THE RED BOOK
OF MATHEMATICAL
PROBLEMS
KENNETH S. "VIILIAMS
KENNETH HARDY
Carleton University, Ottawa
Dover Publications, Inc.
Mineola, New York

PREFACE TO THE FIRST EDITION
Copyright
Copyright @ 1988 by Integer Press.
, All rights re.served under Pan AmeriC'dn and International Copy-
nght Gmventlons.
Published in Canad.a by Genual Publishing Company, LId.. 3f!
Lesmlll Road, Don :\fIlh, Toronto. Ontario.
II has become the fashion for some authors to include li.eraryquotations in
Ihf'ilmathematical texts, presumably with the aim of connecting mathematics
illl<I lilt' humanities. The preface of The Green Book*' of 100 practice problems
,." undergraduate mathematics competitions hinted at connections between
jllllhh-IIJ-solving and all the traditional elements of a fairy tale: mystery,
"';11.-11. discovery, and finally resolution. Althouh The Red Book may s('em 10
1.,1'" polIucal overtones, rest assured, de3r reader, that the quot:Jtions (labelled
IVI,.. x, Pushkin and Trotsky, just for fun) are merely an in'ipiration for your
I' ""IH'V through the enchantl:tl realms of mathematics.
Bibliographical Note
This. Do."er edition, first publishr.d in 1996, is a slightly corrcl'led
republIcauon of he work originally published hy Integer Press,
Ottawa, Canada, In 1988 under the title The Red Book: 100 Practice
Problems for Undergraduate Mathematics Competitions. A section
of the original page 97 has been deleted and all subsequent copy
repaged thereafter.
J'ht! Red Book contains 100 problems for undergraduate swdents training
fill mathematics competitions, particularly the William Lowell Putnam
Mathematical Competition. Along with the problems come useful hints. and
. tllllpicte solutions. The book will also be useful to anyone interested in the
posing and solving of mathematic.al problems at the undergraduate level.
Many of the problems were suggested by ideas originating in a variety of
ources, including Crux Mathematicorum, Mathematics Magazine and the
American Mathematical Monthly, as well as various mathemaLic;. competi.
tiolls. Where possible, acknowledgement to known sources is given at the end
of t he book.
Libraty of Congmss Cataloging-in-Pu.blication Data
Williams, Kenneth S,
The red book of mathematical problems! Kenneth S. Williams,
Kenneth Hardy.
p, em.
. '1\ slightly corncted I'epublication of te work originally pub-
hshed b}' Imeger Press, Ouawa, Canada, 111 1988 under the title:
The. red book: 100 practice problems for undergraduat( mathe-
matIc;. competitions"-Tp. verso.
Includes bibliographical references.
ISBN 0-486-69413-1 (pbk.)
l. Mathemalics-Problems, exercises, elc. I. Ilardy, Kenneth.
n. Title.
QA43.W55 1996
510'. 76-dc20 96-43820
CIP
Once again, we would be interested in your reaction to The Red Book, and
illvitecomments, alternate solutions, and even corrections. We make no claim
thill the solutions are the "best possible" solut.ions, but we trust that. you will
find them elegant enough, and that The Red Book will be a practical tool in
training undergraduate competitors.
We wish to thank our typeseuer and our literary adviser at Integer Press for
their valuable assistanl"C in this project.
Kmneth S. Williams and Kenneth Hardy
Ottawa, Canada
May, 1988
-To be reprinted by Dover Publications in 1997.
Manufactured in the United States of America
Dover Publications, Inc., 31 East 2nd Street, Mineola, KY. 11501

CONTENTS
I-'age
Notation
The Problems
The Hints
The Solutions
The Sources
]X

f
:"{
1 '7"

[x]
lnx
exp x
q,(n)
GCD(a,b)
()
()
deg U(x»
Y.. t
T(n)
1'( x)
dct A
z
Q,R,C
NOTATION
denotes the greatest integer  $, where $ is a real number.
denotes the natural logarithm of x.
denotes the exponential function e"'.
denotes Euler's totient function defined for any naLllra.1 num-
bcr n.
denotes the greatest common divisor of the integers a a,nd b.
denotes the binomial coefficient n!fk! (n - k)!, where nand
k are non-negative integers (the symbol having value zero
when n < k) .
denotes Legendre's symbol which has value +1 (resp. -1)
if the integer n is a quadratic residue (resp. nonresidue)
modulo the odd prime p .
denotes the degree of the polynomial f( x) .
denotp.s the transpose of the row vector Y.. .
denotes the number of distinct prime divisors of the positive
integer n.
denotes the derivative of the function f( x) with respect to
x.
denotes the determinant of the square matrix A.
denotes the domain of rational integers.
denote the fields of rational, real, complex numbers respec-
tively.

