It is easy to see that d_ = 1 - a.d. , so that by (62.2) we have
(62.4)
do < ITT
2 a.+l
Thus, from (62.3) and (62.4), we obtain
al < a2 •
Continuing this process we obtain an infinite sequence of lattice
points i(\ »yi,): k * 1.2,... } , whose vertical distances d,
from L satisfy
1
dk<I
k-1
+ 1
, k> 2 ,
where a, =
k
^J '
k > 1 . Furthermore {el : k = 1,2,
strictly increasing sequence of positive integers.
Finally choose a positive integer N such that tj < e
for all n k N + 1 , we have
d <
n a
1 1^1
n-l N
} if
Then,
and the lattice points (x , y ) (n > N) are as required.

122
SOLUTIONS (63-64)
63. Let n be a positive integer. For k = 0,1,2,.,
define
.,2n-2
(63.0)
r°° k
x
2n. n
x +x +1
dx
Prove that I £ I _1 , k = 0,1,2 2n-2
Solution: For k = 0,1,2,...,2n-2 , (63.0) can be written
I, ■ lim
k
a -*■ oo
b-t-0+ '
2n, n ,
x +x +1
dx .
Applying the transformation x » — , we obtain
I, = lim
k
a -*■ oo
b + 0+
1/b
2n-k-2
y
y +y +1
1/a
dy ,
so that Ik = I2n_k_2 .
Now, using the arithmetic mean-geometric mean inequality, we have
for x > 0 ,
k , 2n-k-2
(63.1)
x + x
> xn 1 , k - 0,1,2 2n-2 .
As x +x +1 > 0 , we may divide (63.1) by x +x +1 , and integrate
the resulting inequality to obtain
Xk + hn-^-Z
> I
n-1 '
from which the desired inequality follows.

SOLUTIONS (63-64)
123
64. Let D be the region in Euclidean n-space consisting of all
n-tuples (x.,x_,...,x ) satisfying
x. * 0 , x, 2 0 , ... , x 20 , x.+x, + ...+X £1.
1 '2 n 12 n
Evaluate the multiple integral
(64.0)
x. x. ...x (1-x -x.-...-x ) dx, ...dx ,
12 n 1 2 n 1 n
where k. ,...,k are positive integers.
Solution (due to L. Smith): We begin by considering
I(r,s)
x (a-x) dx ,
where r and s are positive integers and a > 0 .
Integrating I(r,s) by parts, we obtain
so that
that is
I(r,s) -^j- I(r+1 , s-1) ,
l(r,s).= ^r.f^...^rl(r+s,0) ,
Kr.s) -^fyy Kr+s ,0) .
As l(k,0) ■ t—-r- , for k > 1 , we obtain
(64.1)
T. . rls!ar+s+1
I(r's) " (r+s+1)!
Now, applying (64.1) successively, we obtain

124
SOLUTIONS (64-65)
1 2 n,, \ n+1, ,
x. x, ...x (1-x,-x,-...-x ) dx,...dx
12 n 1 2 n 1 n
1-x,
x,=0 x2=0
1 2 k , n_2
n-1
x i
n-1
xnn (d-x^
-x ,)-x )
n-1 n
n+1
x -0
n-1
x =0
n
dx ...dx,
n 1
k ! k ..!
n n+1
(k +k ,.+l)!
n n+1
1-x,
Kl ■■
1-x.-.. .-x _.
1 k . n L k +k ^.+1
n-1 ,, -, n n+1
Vi (1-Xf--Vi)
x »0 x,=0
x =0
n-1
dx .... dx,
n-1 1
k , ! k ! k .. !
n-1 n n+1
(k ,+k +k ,,+2)!
n-1 n n+1
1-x,
x^O x2-0
1-x,-...-x n k ,+k +
1 , n-3 n-1 n
n-2 ., k Al+2
x . (1-X.-...-X ,,. n+1
n-2 1 n-2)
x ,=0
n-2
dx ....dx,
n-2 1
k. ! k,! ... k ! k _,_, !
1 2 n n+1
(k,+k,+...+k +k ,+n)!
12 n n+1
65, Evaluate the limit
i n
lim I I
n + = n k-1
2
■2^1
L/£J
{%
Solution: We show that L = -= 3 . For any real number r we have
[2r] - 2[r]
0 , if [2r] is even,
1 , if [2r] is odd .
Hence we obtain

SOLUTIONS (64-65)
«■>-!([£]-SI
where f(n) =
and only if
r[2vC] -
2
n
I 1
k-1
[2^] odd
f(n) n
-III.
s=l k-1
[2^]-2e+l
• 1
Next we see t
4n - ,, <- 4n
14\
(2s+2) (2s+l)
and thus
t 1
k-1
4n 4n
[2^]=2s+l
(2s+l)
4n 4n
(2s+2)
+ *, .
(2s+l)2 (2s+2)2 *
where |E.| < 1 . Hence we have
- A(n) = 4
n
s-lH2s+l)
+ E, ,
2 (2s+2)2J 2
where
|E | *I<2>..|e,| <I<£i£^-JL
'2' n ' 1' n n ,m
Letting n-*•">> gives

126
SOLUTIONS (66-67)
*I
1
1
s=l (2s+l) (2s+2)
that is,
L = 4
2
V1
2
24'
2-- 3
3
-j + — + — +
r 3 5
TT 1,1,1 TT
_ and -I + -I + -I+ ... = —
I 4 b
66i Let p and q be distinct primes. Let S be the sequence
consisting of the members of the set
{pmqn: m.n = 0,1,2,... }
arranged in increasing order. For any pair (a,b) of non-negative
integers, give an explicit expression involving a, b, p and q for
the position of p q in the sequence S .
Solution: Without loss of generality we may suppose that p < q .
Clearly p q is the n term of the sequence, where
n is the number of pairs of integers (r,s) such that
prqS £ P3q . r > 0 , s > 0 .
Set
k = a +
so that p is the largest power of p less than or equal to p q .
Then we have
k
n- 1 1
r,s=0
r s , a b
p q £p q

SOLUTIONS (66-67)
127
t 1
r,s - 0
vlnp + slnq £a£np+b£nq
V f f a -fa P + bin q - r In pi ,
i011 ^ J
r=0
k
j> * 1^
-r) In p
+ 1 .
a b
so that the position of p q is
(k+l)(b+l) +
r-0 L tJl
-r) Zn p
67i Let P denote an odd prime and let Z denote the finite
field consisting of the p elements 0,1,2 p-1 . For a an
element of Z , determine the number N(a) of 2x2 matrices X ,
P
with entries from Z , such that
P
(67.0)
X » A , where A
[:°J
Solution: It is convenient to introduce the notation
{1 , if a = b for some non-zero element b of Z ,
0 , if a - 0 , P
-ri, otherwise .
We will show that
(67.1) N(a)
'p2+p + 2 , if k(a) - 1 ,
p2 , if k(a) - 0 ,
2
p - p , if k(a) «-l .

128
SOLUTIONS (67-68)
We note that if k(a) = 1 , so that there is a non-zero element b
2 2
of Z such that a = b , then the only other solution of a = x
P
is x = -b .
I z wj
Let X = J J , where x,y,z and w are elements of Z , be a
2
matrix such that X = A , Then we have
2 2
(67.2) x +yz=yz+w =a,
(67.3) (x + w)y = (x + w)z = 0 .
We treat two cases according as (i) x + w = 0 or (ii) x + w * 0 .
In case (i) the equations (67.2) and (67.3) become
2
(67.4) x + yz = a .
2
If k(a) = 1 , say a = b , b * 0 , then all the solutions of (67.4)
are given by
(x,y,z) = (±b,0,0) , (±b,t,0) , (±b,0,t) , (u,t,t_1(b2-u2)) ,
where t denotes a non-zero element of Z and u denotes an ele-
P
ment of Z not equal to +b . Thus there are 2+2(p-l) +2(p-l)
P 2
+ (p-2)(p-l) = p + p solutions (x,y,z,w) in this case.
If k(a) = 0 , so a = 0 , then all the solutions of (67.4) are
given by
(x,y,z) = (0,0,0) , (0,t,0) , (0,0,t) , (t.u.-tV1) ,
where t and u denote non-zero elements of Z . Thus there are
2 2 p
1 + (p-1) + (p-1) + (p-1) ■ p solutions (x,y,z,w) in this case.
If k(a) = -1 , so that a is not a square in Z , then all
P
solutions of (67.4) are given by
(x,y,z) = (0,t,at_1) , (t,u,(a-t2)u_1) ,
where t and u are non-zero elements of Z . Thus there are
2 2 p
(p-1) + (p-1) = p - p solutions (x,y,z,w) in this case.

SOLUTIONS (67-68)
129
In case (ii) the equations (67.2) and (67.3) become
2
x ■= w * 0 , x ■= a , y * z = 0 ,
which clearly has two solutions if k(a) - 1 , and no solutions if
k(a) = 0 or -1 .
Hence the total number of solutions is given by (67.1).
£$, Let n be a non-negative integer and let f(x) be the
unique differentiable function defined for all real x by
(68.0)
Evaluate the integral
for x 2 0 .
(f(x))2n+1 + f(x) - x = 0
f(t)dt .
Solution: The function y = f(x) defined by (68.0) passes through
the origin and has a positive derivative for all x .
Hence, there exists an inverse function f (x) , defined for all x,
and such that f (0) = 0 . Clearly we have f (x) - x + x .
When x > 0 , the graph of f(x) lies in the first quadrant, as
f(0) = 0 and f is increasing. Thus for all x £ 0 we have
f(t) dt +
Jo
f(x)
f *(t) dt - xf(x)
Jo
Now
f(x)
f_1(t) dt
Jo
JO
tW(t^tt)dt-^^,
so that

130
SOLUTIONS (69)
f(t) dt = xf(x) -
(f(x))2n+2 (f(x))2
2n+2~
2n+l
2n+2
xf(x) -
2n+2
(f(x))'
69i Let f(n) denote the number of zeros in the usual decimal
representation of the positive integer n , so that for example,
f(1009) = 2 . For a > 0 and N a positive integer, evaluate the
limit
In S(N)
lim
N-t-oo
In N
where
S(N) - I a
k=l
f(k)
Solution: Let I be a non-negative integer. The integers between
I £+1
10 and 10 -1 have £+1 digits of which the first is
necessarily non-zero. The number of these integers with i (OSiS-d)
(P~\ P — '.4-1
.9
Choose m to be the unique non-negative integer such that
10m-l £ N < lo"*1-! ,
so that
Then we have
'in (N+l)1
. In 10 J •
s(io«-n = 10f V(k) -1110i"V^
k=l
i=0 k=l
f(k)-i

SOLUTIONS (69) 131_
m-1 . 10m-l
la1 I 1
i=0 k=l
f(k)=i
l+l
m-1 , m-1 l(T -1
e first
S(10m-1)
- la1
i-0
t t 1
f(k)-i
remark we obtain
m-1 .
- la1
i=0
£=0 W
■:trM
■m*€
m;1 /
- 9 l (a + 9)* ,
1-0
that is
where c - ^
S(10m-1) = c((a+9)m - 1) ,
As
we obtain
S(10m-1) £ S(N) < SdO1^1-!) ,
c(a+9)m i S(N)+c < cCa+g)"^1
Taking logarithms and dividing by m , we get

132
SOLUTIONS (70-71)
£"_£ _ £n(a+9) < ±£n(S(N)+c) < ^L£ + MU £n(a+9)
m m
Letting N*'A , so that m-*-"> , we deduce that
lim -£n(S(N)+c) - £n(a+9) .
N*» m
Hence we have
lim -£nS(N)
m
N -» <"
lim -£n(S(N)+c) - -£n|l + C
N + ~ ~ - t S®
and so
- H
£n(a+9) ,
In S(N) £n(a+9)
70. Let n 2 2 be an integer and let k be an integer with
2 £ k £ n . Evaluate
M - max | min (a. -a.)
where S runs over all selections
S - {a..,a,, — ,a } from
{l,2,...,n} such that a < a, < ... < a

SOLUTIONS (70-71)
133
Solution (due to J.F. Semple and L. Smith):
We show that M - wj" * We consider the selection S*
given by
a. - (i-l)[^3 + 1 » 1 S i £ k ,
which has
a.+1 - a, = |^4I , 1 S i £ k-1 .
Thus for S* we have
i = [k-lj '
»»[3|
on S with
-12V*i-1!<*w-*i>£<"'([Fr]
min (a... - a.)
l£i<k-l 1+1 X
so that M £ T~T • In order to prove equality, we suppose that
there is a selection S with
min (a. - a.) 2 |££| + 1
UiSk-1 1+i x
Then we have
k-1
._. . . . ._.. +1l > n"! •
i=l
which is impossible. Hence any selection has min (a. ,-a.) < I —rl+1,
u- u .u • a n. ISiSk-l 1+1 X Lk ll
which proves the required result.
2
71, Let az + bz + c be a polynomial with complex coefficients
such that a and b are non-zero. Prove that the zeros of this
polynomial lie in the region
b
(71.0) z <

134
SOLUTIONS (71-73)
Solution: Note that
/b2-4ac|
FW
£ b i/l +
S b
I4ac|
1 +
+
2ac
b2
2ac
b
so that
_b_ + /b2-4ac
2a ~ 2a
2a 2a b a b
and hence the solutions of az + bz + c = 0 satisfy (71.0).
Second solution (due to L. Smith): Let w (*-l) be a complex number,
The inequality
(71.1)
Iwfll +
wfl
i 1 ,
is easily established, for if |wfl| - 1 then (71.1) clearly holds,
while if 0 < |w+l| < 1 then
wfl +
wfl
> wfl f w > 1
Let z.,z. be the roots of the given quadratic chosen so that
|z,|S|z.| . As a and b are non-zero, z, * 0 and z.+z. * 0
and setting w = zi/zo (*~1) in (71.1) we obtain
|z1z2l
lZ2l £ lzl+z2l + |z1+Z2[ •
b c
The inequality (71.0) follows as z. f z. = and z z. = — .

SOLUTIONS (71-73)
135
72, Determine a monic polynomial f(x) with integral coefficients
such that f(x) = 0 (mod p) is solvable for every prime p but
f(x) =0 is not solvable with x an integer.
2
Solution: If p = 2 or p = 1 (mod 4) the congruence x +1=0
(mod p) is solvable. If p = 3 (mod 8) the congruence
2
x +2=0 (mod p) is solvable. If p ; 7 (mod 8) the congruence
2
x -2=0 (mod p) is solvable. Set
f(x) = (x2+l)(x2+2)(x2-2) .
Clearly f(x) is a monic polynomial with integral coefficients such
that f(x) =0 is not solvable with x an integer.
73i Let n be a fixed positive integer. Determine
M = max £ |x.-x.| .
OSx^l l<i<j<n * J
k-l,2,...,n
Solution: Without loss of generality we may assume that
0 < Xl < x2 £ ... £ xn £ 1 ,
so that
S =
t ix -x i - t (x -x > .
l<i<j<n J l£i<j£n J
The sum S has n(n-l)/2 terms. For each k , 1 < k £ n , x,
appears in k-1 terms in the left position and n-k times in the
right position. Hence we have
n n
S = I x, ((k-1) - (n-k)) = I x (2k-n-l) .
k=l K k-1 K
As 2k-n-l < 0 , for k < —=— , we have

136
SOLUTIONS (73-74)
S < I xk(2k-n-l) < I (2k-n-l) .
Thus, for n even, we have
n
S < I (2k-n-l) = l+3+5+...+(n-l) = n2/4 ,
k-f+1
and for n odd
n
S < I (2k-n-l) = 2+4+6+...+(n-l) = (n2-l)/4 .
.n+1
k=- 2
Thus we have
S £ [n2/4]
For n even, the choice
gives
x- x, - ...» x - 0 , x = ... = x =1
1 l n n n
2 I+1
S - n2/4
and for n odd, the choice
xi 'x2- ••• "Vi = 0 • Vi ' ••• ~\* i
gives
S - (n2-l)/4
This shows that
M » [n2/4] .
"Ii\, Let (x. : i - 1,2, ...,n} and {y. : i - 1,2, ...,n} be two
sequences of real numbers with

SOLUTIONS (73-74)
137
X. £ X. > . . . > X
i I n
How must y.,...,y be rearranged so that the sum
(74.0)
is as small as possible?
X (xi-yi}'
i-1 x
Solution: Suppose the i term y. is smaller than the j term
y. , 1 < i < j < n . Interchanging y. and y. produces
a new sequence
with
Zk =
zl ' z2 zn
yk , if k * i or j ,
y. , if k = i ,
yi , if k = j .
Moreover we have
.2
J/w - &vi + («i^j> + (xj-yi} - (vyi} - <*ny
k
where
((x.-y,)2+(x.-y.)2) - ((x.-y.)2+(x -y )2)
-tj J J. J. J. JJ
' 2(y.-y.)(x.-x.)
J **■ J
£ 0 ,
so that
t (x -z )2 £ I (Vy )2 .
k-1 K K k-1 k