THE PROBLEMS
Mankind always sets itself only sudt problems t1. ii, mn _-ob.'{,; .,. it
will always bt: found lhal tlu lask ilsdf fI1'iM'S only wlu 11 III.( l1W,-
terial conditions for its solution already exist 01' are at least in the
proc.ess of formation.
Earl Marx (1818-1883)
1. Let p denote an odd prime and set w =: exp(27ri/p). Evaluate the
product
(1.0) E(p) (W T1 + w T ' + . . . + d(P-l)!. )(w nt + w n . + . . . + W"{P-l)!2),
where 1'1,.. . ,r(p-I)/2 denote the (p - 1 )/2 quadratic residues modulo p and
n},.. ., n(p-l)/2 denote the (p 1)/2 quadratic nOIlfesidues modulo p.
2. Let k denote a positive integer. Determine the number N(k) of
triples (x,y,z) of integers satisfying
{ Ixl s: k, Iyl s: k, Izi s: k,
Ix yl s: k, Iy zl s: k, Iz xl s: k.
(2.0)
3. Let p 1 (mod 4) be prime. It is known that there exists a unique
integer w == w(p) such that
w 2 == -1 (mod p), 0 < w < p/2.

2
PROBLEMS
(For example, -w(5) 2,w(13) = 5.) Prove that there exist integers a,b,c,d
with ad - be 1 such that
pX 2 + 2wXY + (w 2 + 1) 1'2 == (aX + by)2 + (eX + dY)2.
P
(For example, when 1) =- 5 we have
5.\'2+ 4 xy+y2 ,y21-(2,y+n 2 ,
and wlwn I' ,= I: WI' hay"
13X 2 + lOX}' + 2y2 (3)( + Y)2 + (2X + Y)2.)
4. Let d,,(n), T = 0,1,2,3, denote the number of positive integral
divisors of n which are of the form 4k + T. Let m denote a positive integer.
Prove that
(4.0)
f(d1(n) - ds(n» = f( -lY [ 2 ': 1 ] .
n=l 3=0 J
5.
Prove that the equation
(5.0)
y2 = x 3 + 23
has no solutions in integers x and y.
6. Let f(x, y) -= a.x 2 +2bxy+cy2 be a positive-definite quadratidorm.
Prove that
(flxl,ydf(x2,Y2))l/2f(xl - X2,Yl - Y2)
(6.0)
 (ac- b 2 )(XI1/'l X2Yl?,
PROBLEMS
3
for all real numbers Xl,X2,YI,Y2.
7. Let R, S, T be three real numbers, not all the same. Give a condi-
tion whkh is satisfied by one and only one of the three triples
(7.0)
{ (R,S,1'),
(1',-8 + 21',R .') +1'),
(R 5' + T,2R 8,R).
8. Let ax 2 + bxy + cy2 and Ax2 + Bxy + Cy2 be two positive-definite
quadratic forms, which are not proportional. Prove that the form
(8.0) (aB-bA)x 2 +2(aC cA)xy+(bC cB)y2
is indefinite.
9.
Evaluate the limit
I
(9.0)
L
. n n 2 k
lim - ,,- .
n_OO 2 n L..i k
k=l
I
Prove that there does not exist a constant c  1 such that
10.
(10.0)
n C 4>(n)  mct/>(m),
for all positive integers nand m satisfying n  m.
I
11. Let D be a squarefree integer greater than 1 for which there exist
positive integers AI,A2,Bl,B2 such that
{ D= Ai+Bi =A+B,
(11.0) (AhBd :f:. (A 2 ,B 2 ).