138
SOLUTIONS (75-76)
Hence every such transposition decreases the size of the sum (74.0),
and the smallest
decreasing order.
and the smallest sum is obtained when the y. are arranged in
751 Let p be an odd prime and let Z denote the finite field
consisting of 0,1,2,...,p-l . Let g be a given function on Z
P
with values in Z . Determine all functions f on Z with values
P_ P
in Z , which satisfy the functional equation
P
(75.0) f(x) + f(x+l) » g(x)
for all x in Z
Solution: Replacing x by x+k (k = 0,1,2,...,p-l) in (75.0), we
obtain
(75.1) f(x+k) + f(x+k+l) = g(x+k) .
Hence, using (75.1), we have
p;1 k p;1 k
I (-1) g(x+k) = l (-l)K(f(x+k) + f(x+k+D)
k=0 k=0
I ((-l)kf(x+k) - (-l)k+1f(x+k+D)
k=0
- f(x) - (-l)Pf(x+p)
- 2f(x) ,
so that there is only one such function, namely,
-1 P"1 v
f(x) - 2 I (-1) g(x+k) .
k=0

SOLUTIONS (75-76)
139
76• Evaluate the double integral
drl
(76.0)
f -f-*2
oJo1"
dxdy
xy
Solution: Set S - {(x,y) |0SxSl,0SySl} .
For 0 < £ < 1 set
Rx(e) = {(x,y) |0SxSl-e,0SySl},
R2(e) = {(x,y) | 1-e £ x £ 1 , 0 < y < 1-e } ,
R(e) = R^e) u R2(e) ,
so that
R(e) = S - {(x,y) | 1-e S x S 1 , 1-e S y S 1 } .
Then, for j = 1,2 , we set
I.(e) =
J
dxdy
The function
1
1 - xy
)
is continuous on the square S except for a
R.(e)
J
1 - xy
discontinuity at the corner (x,y) = (1,1) , so that (76.0) becomes
I - lim
£-*-0+
R(e)
Y^-' lim (l,(e) + I (e))
1 xy £-0+ 2
For n a positive integer and for j » 1,2 , we set
J.(e,n) =
r, , ,22 n-1 n-K , ,
(1+xy + xy +...+X y )dxdy,
R.(e)
J
K.(e,n)
R.(e)
J
n n
x y
1 - xy
dx dy

140
SOLUTIONS 06)
2 2
As 1 + xy + x >
2 , , n-1 n-1 , x y
y + . .. + x y + . _\
n n
xy
n n n n
1 - x y + x y
1 - xy 1 - xy
1
1 - xy '
we have, for j=l,2;0<£<l;n>l,
(76.1) J.(e,n) +K.(e,n) = l.(e) .
Next, as the largest value of xy on both R. (e) and R,(e) is
l-£ , we have
x y
1 - xy
(1 - £)'
, on R^e) and R^e) ,
so that
I^Ce.n)! £ {
(1-e)1
(1-e)1
(1-e) , j = 1 ,
•e(l-e) , j = 2 .
Hence for 0 < e < 1 we have
lim K.(e,n) =0 , j - 1 , 2
Next, for j » 1,2 , we have
J.(e,n)
r r rn_1
I.
R.(e)
J
n-1
I
m=0
x y dx dy
mm, ,
X y dx dy ,
R.(e)
J

SOLUTIONS (76)
141
so that
n-1
J^e.n) - I
m-0 <>
1-e
C>
m | m ,
x dx y dy
V (1-e)11*1
m-0 (m+1)
giving
and similarly
lim J
jCe.n) - J
(1-e)
m+1
m=0 (m+1)'
lim J2(.e,n; - I ^ I .—
n-»-o> m-0 (m+1) m-0 (m+1)
Letting n-»-"> in (76.1), we obtain
!,(£) ' t
(1-e)
m+1
so that
m-0 (m+1)
m-0 (m+1) m-0 (m+1)
» ,. .ofl °° ,. _,.2m+2
1.(0+1,00 -2 t ^> - J d-e)
1 m-0 (m+1) m-0 (m+1)
Hence, by Abel's limit theorem, we have
Y=*=- lim (I (e) + 1 (e))
2 I
I -L
2 ' 2
m=0 (m+1) m-0 (m+1)

142
SOLUTIONS (77-78)
= I
1
m=0 (nH-1)
that is I - t /6 .
77• Let a an<i *> be integers and m an integer > 1 .
Evaluate
[51* f
a+b
m
2a+b
+ . . . +
f(m-l)a + b'
L m
Solution: Our starting point is the identity
(77.1)
k-1
I
x=0
k + e
J - [ek]
where k is any positive integer and e is any real number. As
{y} ■= y - [y] , for any real y , (77.1) becomes
I {f +e} = i(k-l) + {ek}
■}
(77.2)
For fixed k and e , i — + ef is periodic in x with period k .
If c is chosen to be an integer such that GCD(c.k) = 1 , the
mapping x"*cx is a bijection on a complete residue system modulo k .
Applying this bijection to (77.2), we obtain
(77.3)
'I [f +e}= i(k-l) +{ek> .
k-1
t
x=0
We now choose c = a/GCD(a,m) and k = m/GCD(a,m) , so that
GCD(c,k) ■ 1 , and e = b/m . Then (77.3) becomes (keeping k in
place of m/GCD(a,m) where convenient)

SOLUTIONS (77-78)
143
V1 fax+bl _ _1 f m .] / b \
in 1 m J ~ 2 iGCD(a.m) J \GCD(a,m)J *
k-1
I
x=0
and so
m-1 r ,..> GCD(a.m)-l k-1
"y /«!*\ ^*f V fa(y+kz) + bl
K:0 l m ■> z-0 y-0 \ m /
a.m)-l k-1 t ,.■»
*=0 y-0 I m /
= y(m - GCD(a.m)) + GCD(a,
l,m)lGCD(a,m)}
Finally, we have
m-1,
x
that is
Iffl ' I^0 + b " i^™^) - ^^{asfa?} '
m-1
]Jrr\ =!(—a-m + GCD(a,m)) + GCD(a>m){G^^y } .
Remarks: The identity (77.1) is given as a problem (with hints) on
page 40 in Number Theory by J. Hunter, Oliver and Boyd, 1964.
78. Let a.,...,a be n (>1) distinct real numbers. Set
1 n
S = a, + ... + a , M = min (a. - a.)
1 n ■**•,■* i J
lai<jan J
Prove that
i > n(n-l)(n+l)
M 12
Solution: Relabeling the a's . so that a, < a, < ... < a
12 n
preserves the values of S and M .

144
SOLUTIONS (78-79)
2 2
Let mm a. = a. , where j is a fixed subscript. Then, we have
l<i*n x J
a. > 0 , for i > j ,
l
a. < 0 , for i < i .
l J
Next, rain (a. ,, - a.) = J min (a. ,, - a.)2 = / min (a. - a.)2
l<i£n-l 1+1 X UiSn-1 1+1 x l<i<j<n x J
= S& .
Define b. = a. + S& (i-i) , i = 1,2,...,n , so that
l j
b. = b. + M (i- 1) .
l 1
Then, for i > j , we have
a. = (a. - a. .) + (a. - a. .) + ... + (a.. - a.) + a.
l l l-l l-l 1-2 j+1 j j
i S& (i - j) +
a.
b.
l
> a. > -
> a. 1 - a. ,
J i
that is a. > b. > - a. , i > i
11 l J
Similarly we have
a. £ b. < - a. , i < i
11 l J
2 2
Thus, we obtain a. > b. (i = 1,2 n) , and so
11
s= I a?s x b? - Kb +m (i-D)z
i=i x i-i x i=i x
(n-l) ^2 _,_ (n-l)n(n+l)
= n(bi+^^Zil)^+12ZUi
> (n-l)n(n+l) M
12

SOLUTIONS (78-79)
145
79. Let x,,...,x be n real numbers such that
' •" 1 n
\ U I - 1 , fxfc-0 .
k«l K k«l K
Prove that
(79.0)
!*
k=l
a 2 2n
Solution: For 1 S k £ —-r we have
n+1
osl-i-isi-i,
k n n
and for —-?■ S k £ n we have
n+1
0S1+1-1*1-1,
n k n
so that
(79.1)
k n
SI--, 1 < k £ n
Thus, as J x, * 0 , we have
k=l
n x.
V
k=l
1 -
'- k
Hf-L-n-K
k-1
ji i-l-i
£ T^-iH Kl . by (79.1) .
k=l
The inequality (79.0) now follows as J |x | ■ 1 .
k-1

146
SOLUTIONS (80-81)
80, Prove that the sum of two consecutive odd primes is the
product of at least three (possibly repeated) prime factors.
Solution: Let p denote the n prime number, so that p. = 2 ,
p~ = 3 , p, = 5 , ... . For n i 2 , we consider
q = p + p , . Clearly q is even. If q has exactly one
Hn *n *n+l k
prime factor then q = 2 for some positive integer k , and as
3 n
q 5 3+5=2 we have k £ 3 proving the result in this case.
If q has exactly two distinct prime factors, then we have
n
k Z
q = 2 p for positive integers k, I and an odd prime p . If
n
k £ 2 or I i 2 the result holds. If k = I = 1 then, as
P ,, > P , we have
n+1 n
Pn < I (pn + Pn+1} = P < Pn+1 '
which is impossible as p and p , are consecutive primes. This
n n+1
completes the proof.
81, Let f(x) be an iategrable function on the closed interval
[it/2,it] and suppose that
[*
%/2
(81.0) | f(x)sinkxdx = |J | k = n ^ n 1 '
Prove that |f(x)| £ —j—~ on a set of positive measure.
Solution: From (81.0) we have
(81.1) I
k=l
f (x) s in kx dx =1 .
%/2

SOLUTIONS (80-81)
147
Interchanging the order of summation and integration, and using the
identity
k=l
(81.1) becomes
n cost - cos(a+-r)x
£ sinkx » , 0 < x < 2ir ,
, • x
2 sin7
(cos-y - cos (n+ ■*•))
f(x) : — dx - 1
hll
, . x
2 sin-?
Suppose |f(x)| < —;—tr on [-=■ , tt] , except for a set of measure 0
Then we have
1 i
cos ■=• - cos(n+-=0x
|f(x)|! - i-'dx ,
kll
2 sin-;
giving
(81.2)
1 <
1
ir£n2
dx
. x
i ,, suit
%/2 2
Now Jordan's inequality implies that
- £ sin ^ (0 < x <, tt) .
TT 2
and so, on [•? , it] , we have
(81.3)
1
sin
X
2
S
TT
X
Using (81.3) in (81.2) we obtain
1 <
Inl
& -l
h/2

148
SOLUTIONS (82-84)
which is impossible. Hence |f(x)| i » - on a set of positive
82. For n - 0,1,2,... , let
(82.0) s = 3/a + Va~ , . .„ ,
n n n-1 n-2 u
where a = r- • Show that lim s exists and determine its
n n+1 n
value.
Solution: First we show by mathematical induction that the sequence
(s : n » 0,1,2,... } is non-decreasing. Note that
3/9 n
s„ = 1 < \-r = s, . Assume that s , - s . Then we have
0 V 2 1 n-1 n
s ■ 3 /a + s T £ 3/a , + s = s , .
n n n-1 n+1 n n+1
Next we show, also by induction, that the sequence
{s : n = 0,1,2,... } is bounded above by 2 . Clearly s. = 1 < 2 .
Assume that s < 2 . Then we have
n-1
s = 3/a +~s ~ < V6 + 2 = 2 .
n n n-1
Thus L - lim s exists. Letting n-»-=> in s ■ a + s
n-i-o> n n n n-1
3
we obtain L = 6 + L , so that L - 2 .
83, Let f(x) be a non-negative strictly increasing function on
the interval [a,b], where a < b . Let A(x) denote the area below
the curve y = f(x) and above the interval [a,x], where a £ x £ b ,
so that A(a) - 0 .
Let F(x) be a function such that F(a) = 0 and
(83.0) (x' - x)f(x) < F(x*) - F(x) < (x* - x)f(x')

SOLUTIONS (82-84)
149
for all a £ x < x' i b . Prove that A(x) « F(x) for a £ x £ b .
Solution: Clearly A(x) satisfies the inequality (83.0). Assume
that A(x) and F(x) are not identical on [a,b]. Then
there exists c with a < c S b such that A(c) * F(c) . We
partition the interval [a,c] by
. k(c-a)
xk*a + —n~ '
where n is a positive integer and k - 0,1,2,...,n . Then we have
^f(Vi>< A(V - A(vi>< -^V (k -1-2 n> •
Summing from k ■= 1 to k ■= n , we obtain
(83.1) ^i n\ f (O < A(c) < Z*. I f (, ) .
n k-0 k-1 K
Similarly we have
(83.2) £=£ °J f(x. ) < P(c) < ^2. I f(, ) .
n k-0 K n k-1 K
From (83.1) and (83.2) we obtain
|A(c) - F(c)| < ((c-a)/n)(f(c) - f(a))
so that (as A(c) * F(c) )
„ . (c-a)(f(c) - f(a))
n <_-|aW -f(c>|
This is a contradicton for sufficiently large positive integers n .
84, Let a and b be two given positive numbers with a < b .
How should the number r be chosen in the interval [a,b] in order
to minimize
(84.0) H(r) * f*
aixib

150
SOLUTIONS (84-85)
Solution: For a < r £ b we have
- - 1 , if a < x < r ,
x
1--, if r£x£b,
x
and so
M(r) = max (j - 1 , 1 - £) .
Thus, for any c in the interval [a,b], we have
min M(r) = min max ( 1 , 1-t-)
aSrSb a<r<b a °
mm max
(1-1.1-1).
mm max f 1 , 1 -t-1
cSrSb
Choosing c to be the point in [a,b] such that
that is
we have
a L L b '
2ab
a+b '
and
so that
mm max
a<r£c
mm max
cSrSb
(I-l.l-f)- min (1-f) -1-|
aSrSc
(l- 1.1-1)- -1- (f-1) =7-1
c£r<b
c r c
min M(r) = 1 - £ (= - - l) =
a£r£b
b ^ a
and the required r is (2ab)/(a+b)
b-a
a+b '

SOLUTIONS (84-85)
151
85, Let {a : n = 1,2,... } be a sequence of positive real
numbers with lim a - 0 and satisfying the condition
n -»■<» n
an - an+1 > an+1 - a^+2 > 0 . For any e > 0 , let N be a positive
00
t* k+1
integer such that a^, £ 2e . Prove that L - £ (-1) a, satisfies
k-1
the inequality
N
(85.0) |L - t (-I)**1 a. | < e .
k-1 K
n ,k+l
Solution: For n a positive integer, we define S = t (-1) a,
n k=l
We have
00
L - S + (-l)n I (a _,_ - a _,____,)
n »■, n+r n+r+1
r-1
and
00
L - S . + (-l)n_1 J(a_-aJ .
n-1 «•, n+r+1 n+r
r»l
As a ,-a, >a, - a , ,, , we have
n+r-1 n+r n+r n+r+1
(85.1) |S - L| < IS . - Ll .
1 n ' ' n-1 '
Since a = |S - S ,1 and L lies between S , and S , we
n ' n n-1' n-1 n
have
(85.2) a -IS - Ll + |S . - Ll .
n ' n ' ' n-1 '
Taking n = N , where a^, < 2e , we obtain
lsN " L| < e
as required.

152
SOLUTIONS (86-87)
36, Determine all positive continuous functions f(x) defined
on the interval [0,ir] for which
(86.0)
f(x)cosnxdx = (-1) (2n+l) ,
0,1,2,3,4
Solution: We begin with the identities
cos 2x = 2 cos x - 1 ,
Hence
and so
cos 3x = 4 cos x - 3 cos x ,
4 2
cos 4x = 8 cos x - 8 cos x + 1
cos 4x + 4 cos 3x + 16 cos 2x + 28 cos x + 23
4 3 2
* 8 cos x + 16 cos x + 24 cos x + 16 cos x + 8
2 2
* 8(cos x + cos x + 1) ,
2 2
f(x)(cos x + cos x + 1) dx
= 9 + 4(-7) + 16(5) + 28(-3) + 23(1)
"* 0 ,
which is impossible as f(x) is positive on [0,tt] . Hence there are
no positive functions f(x) satisfying (86.0).
87i Let P and P' be points on opposite sides of a non-
circular ellipse E such that the tangents to E through P and
P' respectively are parallel and such that the tangents and normals
to E at P and P' determine a rectangle R of maximum area.
Determine the equation of E with respect to a rectangular
coordinate system, with origin at the centre of E and whose y-axis is
parallel to the longer side of R .

SOLUTIONS (86-87)
153
Solution: We choose initially a coordinate system such that the
x2 y2
equation of E is — +^5=l,a>b>0. The points
Q » (a cos t , bsint) (0 £ t £ 2ir) and Q' = (-a cos t , -b sin t)
lie on E and the tangents to E through Q and Q' are parallel.
We treat the case 0 £ t £ tt/2 as the other cases tt/2 £ t £ tt ,
tt £ t < 3ir/2 , 3tt/2 £ t £ 2tt can be handled by appropriate
reflections .
Let the normals through Q and Q' meet the tangents through
Q' and Q at T and T' respectively. Our first aim is to choose
t so that the area of the rectangle QTO.'T' is maximum. The slope
—b cos t
of the tangent to E at Q' is :—— , and so the equations of
the lines Q'T and QT are respectively b cos tx + a sin t y + ab = 0
2 2
and a sin t x - b cos t y - (a -b ) sin t cos t = 0 . Thus the lengths
|QT| and |Q'T| are given by
lOTl = 2ab lo'Tl - 2(*2-b2)sintcost .
>*' /a2sin2t + b2cos2t ' |4 ' /a2 sin21 + b2 cos21
The area of the rectangle QTQ'T' is clearly
2 2
4ab(a -b ) tan t
2 2.2'
a tan t + b
2 2 b
whose maximum value 2(a -b ) is attained when tan t = —. In this
a
iWl a2_b2
a2 b2 1
V'a*+ba ' A2+bsj
and the slope of the tangent at P is -1 . Rotating the axes
so that R is not a square. Thus P is the point
and the slope of the tangent at P is -1 . Rotat
through it/4 clockwise by means of the orthogonal transformation
(x,y) ■* (X,Y) , where X = ^ (x-
tion of the required ellipse is
(x,y) ■* (X,Y) , where X == (x-y) , Y == (x+y) , we find the equa-
v2 /2