4
PROBU'}MS
PROBLEMS
!)
Prove that neither
2D(D + il]A 2 + B]B 2 )
nor
(15.0)
2D(D + A]ft 2 RIBz)
is the sqllare of an integ;er.
12. [",t Q and R deno!." the fields of rational and re<!l numhers
rl?f>pcctivf'ly, Lf't K ;md L 1)(' 1.11f' smallf'st SII bfif'lds of n. which contain both
Q a.nd the real numbers
(16.0)
J lf)85 + 31 Y1985 and J 3970+ 64V1985 ,
respectively. Prove that K L.
13.
Let k a.nd 1 be positive integers such that
15.
Evaluate the integral
I follnXlIl(1 x)dx.
16. Solve the nXUfn'nce r(']ation
E (;)a(k)
'II
11 + 1 '
n
1,2,. '.
17.
Let n alld k be positive integers. Let p be a prime such tha.t
p> (n2+n+k?+k.
GCD(k,5) GCD(l,5) GCD(k,l) 1
(17.0)
Prove that the sequence
n 2 , n 2 + 1, n 2 + 2, . . . , n 2 + 1 ,
and
where 1
that
_k 2 + 3kl- t
p2
,
where GCD(}5)
1.
Prove that tllt pair of equations
(13.0)
{ k x 2 + y2,
1 = x 2 + 2xy + 2y2,
has exa.ctly two solutions in integers x and y.
(n2+n+k)2
n 2 + k, conta.ins a pair of integers lm., m. + k) such
( : ) ( m; k) = 1.
18.
Let
14.
Let rand B be non-zero integers. Prove that the equation
Docs the infinite series 2::;:'=0 an converge, and if so, what is its sum?
an
1 1
-+-
4n + 1 4n + 3
n=O,l,... .
(14.0)
(r 2 _ s2)x 2 4rsxy (r 2 _ B2)y2 1
has no solutions in integers x a,nd y.
1
2n+2 '
19.
Let a],..., am be m ( 2) real numbers. Set
An 0,1 + 0,2 + . . . + an, n = 1,2,.., , m. .

6
PROBLE:M.8
PROBLEMS
7
Prove that
Let P be a real-valued function of the n(n - 1)/2 variables Xij such that the
inequality
(19.0) f ( n r  12fa;.
n=2 n=1
(no)
n
F(:CIl,:f12,..' ,:1'" 111)  2::x%
1.:=1
s -. 'n ()
- L-. ( 2n-l )
k=O k
holds for all Xl,"" Xn.
Prove lItal. equality eaIJllot hold ill (23.0) jf 7::k=1 :1:).. f O.
20. Evaluate the sum
for all positive integers n.
24. Let. Uh''',U", he 11! (;2: 1) real 11111111)('1'" whidt an' "uelr that
2':::.;'=1 an f O. Prove tILe inequality
21. Let a and b be coprime positive integers. For k a positive integer,
let N(k) denote the number of integral solutions to t,he equation
(24.0)
( t na. ) / ( t 0", ) 2 > 2:m
1't_1 'n_l
(21.0)
ax + by ;;::; k, x 2': 0, y 2': 0 .
Evaluate the limit
L = lim N(k) .
k-+oo k
25. Prove that there exist infinitely many positive integers which are
not expressible in the form n 2 + p, where 'II is a positive integer a.nd p is a
prime.
22.
Let a, d and 'I' be positive integers. For k = 0,1,. .. set
26.
Evaluate the infinite series
(22.0)
Uk
1
Uk( a d T) ;;::; .
" (a+kd)(a+(k+1)d)...(a+(k+r)d)
s = farctan ( :2 )'
n=1
Evaluate the sum
"
s 2:: 1Lk ,
k=O
27. Let Pl"",Pn denote 'II (2': 1) distinct integers and let fn(x) be
the polynomial of degree 'II given by
where n is a positive integer.
fn(x):::: (x - pd(x - 112)...(x - Pn).
23.
Let Xl,''',X n be 'II (> 1) real numbers. Set
Xij Xi - Xj (1  i < j  n).
Prove that the polynomial
gn(X) (fn(x))2 + 1