154
SOLUTIONS (88-89)
(X+Y)2 , (X-Y)2 = 1
2 2
2a 2b
83, If four distinct points lie in the plane such that any three
of them can be covered by a disk of unit radius, prove that all four
points may be covered by a disk of unit radius.
Solution: We first prove the following special case of Helly's
theorem: If D. (i = 1,2,3,4) are four disks in the
plane such that any three have non-empty intersection then all four
have non-empty intersection. Choose points W,X,Y,Z in D^iDyiD, ,
D.lDyiD , D .~D,->D, , DjiD nD respectively. We consider two cases
according as one of the points W,X,Y,Z is in or on the (possibly
degenerate) triangle formed by the other three points, or not.
In the first case suppose that Z is in or on triangle WXY.
Then the line segments WX.WY.XY belong to D oD , DuD, , D.^D,
respectively, so that triangle WXY belongs to D. , and thus Z
belongs to D. . Hence Z is a point of DrtD.nD.nl, .
In the second case WXYZ is a quadrilateral whose diagonals
intersect at a point C inside WXYZ . Without loss of generality
we may suppose that C is the intersection of WY and XZ . Now
the line segments WY and XZ belong to D n D, and D.n D
respectively. Thus C is both in D n D, and in D.oD , and so
in D..0 D.lD.oD, .
I 2 3 4
To solve the problem let A,B,C,D be the four given distinct
points. Let G be the centre of the unit disk to which A,B,C
belong. Clearly the distances AG,BG,CG are all less than or equal
to l , and so G belongs in the three unit disks U.,U_,U„ centred
ABC
at A,B,C respectively. Thus any three of the four disks U ,U .U-.IL.
have a non-empty intersection, and so by the first result there is a

SOLUTIONS (88-89)
155
point P in U n U_n U flU_ . The unit disk centred at P contains
A, B, C and D .
89■ Evaluate the sum
00 00
m"l n"l m - n
m*n
Solution: For positive integers m and N with N > m , we have
N , . N
A(m,N) = I 1 . -± I
n"=l m -n n=l
n*m n*m
_1 1
n-m n+m
= -^ (s(N+m) - S(N-m)) - -^ ,
4m
where for r ■= 1,2,3,... we have set
S(
r)= I -j-=£nr + c+ E(r) ,
k-1
c denoting Euler's constant and the error term E(r) satisfying
|E(r)| S^ ,
for some absolute constant A . Then
lim (S(N+m) - S(N-m))
lim
N-i-oo
In
N+m
N-m
+ E(N+m) - E(N-m)
and so
lim A(m,N)
N-i-oo
4m
2 '

156
SOLUTIONS (90-92)
and thus
M
lim £ lim A(m,N)
M + oo m-i N-*-">
- -r lim > 2
4 m
M-*'00 m"=l
*2
90, If n is a positive integer which can be expressed in the
2 2 2 ...
form n = a + b + c , where a,fa,c are positive integers, prove
2k
that, for each positive integer k , n can be expressed in the
2 2 2 ...
form A + B + C , where A,B,C are positive integers.
2 2 2
Solution: We begin fay showing that if m ■= x +y + z , where
2 2 2 2
x,y,z are positive integers, then m » X + Y + Z ,
where X,Y,Z are positive integers. Without loss of generality we
may choose x £ y £ z • Then the required X,Y,Z are given fay
2 2 2
X - x + y - z , Y = 2xz , Z - 2yz .
Letting 2k - 2 (2s+l) , where r i 1 , s £ 0 , we have
2s+l ,Sx2, ,2, ,.2, /Sv2
n = (n a) + (n fa) + (n c) ,
and applying the above argument successively we obtain
2k 2s+K2r v2 _,_ v2 _,_ _2
n » (n ) =X+Y+Z,
where X,Y,Z are positive integers.
91, Let G be the group generated by a and b subject to the
3 5 . .
relations aba = fa and fa = 1 . Prove that G is afaelian.

SOLUTIONS (90-92)
157
Solution: It suffices to show that a and b commute. The relation
aba = b gives b ab ■= b a , and so
b~2ab2 - b-1(b_1ab)b
- b_1(b2a_1)b
- b2(b_1a_1b)
- b2(b_1ab)_1
,2,, 2 -K-l
= b (b a )
2 -2
- b ab L '
giving
4 4
ab ■= b a .
Hence, as b = 1 , we obtain ab » b ab » b(b a)b » ba
92■ Let {a : n = 1,2,3,... } be a sequence of real numbers
n
satisfying 0 < a < 1 for all n and such that J a diverges
t 2 n=1 n
while I a converges. Let f(x) be a function defined on [0,1]
such that f"(x) exists and is bounded on [0,1]. If £ f(a )
converges, prove that J |f(a )[ also converges.
n"l
Solution: Applying the extended mean value theorem to f on the
interval [0,a ], there exists w such that 0 < w < a
L ' n4 n n n
and
2
a
f(a ) - f(0) + a f'(0) + -£f"(w ) .
n n 2 n
00
If I f(a ) converges, then we must have lim f(a ) - 0 , and so
n=l n-»-">

158
SOLUTIONS (93-94)
by continuity f(0) = 0 . Next, as |f"(x)| £M,0£x£l,we have
t
n=l
a2f"(wn)
n n
2
M » 2 ™ a f"^w ^ ™
$t J a , so that both 5! 7—~ and ^
n=l
n=l
n-1
a2f"(w )
00 00
» a2f"(w )
converge. Hence f'(0) J a ■ J f(a ) - J r converges, and
n=l n-1 n=l
00
so as J a diverges, we must have f'(0) = 0 . Thus
n-1
I |*<OI - t
n-1
n-1
a2f"(w )
converges.
93■ Let a>b>c t>e real numbers such that the roots of the cubic
equation
3 2
(93.0) x +ax +bx + c = 0
are all real. Prove that these roots are bounded above by
(2^a2-3b - a)/3 .
Solution: Let p,q,r be the three real roots of (93.0) chosen so
3 2
that p > q £ r . Then, as p + ap + bp + c - 0 , we
have
3 2 2 2
x + ax + bx + c = (x-p)(x + (p+a)x + (p +ap+b))
2 2
The quadratic polynomial x + (p+a)x + (p +ap+b) has q and r as
its two real roots, and hence its discriminant is non-negative, that
is
(93.1) (p+a)2 - 4(p2+ap+b) £ 0 .
Solving (93.1) for p we obtain
p £ (2/a~-3b - a)/3 , which completes the proof.

SOLUTIONS (93-94)
159
94, Let Z » {0,1,2,3,4} denote the finite field with 5
elements. Let a,b,c,d be elements of Z. with a * 0 . Prove that
the number N of distinct solutions in Z, of the cubic equation
2 3
f(x) - a + bx + ex + dx =0
is given by N » 4 - R , where R denotes the rank of the matrix
abed
b c d a
c d a b
d a b c
Solution: Define B to be the Vandermonde matrix
,2
so that
BA =
1
1
1
1
1
2
3
4
1
1"
f(D i_1f(i) r2f(o r3f(D
f(2) 2_1f(2) 2_2f(2) 2~3f(2)
f(3) 3_1f(3) 3~2f(3) 3~3f(3)
f(4) 4_1f(4) 4~2f(4) 4"3f(4)
As a * 0 , the matrix BA has N zero rows, so that rank BA £ 4-N.
Let
[f(r.) r^fCrj,) r^2f(r.) r~3f(r.)] (i - 1,... ,4-N)
be the 4-N non-zero rows of BA , where 1 £ r, < ... < r, .. < 4
1 4-N

160
SOLUTIONS (95-96)
Clearly
f(ri) rl f(ri>
f(r4_N) VNf(r4_N)
r-(3-N)f(ri)
"(3-N) , .
Vn f(r4-N}
f(rx) ... f(r4_N)
1 r
-1
1 r
-1
4-N
-(3-N)
-(3-N)
4-N
* 0
so that the rank of BA is exactly 4-N . Finally, since B is
invertible, we have
R » rank A = rank BA = 4-N ,
that is, N = 4 - R .
95. Prove that
(95.0)
is a rational number.
m,n»l (am)
(m,n)=l
Solution: We notice that
00 n 2 °° 00 OO
I -2} - I TV t I -T2
r=l r > r,s=l (rs) d=l r,s=l (rs)
GCD(r,s)»d
Setting r = dm , s = dn , so that GCD(m,n) = 1 , we obtain