8
PROBLEMS
PROBLEMS
9
cannot be expressed as the product of two non-constant polynomials with
integral coefficients.
33. Let I denote the closed interval [a,b], a < b. Two functions
f( x), g( x) are said to be completely different on I if f( x) =I g( x) for all x in I.
Let q(x) and rex) be functions defined on I such that the differentia] equation
28. Two people, A and B, playa game in which the probability that
A wins is p, t.he probability that B wins is q, and the probability of a draw is
r. At the beginning, A ha.<; m dollars and B has 11 dollars. At the end of each
game th(' winn('r takes a dollar from the loser. If .1 and 11 agree to play until
one of them Inses aU his/her money, what is the probabilty of A winning all
the money'!
dx
y2 + q(x)y + rex)
ha.<; three solutions Yl (:1'), Y2t.:r), Y3(X) which ar(> pa.irwise completely different
on .T. If z(x) is a fourth solution such that the pairs offunctions z(:r.), y.(x)
a.re completely different for ii, 2, :1, prove that then' exists a wl1st.ant
K (cf 0,1) such that
29. Let f(x) be a monic polynomial of degree n  1 with complex co-
efficients. Let x}, . .. ,X n denote the n complex roots of f( x). The discriminant
DU) of the polynomial f( x) is the complex number
(33.0)
Yl(I(Y:/. Y3) + (1- J()Y2Y3
(I( 1)Yl+(Y2-J(Y3)
(29.0)
DU) = n (Xi - Xj)2.
l$i<j$n
34. Let an, n = 2,3,..., denote t.he number of ways tlw product
b 1 b 2 ... b ll can be bracketed so that only two of the bi are multiplied together
at a.ny one time. For e.xample, a2 = 1 since b 1 b 2 can only be bracketed
as (hb 2 ), whereas a3 = "l. as blb2b; can be bracketed in two ways, namely,
(b l (b 2 b 3 )) and «b 1 b 2 )b 3 ). Obtain a formula for an.
Express the discriminant of f(x 2 ) in terms of DU) .
30. Prove that for each positive integer n there exists a circle in t.he
xy-plane which contains exactly '/1. lattice points.
31. t '(). be a given non-negative integer. Determine the number
S( n) of solut.ions of the equation
(35.0)
35. Evaluate the limit
L = lim  r tan(ysinx) dx.
!I_ O Y 10
(31.0)
x + 2y + 2z = n
in non-negative integers x, y, z.
36. Let £ be a real number with 0 < £ < 1. Prove that there are
infinitely many integers n for w.hich
32. Let n be a fixed integer  2. Determine all functions f(x), which
ItffJ bounded for 0 < x < a, and which satisfy the functional equation
(36.0)
cosn 2: 1- £ .
(:12.0)
1 ( ( X ) ( x+a ) ( x+(n-1)a ))
f(x) = n 2 f ;; + f --;;- +... + f n .

10
PROBLEMS
PROBLEMS
11
37. Determine a.1I the functions f, which a.re everywhere differentiable
a.nd satisfy
41. A,B,C, D are four points lying on a circle such that ABCD i5 a
convex quadrilateral. Determine a formula for the radius ofthe circle in terms
of a = 1.4.BI, b = IBGI, c ,CDI and d = IDAI.
( X+Y )
(37.0) f(:I:) + fey) = f 1- :r.y
for aU real x a.nd y with X.1/ f L
42. Let ABCD be a convex quadl'ila.l.eral. Let P be the point outside
ABCD slIch that j.4.PI IPHI a,nd LAPB 90°. The points Q,1l,8 are
similarly defmed. Prove that the lines P Rand Q,<; an oJ e!luallcngth and
perpendi cular.
38. A point X is chosen inside or on a circle. Two perpendicular
chords .4C and BD of the circle are drawn through X. (In I;he case when X
is on the circle, the degenerate case, when one ('.hord is a diameter and the
other is reduced to a point, is allowed.) Find the greatest and least values
which the sum S = IACI + IBDI can take for all possible choices ofthe point
X.
43. Determine polynomials p(x,y,z,w) and q(x,y,z,w) with real
coefficients such that
(43.0)
(xy + z + w? - (x 2 - 2z)(y2 - 2w)
(p(x, y, z, w»)2 - ($2 _ 2z)(q(x, y,z, w))2 .
39. For n
An == {
1,2,. .. define the set An by
{0,2,4,6,8,...}, ifn
{0,3,6,...,3(n- 1)f2}, if n
o (mod 2) ,
1 (mod 2) .
nQ1 C01 An+k) nQ1 CQ An+k) ?
44. Let C denote the field of complex numbers. Let f ; C --+ C be a
function satisfying
Is it true that
( 44.0)
{ f(O) == 0,
If(z) f(w)1 = Iz
wi,
40. A sequence of repeated independent trials is performed. Each
trial has probability p of being successful and probability q ] - p of failing.
The trials are continued until an uninterrupted sequence of n successes is
obtained. The variable X denotes the number of trials required to achieve
this goal. If Pk Prob(X == k), determine the probability generating function
P( x) defined by
for all z in C and 1lJ = 0,1, i. Prove that
fez) = f(l)z or f(l)z,
where If(l)1 1.
( 40.0)
00
P(x) L:l)kXk.
k=O
45. If $ and y are rational numbers such that
(45.0) tan 1!'X = y,

12
PROBLEMS
PROBLEMS
13
prove that x = k/1 for some integer k not congruent to 2 (mod :1).
has a solution in integers Xl, X2,..., x"' not all zero, satisfying
46. Let I' be a point inside the triangle ABC. Let AI' mcct BC at
0, Hi' llleet C A at ,f;, and CJ> meet :1.B at P. prove that
IXjl ::; [(2na) nn' ], 1::; j ::; n .
( 46.0)
IPAi IPB: IPBIIPCI IPCIIPAI
tpDllPHI + j PEI TpFI + iPFl lPDI 2': 12.
19. Liouville proved tIwl; if
1:5: 1 < n, GCD(l,n) 1.
J f!:r)t '1(.r)d.r.
is an elementary fUllction, where f(x) and g(x) are rational functions with
degree of g(x) > 0, then
47.
Let I a.nd n be positive integers such that
Define the integer J;; uniquely by
l::;k<n, kl=-1(modn).
Let M be the k X 1 matrix whose (i,j)-th entry is
J f(x)(!I(x)dx h(x)e 9 (x) ,
where h(x) is a rational function. Use Liouville's result to prove that
( i 1 )l + j .
J e- x2 d:r
Let N be the k x I matrix formed by taking the columns of M hI reverse order
and writing the entries as the rows of N. What is the relationship between
the (i,j)-th entry of M and the (i,j)-th entry of N modulo n?
is not an elementary function.
48. Let m and n be integers SUdl that 1 ::; m < n. Let a'ij, i =
1,2,.. ., fn.; j = 1,2, . . . , n, be mnintegers which are not all zero, and set
50.
The sequence Xo, Xl>' .. is defined by the conditions
a max laijl.
l<'<m
l$.1$n
(50.0)
Xo
0,
Xl = 1,
Xn+l =
X" + nX n -1
n+1
n 2': 1.
Determine
Prove that the system of equations
L = lim X n .
n-+oo
(48.0)
{ anXl + a12 x 2 + ... + alnX n
0,21 Xl + a22 X 2 + ... + a2nXn
0,
=0,
51.
Prove that the only integers N 2': 3 with the following property:
amI Xl + a m 2 X 2 + . . . + amnX n
0,
(,) 1.0)
if 1 < k::; Nand GCD(k,N) = 1 then J.. is prime,

14
PROBLEMS
PROBLEMS
15
are
Prove that
N 3,4,6,8,12,18,24,30.
IZjl ::; 1 + A, j = 1,2,. .., n.
52.
Find the sum of the infinite series
56.
If m. and n are positive integers with m odd, determine
s
1
1 1
.1 + 6'
1 1
9' + 11
1
'j- .. .
1--1
it = GCD('.?!" - 1, 2'n T ]) .
53. Semicircles are drawn externally to the sides of a. given triangle.
The lengths of the common tangents to these semicircles are 1, m, and n.
Relate the quantity
57. If f( x) is a polynomial of degree 2m + 1 with integral coefficients
for which there are 2m. r 1 integers k 1 ,. . . ,k 2m +1 such that
1m mn nl
-+ +-
n 1 m
to the lengths of the sides of the triangle.
(57.0)
f(k 1 ) = .., = f(k 2m + 1 ) = 1,
prove that f( x) is not the product of two non-constant polynOInials with
integral coefficients.
54.
Determine all the functions H : R-I -> R having the properties
58.
that
Prove that there do not exist integers a, b, c, d (not all zero) such
(i)
( ii)
( iii)
(iv)
H(l,O,O,l) 1,
H(>'a,b,Ac,d) >'H(a,b,c,d),
Hla,b,c,d) = -H(b,a,d,c),
H(a + e,b,c or J,d) H(a,b,c,d) + H(e,b,J, d),
(58.0)
a 2 + 5b 2 - 2c 2 - 2cd - 3d 2 = 0 .
where a,b,c,d,e,J,A are real numbers.
59. Prove that there exist infinitely many positive integers which are
not representable as sums offewer than ten squares of odd natural numbers.
55. Let Z1,..., Zn be the complex roots of the equation
zn +alz n - l +.... + an 0,
60.
Evaluate the integral
wllCre 0,1,... ,an are n (?:: 1) complex numbers. Set
(60.0)
I(k)
L oo sin kx cosh x
dx
ox'
A = max lakl .
1 :$ksn
where k is a positive integer.