SOLUTIONS (95-96)
161
V " I \ t 1
d»l d m,n=l (mn)
GCD(m,n)=l
that is
4 4 .
2!_ - 2!_ s s - 1
36 90 & ' b 2 •
96i Prove that there does not exist a rational function f(x)
with real coefficients such that
(96.0) fjij.] = p(x) ,
where p(x) is a non-constant polynomial with real coefficients.
Solution: Suppose there exists a rational function f(x) and a
non-constant polynomial p(x) (both with real coefficients)
such that (96.0) holds. As f(x) is the quotient of two polynomials,
there exist complex numbers a (*0) , a , ... , a , b. , ... , b
with a. * b. such that
i J
a(x-a )...(x-a )
(96.1) f(x) - 7—A 7—Ar .
(x-b.)...(x-b )
1 s
Since p(x) is a non-constant polynomial, f(x) can neither be
constant nor a polynomial, and so s £ 1 .
From (96.0) and (96.1), we obtain
(x -a. x-a ; .. . (x -a x-a )
(96.2) a(x+l)sr—j—i 1 g—E L. - p(x) .
(x -b, x-b. ) — (x -b x-b )
11 s s
2
If s-r < 0 , x+1 divides x -a.x-a. for some i , 1 i i $ r , and
2 X x
so (-1) - a.(-l) - a. = 0 , that is, 1 + a. - a. = 0 , which is
clearly impossible, and thus s £ r

162 SOLUTIONS (97-98)
2
Now let x-c be a factor of x -b x-b1 , so that
(96.3) c2 - blC - b1 » 0 .
Clearly c * -1 . As the left side of (96.2) is a polynomial, we
must have r i 1 and x-c | x -a.x-a. , for some i , 1 £ i £ r ,
that is
2
(96.4) c - a.c - a. = 0 .
From (96.3) and (96.4) we obtain
2
c ,
1 C+l 1
which is a contradiction. Hence no such rational function f(x)
exists.
97 • For n a positive integer, set
S(n) = I JL .
*0
Prove that
„ n+1 ,k
s(n) -^It-
2n+i k-1 K
Solution: For n i 2 , we have
-n+1 _n
T5T s(n) - T s<-"
-n+1 n , ^n n-1
"■ . — 11 11 J. -
n+i *=° w" ~ ^ r?i

SOLUTIONS (97-98)
163
and so
2n+l
"n+T
2n+l
n+1
,n+l
n-1
k-0
n-1
k=0
(ttH)(3 nf?)
2k!(n-k)! k!(n-k-l)!
(n+1)!
^n n-1
^r + x^rrrj^Kn-k-DKn-zk-D
-n+1 _n n-1
^T + T^ryrkyk!(n-k)! " oh-d'^-dO
-n+1 _n
^r+X^yyCOini-n-O!)
,n+l
n+1
,n+l ,2 n+1 _k
T3T s(n) " - s(1) " J3 V •
which gives the required result as
,2
S«»-i*V
98^ Let u(x) be a non-trivial solution of the differential
equation
u" + pu » 0 ,
defined on the interval I = [l,00) , where p = p(x) is continuous
on I . Prove that u has only finitely many zeros in any interval
[a,b], 1 £ a < b .
(A zero of u(x) is a point z , 1 £ z < => , with u(z) = 0 ) .

164
SOLUTIONS (98-99)
Solution; Let S denote the set of zeros of u(x) on the interval
[a,b], 1 £ a < b . We will assume that S is infinite
and derive a contradiction. The Bolzano-Weierstrass theorem implies
that S' has at lease oneaccumulation point, say c , in [a,b]. Hence,
there exists either a decreasing or increasing sequence of zeros
{x : n = 1,2,3,... } converging to c . As u is continuous we
n
have u(c) "0 . Applying the mean value theorem to u on the
intervals with end-points x and x .. (n ■= 1,2,3,...) , there
exists a sequence {y : n » 1,2,3,... } with y lying between
x and x ,, and u'(y ) "0 for n » 1,2,3,... . By the contin-
n n+1 n
uity of u' we see that u'(c) » 0 since the y 's converge to c.
Now define
q = q(x) »u+u' , c£x£b.
Then q(c) » 0 and
q'(x) - 2uu'(l-p)
so that
(98.1) q' £ (l+|p|)(u2 + u'2) £ Kq ,
where |p| £ K-l on [c,b]. From (98.1) we deduce that
q(x) i q(c)eK(x"c) , c £ x £ b .
However q(x) sJ 0 so that q(c) » 0 implies that q(x) =0 on
[c,b], that is u(x) = 0 on [c,b].
The proof will be completed by showing that u(x) =0 on [a,c].
We set
v(x) ■ u(a+c-x)
and
r(x) « p(a+c-x)
for a £ x £ c . Then v is a solution of the differential equation

SOLUTIONS (98-99)
165
v" + rv = 0
satisfying v(a) » 0 , v'(a) ■= 0 , By the above argument we deduce
that v(x) = 0 on [a,c], and thus u(x) = 0 on [a,c].
This shows that u(x) = 0 on [a,b], for any b > a , and so
u(x) =0 on [l,00), contrary to assumption.
99, Let P. (j - 0,1,2,...,n-l) be n (£2) equally spaced
points on a circle of unit radius. Evaluate the sum
S(n) « t |P P
0£j<k<n-l J
2
where |PQ| denotes the distance between the points P and Q
Solution: Without loss of generality we may take P. (j = 0,1,2,. .. ,n-l)
to be the point exp(2irji/n) on the unit circle |z| ■ 1
in the complex plane. Then, for 0 £ j < k £ n-1 , we have
|P.P |2 - |exp(2irji/n) - exp(2irki/n)|2
= 2 - exp(2ir(k-j)i/n) - exp(-2ir(k-j) i/n)
and so
n-2 n-1
S(n) 'I 1 (2 - exp(2ir(k-j)i/n) - exp(-2ir(k-j) i/n))
j-0 k-j+1
n;2
- 2 > (n-l-j) - A - A ,
j»0
where
n-2 n-1
A„ ' I I exp(2Tr(k-j)i/n)
j-0 k-j+1
and A denotes the complex conjugate of A . Now

166
SOLUTIONS (99-100)
A - V exp(2iri/n) - exp(2iri(n-j)/n)
n ~ .£_ 1 - exp(2iri/n)
_ (n-l)exp(2iri/n) 1 r /,_••/ \
1 - exi»(2iri/n> ~ 1 - exp<2iri/n> .yxpC^ij/n)
- (n-l)exp(2iri/n) 1
1 - exp(2iri/n) 1 - exp(-2iri/n)
n exp(2irj/n)
1 - exp(2iri/n) '
and so
A + A = -n .
n n
Hence we obtain
giving
S(n) - 2(n-l)2 - 2 ^~2\^~^ + n
S(n) - n2 .
100, Let M be a 3x3 matrix with entries chosen at random from
the finite field Z. - {0,1} . What is the probability that M is
invertible?
Solution: Let M = (a..) , 1 £ i , j £ 3 , so that
M - (a.J , 1 <, i , j
D « det M - auAu + a^A^ + a13A13 ,
where
All * a22a33 a23a32 ' A12 * a23a31 a21a33 ' A13 * a21a32 a22a31"
If (A .A..,A.,) ■ (0,0,0) , the number of corresponding triples
(a .,a.,,a -) such that D - 0 is 8 . For each triple
(A..,A 2,A.,) * (0,0,0) , the number of corresponding triples
(a..,a.,,a ,) with D = 0 is 4 . Hence the number N of matrices

SOLUTIONS (99-100)
167
M with D = 0 is
(100.1) N = 8n + 4(64-n) = 4n + 256 ,
where n is the number of sextuples (a ..a.-.a.-.a, a.-.a, ) with
All = A12 = A13 = ° •
If a,. = a = a - 0 there are 8 triples (a >a32»a33^
with Au - A 2 - A - = 0 . For each triple (a2i>a22,a23^ * (°>°»°)>
there are 2 triples (a,,.a.j.a,,) with A.. = A., = A., = 0 .
Hence we have
n = ix8 + 7x2 - 22 ,
and so N = 344 .
The required probability is
512 - 344 168
512 512
0.328