16
PROBLEMS
61.
Prove that
1 ( 2n )
n+ 1 n
is an integcr for n 1,2,3,. .. ,
62.
Find the sum of the infinite scries
.'i
"" 2 n
)'--
 a" + 1 .
whcff a > 1.
63.
Let k be an integer. Prove that the formal power series
VI + kx == 1 + alx + a2x2 + ...
has integral coefficients if and only if k 0 (mod 4).
64. Let m be a positive integer. Evaluate the determinant of the
m X m matrix Urn whose (i,j)-th entry is GCD(i,j).
65. Let land m be positive integers with 1 odd and for which there
arc integers x and y with
{
x2 + y2,
x 2 + 8xy + 17y2.
Prove that there do not exist integers u and v with
(65.0)
{ 1 == u 2 + v 2 ,
m == 5u 2 + 16uv + 13v 2 .
r
PROBLEMS
17
66.
Let
an
1
1 1
-+ -...+
2 3
n
-ln2.
Provc that
an cOIlverges and determine its sum.
67.
Let .4 :::- {a.i I (J <: i :S; G} he a. S(qllence of sevell iutep;ers satisfying
o 'CC Ill) :::: (1.1 :S; _ . . :S; (Iii  6.
For i
0,1,...,61et
Ni numbcr of aj (0::: j :S; 6) S]Jch that aj .
Determine a.U sequences A stich that
(67.0)
Ni , ==a6-i i==0,1,...,6.
68. Let G be a finite group with identity e. If G contains elements g
and h such that
(68.0) g5 = e, ghg- l h 2 ,
determine the order of h.
69.
Let a and b be positive integers such that
GCD(a,b) == 1, a:;t b (mod 2) .
If the set S has the following two properties:
(i) a,b E S,
(ii) x,y,z E S implies x + y+ z E S,

18
PROBLEMS
PROBLEMS
19
prove that every integer > 2ab belongs to S.
70. Prove that every integer can be expressed in the form x 2 +y2_5z2,
where x, y, z are integers.
75.
Evaluate the sum of the infinite series
00
1
mn( m + n) .
S
I:
71. [<;.valuate the SUIIl of the iuIiuit(' serie:--
m>T=J
r;CD(m,u=l
In2
"
L,
In3 In4 In 5
+ -  +...
, ,I ,)
76. A cross-country racer runs a IO-mile race in .')0 minutes. Prove
that somewhere along the course the racer raIl 2 miles in exactly 10 minutes.
72. Determine constants a, b aIld c such that
Vii 'f ,/Jak 3 + bk 2 + ck + 1 - vak 3 + bk 2 + ck ,
k=O
77. IJet AB be a line segment with midpoint O. I.J€t R be a point on
AB between A and O. Three semicircles are constructed on the same side of
AB as follows: S} is the semicircle with centre 0 and radillsl0AI = lOBI; S2
is the semicircle with centre R and radius IARI, meeting RB at C; S3 is the
semicircle with centre S (the midpoint of CB) and radius ICSI = ISBI. The
common tangent to S2 and S3 touches Sz at P and S3 at Q. The perpendicular
to AB through C meets S} at D. Prove that PCQD is a rectangle.
for n 1,2, . . . .
73. J.J€t n be a positive integer and a,b integers such that
GCD(a,b,n) 1.
78.
Determine the inverse of the n x n matrix
a} == a (mod n), b 1 == b (mod n), GCD(al,b}) = 1 .
(78.0)
[   
S 1 1 0
1 1 1
1
1
1
Prove that there exist integers a}, b 1 with
o
where n 2:: 2.
74. For n = 1,2,... let o5( n) denote the sum of the digits of 2 n . Thus,
for example, as 2 8 256 we have 05(8) 2+5+6 = 13. Determine all positive
Integers n such that
79.
Evaluate the sum
(71.0)
o5( n) = o5( n + 1) .
(79.0)
n-l
S(n) = I:(-llcosn(k1rln) ,
k=O

20
PROBLEMS
PROBLEMS
21
where n is a positive integer.
85.
(0.0)
[ -1 1 J B'J + C'J .
o -2
Let G be a group which has the following two properties:
(i) G has no element of order 2,
(ii) (xy)2 (yx?, for all x,y E G.
Prove that G is ablian.
80.
Determine 2 x 2 matrices [] and C with integr;J entries such that
(85.0)
86. Let A [a;j] be an n X n real symmetric matrix whose entries
satisfy
81. Find two non-congruent similar triangles with sides of integral
length having the lengths of two sides of OIle triangle equal to the lengths of
two sides of the other.
(86.0) ai. = 1, L laijl S 2 ,
j=l
for all iI, 2, . . . ,n. Prove that 0 S det A S 1.
82. Let a,b,c be three real numbers with a < b < c. The function
I(x) is continuous on [a,c] and differentiahle on (a,c). The derivative f'ex)
is strictly increasing on (a, c). Prove that
87. Let R be a finite ring containing an element T which is not a
divisor of zero. Prove that R mllst have a multiplicative identity.
(82.0)
(c - b)f(a) + (b - a)f(c) > (c - a)f(b).
88.
Set I n = {1,2,..., n}. For each non-empty subset S of I n define
w(S) = maxS - minS.
sES sES
Detennine the average of w(S) over all non-empty subsets S of I n .
83. The sequence {a".. I Tn 1,2,...} is such that am > am+! >
o , m = 1,2" .. , and :L=l am converges. Prove that
=
L m(a m - am+IJ
m=1
89. Prove that the number of odd binomial coefficients in each row
of I'ascal's triangle is a power of 2.
converges and determine its sum.
90. From the n X n array
1 2 3 2: ]
n+l n+2 n+3
2n+ 1 2n+2 2n+3 3n
(n-l)n+ 1 (n-l)n+2 (n-l)n+3 n 2
84. The continued fraction of VD, where D is an odd nons quare
integer > 5, has a period of length one. What is the length of the period of
the continued fraction of t(1 + .JJ5)?

22
PROBLEMS
PROBLEMS
is irreducible ov('r Z for n ;::: 4.
a number Xl is sdected. The row and column containing Xl are then deIcted.
From the resulting array a number X2 is selected, and its row and column
deleted as before. The sclection is continued until only onc number X T . remains
available for selection. Determine the sum .r-l + :e,t + . .. + Xn.
23
95. Let aI:"', an b(' n (;::: 4) distinct real numbers. Determine the
eneral solution of the system of n 2 !inl'ar equations
91. Suppose that p X's and q O's a.re plac(>d on tIle drcumferenc r . of
a circle. The number of occurrences of two a.djacent X's is tI and the number
of occurrences of two adja.cent O's is b. D(tennille a b in terms of J1 and q.
(95.0)
92.
In the triangular array
1
1 1 1
1 2 3 2 1
1 3 6 7 6 3 1
1 4 10 16 ]9 16 10 4 1
96.
1 Xl+ X2+"'+ x"
tl] "1 + {/2:e:l + . .. + anx"
2
a'Ix] + aX2 +... + (l.,."X",
3 n-3 + + n-:J x
a- Xl + lL2 X2 . . . an n
in the n Ullknowns Xl,'" ,X. n .
(92.0)
every entry (except Hie top 1) is the slim of the entry a immediately above it,
and the entries band c immediately to the left and right of a. Ab5ence of an
entry indicates zero. Prove that every row aft('r the second row contains an
entry which is even.
97.
93.
A se<luence of n real numbers Xl, . . . , X n satisfies
(93.0)
{ Xl = 0,
IXil = IXi-l + cl (2:::; i:::; n),
(97.0)
where c is a positive real number. Determine a lower bound for the average
of Xl,'" , X n as a function of conly.
98.
94.
(94.0)
(98.0)
Prove that the polynomial
f(x) x"'+x3+x2+X+5
Evaluate the sum
1
0,
0,
c tJ,
0,
N.:= 2,3,....
B(N) =
mn
l:::;m<n:::;N
m+n>N
GGD(m.n)=1
Evaluate the limit
I r, n j
L = lim - L L '2 + k2 .
n-oo 11 j=l k=l J
Prove that
311" . 211" r;<
tan +4sm = v11.
11 11

21
99.
For 11 =: 1,2,.. . let
Ewtluatc the sum
C n
1 1 1 1
+-+-+...+-
2 3 n'
OG
S'- L C n
 - r.=l n(n+ 1) .
100.
PROBLEMS
For x > .I determine the f;um of the infinite series
X x 2 x4
.1:+1 + (x+1}(x 2 +1) + (x +1)(x 2 + 1)(x 4 + 1) +....
THE HINTS
I
H/.ill shrou(led ,in llu fiaY'kt, $1 niyht, me look to lh f)oBI lI,ilh .'2:]1('('-
tation: a hint of a bright new day.
Aleksander Sergeevich Pushkin (1799-1837)
1. Let
(p-I)/2
N(k) L 1 ,
i,j=l
r,+n, =.k (mod p)
k = O,l,...,p-l,
and prove that
N(k) N(l), k l,2,...,p 1.
Next, evaluate N(O) and N(l), and then deduce the value of E(p) from
p-I
E(p) = Lw k N(k) .
k=O
2. Prove that
N(k) LLL 1,
x 'Y Z
where the variable x is summed from -k to k; the variable y is summed
from max(-k,x k) to min(k,x + k)j and the variable z is summed from