ABBREVIATIONS
ADM Archiv der Mathematik
AI Analytic Inequalities by D.S. Mitrinovid,
Springer-Verlag (1970)
AMM American Mathematical Monthly
BLMS Bulletin of the London Mathematical Society
CF Convex Figures by I.M. Yaglom and V.G.
Boltyanskii, Holt, Rinehart and Winston (1961)
CM Crux Mathematicorum (formerly Eureka)
CMB Canadian Mathematical Bulletin
CN Course Notes for Mathematics 6 9.112,
Carleton University (1984)
ETN Elementary Theory of Numbers by W. Sierpiriski,
Warsaw (196 4)
GCEA Oxford and Cambridge Schools Examination Board,
General Certificate Examination, Scholarship Level,
Mathematics and Higher Mathematics
GCEB Oxford and Cambridge Schools Examination Board,
General Certificate Examination, Scholarship Level,
Mathematics for Science
HCM Oxford and Cambridge Schools Examination Board,
Higher Certificate Mathematics (Group III)
IMO International Mathematical Olympiad
JUM Journal of Undergraduate Mathematics
MM Mathematics Magazine
NMT Nordisk Matematisk Tidskrift
PMA Principles of Mathematical Analysis by W. Rudin,
McGraw-Hill (196 4)
PSM Publicacions, Secci<5 de Matematiques, Universitat
Autbnoma de Barcelona
TN Theory of Numbers by G.B. Mathews, Chelsea N.Y.
(1961)
WLP William Lowell Putnam Mathematical Competition
169

REFERENCES
Problem
suggested by Problem 3, IMO (1970).
see Problem 382, CM 4 (1978), p.250.
see Problem 304, CM 4 (1978), p.11.
see Problem 138, CM 2 (1976), p.68.
suggested by Problem 161, CM 2 (1976), p.135.
see Problem 19, PMA, p.129.
suggested by Problem 162 CM 2 (1976), p.135.
suggested by Problem 207, CM 3 (1977), p.10.
suggested by Problem 5, GCEA (1954), Paper I.
see Problem 1, HCM (1947), Paper 5.
see Problem 3, IMO (1983).
see Problem 4, GCEA (1952), Paper V.
see Problem 1, GCEA (1952), Paper V.
see Problem 9, GCEA (1955), Paper V.
see Problem 4, GCEA (195 5), Paper V.
suggested by Problem 2, HCM (1949), Paper 7.
based on Problem 6, GCEA (1952), Paper V.
see Problem 2(i), GCEB (1955), paper V.
suggested by Problem 18, TN, p.318.
suggested by Problem J-2, CM 6 (1980), p.145.
suggested by Problem A-3, WLP (1980).
see Problem 3, HCM (1947), Paper 5.
suggested by Problem 2(ii), GCEB (1955), Paper V.
suggested by Problem 916, CM 10 (1984), p.54.
see AMM 76 (1969), pp.1130-1131.
see Problem 528, CM 7 (1981), p.90.
based on Problem 4, GCEA (1953), Paper VII.
see Problem 3, HCM (1948), Paper 7.
suggested by Problem 124, CM 2 (1976), p.119.
based on Problem 8, GCEA (1952), Paper V.
based on Problem 4, HCM (1943), Paper 5.
see Problem 10(ii), GCEA (1954), Paper VI.
based on Problem 5, HCM (1944), Paper 5.
171

42: see Problem 853, CM 9 (1983), p.178.
43: see Problem 3(ii), HCM (1949), Paper 5.
44: see Problem 767, CM 9 (1983), p.281.
45: see Problem 5, GCEA (1955), Paper V.
48: suggested by Problem A-6, WLP (1973).
49: based on Problem B-6, WLP (1974).
50: suggested by Problem 4, HCM (1949), Paper 7.
53: based on MM 44 (1971), pp.9-10.
54: see Problem 602, CM 8 (1982), p.17.
55: based on ETN, pp.124-125.
57: see Problem 7, HCM (1948), Paper 5.
58: suggested by Problem 6, CM 7 (1981), p.236.
59: see Problem 5, HCM (1943), Paper 7.
61: suggested by Problem B-2, WLP (1984).
63: suggested by Problem A-3, WLP (1978).
64: based on Problem A-5, WLP (1984).
65: suggested by Problem B-l, WLP (1976).
66: suggested by Problem 326, CM 4 (1978), p.66.
67: suggested by Problem B-5, WLP (1968).
69: suggested by Problem 3, CM 7 (1981), p.268.
71: see AI, p.221.
72: for a related result see AMM 76 (1969), p.1125.
73: see Problem 4 (Afternoon Session), WLP (1961).
74: suggested by Problem 1, IMO (1975).
75: suggested by Problem B-2, WLP (1971).
76: for more general results see BLMS 11 (1979),
pp.268-272.
77: based on ADM 25 (1974), pp.41-44.
78: see Problem E2032, AMM 75 (1968), p.1124.
79: see AI, pp.346-347.
80: see Problem 21, CN, p.A72*3/4.
81: suggested by Problem a-6, WLP (1972).
82: for related results see JUM 15 (1983), pp.49-52.
83: based on AMM 73 (1966), pp.477-483.
84: based on AMM 57 (1950), pp.26-28.
85: based on AMM 69 (1962), pp. 215-217.
86: suggested by Problem 2 (Morning Session), WLP (1964),
172

see CP, Chapter 2.
based on PSM, 28 (1984), pp.75-80.
based on Problem 83, CMB 7 (1964), p.306.
suggested by Problem 1060, MM 52 (1979), p.46.
see Problem 105, CMB 9 (1966), p.532.
see Problem 171, CMB 15 (1972), p.313.
see Problem 178, CMB 15 (1972), pp.614-615.
see NMT 29 (1947), pp.97-103.
suggested by Problem 1104, MM 53 (1980), p.244.
173

A CATALOG OF SELECTED
DOVER BOOKS
IN SCIENCE AND MATHEMATICS
LP

A CATALOG OF SELECTED
DOVER BOOKS
IN SCIENCE AND MATHEMATICS
QUALITATIVE THEORY OF DIFFERENTIAL EQUATIONS, V. V. Nemytskii
and V.V. Stepanov. Classic graduate-level text by two prominent Soviet
mathematicians covers classical differential equations as well as topological dynamics
and ergodic theory. Bibliographies. 523pp. b% x m. 65954-2 Pa. $14.95
MATRICES AND LINEAR ALGEBRA, Hans Schneider and George Phillip
Barker. Basic textbook covers theory of matrices and its applications to systems of
linear equations and related topics such as determinants, eigenvalues and
differential equations. Numerous exercises. 432pp. b% x 8& 66014-1 Pa. $10.95
QUANTUM THEORY, David Bohm. This advanced undergraduate-level text
presents the quantum theory in terms of qualitative and imaginative concepts,
followed by specific applications worked out in mathematical detail. Preface.
Index. 655pp. b% x m. 65969-0 Pa. $14.95
ATOMIC PHYSICS (8th edition), Max Born. Nobel laureate's lucid treatment of
kinetic theory of gases, elementary particles, nuclear atom, wave-corpuscles, atomic
structure and spectral lines, much more. Over 40 appendices, bibliography. 495pp.
b%*m. 65984-4 Pa. $12.95
ELECTRONIC STRUCTURE AND THE PROPERTIES OF SOLIDS: The
Physics of the Chemical Bond, Walter A. Harrison. Innovative text offers basic
understanding of the electronic structure of covalent and ionic solids, simple
metals, transition metals and their compounds. Problems. 1980 edition. 582pp.
6X5x9"/,. 66021-4 Pa. $16.95
BOUNDARY VALUE PROBLEMS OF HEAT CONDUCTION, M. Necati
Ozisik. Systematic, comprehensive treatment of modern mathematical methods of
solving problems in heat conduction and diffusion. Numerous examples and
problems. Selected references. Appendices. 505pp. b% x 8!4. 65990-9 Pa. $12.95
A SHORT HISTORY OF CHEMISTRY (3rd edition), J.R. Partington. Classic
exposition explores origins of chemistry, alchemy, early medical chemistry, nature
of atmosphere, theory of valency, laws and structure of atomic theory, much more.
428pp. b% x m. (Available in U.S. only) 65977-1 Pa. $11.95
A HISTORY OF ASTRONOMY, A. Pannekoek. Well-balanced, carefully
reasoned study covers such topics as Ptolemaic theory, work of Copernicus, Kepler,
Newton, Eddington's work on stars, much more. Illustrated. References. 521pp.
5Xx8K. 65994-1 Pa. $12.95
PRINCIPLES OF METEOROLOGICAL ANALYSIS, Walter J. Saucier. Highly
respected, abundantly illustrated classic reviews atmospheric variables,
hydrostatics, static stability, various analyses (scalar, cross-section, isobaric, isentropic,
more). For intermediate meteorology students. 454pp. 6% x 9!4. 65979-8 Pa. $14.95

CATALOG OF DOVER BOOKS
ASYMPTOTIC METHODS IN ANALYSIS, N.G. de Bruijn. An inexpensive,
comprehensive guide to asymptotic methods—the pioneering work that teaches by
explaining worked examples in detail. Index. 224pp. 5% x 8!4. 64221-6 Pa. $7.95
OPTICAL RESONANCE AND TWO-LEVEL ATOMS, L. Allen and J.H. Eberly.
Clear, comprehensive introduction to basic principles behind all quantum optical
resonance phenomena. 53 illustrations. Preface. Index. 256pp. 5% x 8H.
65533-4 Pa. $8.95
COMPLEX VARIABLES, Francis J. Flanigan. Unusual approach, delaying
complex algebra till harmonic functions have been analyzed from real variable
viewpoint. Includes problems with answers. 364pp. 55* x 8H. 61388-7 Pa. $9.95
ATOMIC SPECTRA AND ATOMIC STRUCTURE, Gerhard Herzberg. One of
best introductions; especially for specialist in other fields. Treatment is physical
rather than mathematical. 80 illustrations. 257pp. 5% x 8!4. 60115-3 Pa. $6.95
APPLIED COMPLEX VARIABLES, John W. Dettman. Step-by-step coverage of
fundamentals of analytic function theory—plus lucid exposition of five important
applications: Potential Theory; Ordinary Differential Equations; Fourier
Transforms; Laplace Transforms; Asymptotic Expansions. 66 figures. Exercises at
chapter ends. 512pp. 55s x m. 64670-X Pa. $12.95
ULTRASONIC ABSORPTION: An Introduction to the Theory of Sound
Absorption and Dispersion in Gases, Liquids and Solids, A.B. Bhatia. Standard reference
in the field provides a clear, systematically organized introductory review of
fundamental concepts for advanced graduate students, research workers. Numerous
diagrams. Bibliography. 440pp. 55s x 8!4. 64917-2 Pa. $11.95
UNBOUNDED LINEAR OPERATORS: Theory and Applications, Seymour
Goldberg. Classic presents systematic treatment of the theory of unbounded linear
operators in normed linear spaces with applications to differential equations.
Bibliography. 199pp. b% x $%. 64830-3 Pa. $7.95
LIGHT SCATTERING BY SMALL PARTICLES, H.C. van de Hulst.
Comprehensive treatment including full range of useful approximation methods for
researchers in chemistry, meteorology and astronomy. 44 illustrations. 470pp.
5Xx8Si 64228-3 Pa. $11.95
CONFORMAL MAPPING ON RIEMANN SURFACES, Harvey Cohn. Lucid,
insightful book presents ideal coverage of subject. 334 exercises make book perfect
for self-study. 55 figures. 352pp. 5S x 8'/«. 64025-6 Pa. $11.95
OPTICKS, Sir Isaac Newton. Newton's own experiments with spectroscopy,
colors, lenses, reflection, refraction, etc., in language the layman can follow.
Foreword by Albert Einstein. 532pp. b% x 8^. 60205-2 Pa. $ 11.95
GENERALIZED INTEGRAL TRANSFORMATIONS, A.H. Zemanian.
Graduate-level study of recent generalizations of the Laplace, Mellin, Hankel, K.
Weierstrass, convolution and other simple transformations. Bibliography. 320pp.
b% x 8!4. 65375-7 Pa. $8.95

CATALOG OF DOVER BOOKS
SPECIAL FUNCTIONS, N.N. Lebedev. Translated by Richard Silverman.
Famous Russian work treating more important special functions, with applications
to specific problems of physics and engineering. 38 figures. 308pp. 5% x 8!4.
60624-4 Pa. $9.95
OBSERVATIONAL ASTRONOMY FOR AMATEURS, J.B. Sidgwick. Mine of
useful data for observation of sun, moon, planets, asteroids, aurorae, meteors,
comets, variables, binaries, etc. 39 illustrations. 384pp. 5X * S'i. (Available in U.S.
only) 24033-9 Pa. $8.95
INTEGRAL EQUATIONS, F.G. Tricomi. Authoritative, well-written treatment
of extremely useful mathematical tool with wide applications. Volterra Equations,
Fredholm Equations, much more. Advanced undergraduate to graduate level.
Exercises. Bibliography. 238pp. b% x m. 64828-1 Pa. $8.95
POPULAR LECTURES ON MATHEMATICAL LOGIC, Hao Wang. Noted
logician's lucid treatment of historical developments, set theory, model theory,
recursion theory and constructivism, proof theory, more. 3 appendixes.
Bibliography. 1981 edition, ix + 283pp. 5% x $'A. 67632-3 Pa. $8.95
MODERN NONLINEAR EQUATIONS, Thomas L. Saaty. Emphasizes practical
solution of problems; covers seven types of equations. ". . . a welcome contribution
to the existing literature. . . ."—Math Reviews. 490pp. 5% x 8!4. 64232-1 Pa. $11.95
FUNDAMENTALS OF ASTRODYNAMICS, Roger Bate et al. Modern approach
developed by U.S. Air Force Academy. Designed as a first course. Problems,
exercises. Numerous illustrations. 455pp. 5% x 8!4. 60061-0 Pa. $9.95
INTRODUCTION TO LINEAR ALGEBRA AND DIFFERENTIAL
EQUATIONS, John W. Dettman. Excellent text covers complex numbers, determinants,
orthonormal bases, Laplace transforms, much more. Exercises with solutions.
Undergraduate level. 416pp. b% x $'A. 65191-6 Pa. $10.95
INCOMPRESSIBLE AERODYNAMICS, edited by Bryan Thwaites. Covers
theoretical and experimental treatment of the uniform flow of air and viscous fluids past
two-dimensional aerofoils and three-dimensional wings; many other topics. 654pp.
5y8x8^. 65465-6 Pa. $16.95
INTRODUCTION TO DIFFERENCE EQUATIONS, Samuel Goldberg.
Exceptionally clear exposition of important discipline with applications to sociology,
psychology, economics. Many illustrative examples; over 250 problems. 260pp.
b% x m. 65084-7 Pa. $8.95
LAMINAR BOUNDARY LAYERS, edited by L. Rosenhead. Engineering classic
covers steady boundary layers in two- and three-dimensional flow, unsteady
boundary layers, stability, observational techniques, much more. 708pp. 5X x 8!4.
65646-2 Pa. $18.95
LECTURES ON CLASSICAL DIFFERENTIAL GEOMETRY, Second Edition,
Dirk J. Struik. Excellent brief introduction covers curves, theory of surfaces,
fundamental equations, geometry on a surface, conformal mapping, other topics.
Problems. 240pp. 5% x W. 65609-8 Pa. $8.95

CATALOG OF DOVER BOOKS
CHALLENGING MATHEMATICAL PROBLEMS WITH ELEMENTARY
SOLUTIONS, A.M. Yaglom and I.M. Yaglom. Over 170 challenging problems on
probability theory, combinatorial analysis, points and lines, topology, convex
polygons, many other topics. Solutions. Total of 445pp. 5% x 8!4. Two-vol. set.
Vol. I 65536-9 Pa. $7.95
Vol. II 65537-7 Pa. $7.95
FIFTY CHALLENGING PROBLEMS IN PROBABILITY WITH
SOLUTIONS, Frederick Mosteller. Remarkable puzzlers, graded in difficulty, illustrate
elementary and advanced aspects of probability. Detailed solutions. 88pp. 5% x 8!4.
65355-2 Pa. $4.95
EXPERIMENTS IN TOPOLOGY, Stephen Barr. Classic, lively explanation of
one of the byways of mathematics. Klein bottles, Moebius strips, projective planes,
map coloring, problem of the Koenigsberg bridges, much more, described with
clarity and wit. 43 figures. 210pp. b% x 8!i. 25933-1 Pa. $6.95
RELATIVITY IN ILLUSTRATIONS, Jacob T. Schwartz. Clear nontechnical
treatment makes relativity more accessible than ever before. Over 60 drawings
illustrate concepts more clearly than text alone. Only high school geometry needed.
Bibliography. 128pp. 6% x 9!4. 25965-X Pa. $7.95
AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS, Earl
A. Coddington. A thorough and systematic first course in elementary differential
equations for undergraduates in mathematics and science, with many exercises and
problems (with answers). Index. 304pp. 5% x 8Si 65942-9 Pa. $8.95
FOURIER SERIES AND ORTHOGONAL FUNCTIONS, Harry F. Davis. An
incisive text combining theory and practical example to introduce Fourier series,
orthogonal functions and applications of the Fourier method to boundary-value
problems. 570 exercises. Answers and notes. 416pp. b% x m. 65973-9 Pa. $11.95
AN INTRODUCTION TO ALGEBRAIC STRUCTURES, Joseph Landin.
Superb self-contained text covers "abstract algebra": sets and numbers, theory of
groups, theory of rings, much more. Numerous well-chosen examples, exercises.
247pp. b% x W. 65940-2 Pa. $8.95
Prices subject to change without notice.
Available at your book dealer or write for free Mathematics and Science Catalog to Dept. GI,
Dover Publications, Inc., 31 East 2nd St., Mineola, N.Y. 11501. Dover publishes more than 175
books each year on science, elementary and advanced mathematics, biology, music, art,
literature, history, social sciences and other areas.