26
HINTS
max(-k,x k,y k)tomin(k,x+k,y+k). Thencxpressthctriplesumas
the iHl1ll of six sums specified aecording to the relative sizes of O,x and y.
3. First U8e the fact that w 2 - 1 (mod p) to prove that there are
integNs a and c sucll tha.t p = (2 + c 2 . Then lei, sand t be integers such tha.t
at -.. cs 1. Prow tha.t as + ct f/J1 (mod p), whr(' f :1:1, and deduce
that a.n intt'!!;c1' !J can be fi.>und o that b (-' S a.fJ) and d (=- t - tg) satisfy
ll.b+ed =- /117, nd be = 1 and b + d 2 c. (w 2 + 1)/p.
4.
Prove that
m m
L (d1(n) - d 3 (n)) = L L (_1)<d-1)/2,
n=l n=l din
dodd
and then intercha.nge the order of summation of the SIl1llS on the right side.
5. Rule out the possibilities x ° (mod 2) and x 3 (mod ,1) by
congruence considera.tions. If x 1 (mod 4), prove that there is at least one
prime p == 3 (mod 4) dividing x 2 - 3x + 9. Deduce that p divides x 3 + 27, and
then obtain a contradiction.
6. "Gse the identity
f(Xbydf(X2,yJ ==
(axlx2 + bXIY2 + 1JX2YI + CYlY2)2 + (ae - b)2(X1Y2 - X2Y1?
together with simple inequalities.
7.
Prove that exactly one of the triples
(a,b,c) = (R, S,T), (T,-S + 2T,R S + T), (R - S + T,2R - S,R),
satisfies
a ::; b < c, or a 2: b> c ,
I lINTS
27
hy considering cases depending upon the relative sizes of R, Sand T.
8.
C..msilkr the sign of the discriminant of
(a11 bA)x 2 + 2(aC - cA)xy + (bG - clJ)y2 .
9. Prove th"l the quantity
I nn 2"
2 n L T
.k=1
n-1 1 I
L2 k
.k=0
tends to zero as n -> 00.
10. Consider the Iase when n = p+ 1 and m = p, where p is a prime
suitably large compared with c.
11. Assume that 2D( D + Al A2 + f.B 1 B2) is a square, where ( = :1:1.
If D is odd, show that
{ D + Al , A2 + f.B 1 B2 == 2D1l 2 ,
D - AI A2 tBIB2 2D\.r2
A 1 B2 - f.A2Bl = 2DllV
Deduce that 112 + V 2 = 1. Then consider the four possibilities (If, V)
(:1:1,0), (0,:1:1). The case D even ca.n be treated similarly.
12.
Set
a::l:: = V 1985:1: 31 V'1985 , {1::1:: = V 3 970:1: 64 V'1985 ,
and prove that
a+ + a_ = fJ+, a+ a_ ;3-.

28
HINTS
13.
If (x, y) is a :solution of (13.0), prove that
x 2 + xy y2 = :l::F ,
and then solve tIJe system of equations
{ x2 + y2
x 2 +2xy +2y2 =
x 2 + xy _ y2
k,
I,
:l::F,
for x 2 , xy and y2.
14. F'actor the left side of (14.0).
15.
Make the following argument mathematically rigorous:
L\nx In(1- x) dx =
L 1 00"
In x Ldx
o ";:1 k
00 1 1
- L k { x" In x dx
";:1 10
00 1
t; k(k + 1)2
00 1 00 1
t; k(k+ 1) - t; (k + 1)2
1 _ ( 2 1)
=
71"2
2 - 6'
IllNTS
29
16. Taking n = 1,2,. . . ,6 in (16.0), we obtain
0,(1) = 1/2, a(2) = -1/3, a(3) 1/4,
a(4) = -1/5, a(5) = 1/6, a(6) = -1/7 .
This suggests that a( n) ( -1 )n+1/( n + 1), which can be proved by induction
on n.
17. Consider three cases accordin to thf' foJ1owio!!; va.1ues of tlH
Legendre symbol:
( n 2 : k ) = 1 or ( n + ;2 + k ) = 1
or ( n 2 : k ) = ( n + ;2 + k ) = -1.
In the third case, the identity
(n 2 + n + k)2 + k = (n 2 + k) (n + 1)2 + k)
is useful.
18. Rearrange the terms of the partial sum
N ( 1 1 1 )
 4n + 1 + 4n + 3 - 2n + 2 '
and then let N --> 00.
19. Use
( An ) 2 ( An ) 2 2 2 ( An
-;;- = an + -;- - an S; 2a n + -;;-
anr

30
to prove that
 ( An ) 2 m
 n 5 4 E a + 2 E ( n r - 4 , a:An .
Then use
-2a" It,,:= ( .112 , . " A2 )
n-L
., "
a;' 5 -(rt;.
A;'_d
1.0 pJ'O\" t.bat
-2 ;,;"'- u"A1I .  .IF
L., < -, L., n
n=l n - n=t n(n + 1) .
Putting these two incquaHties together, deduce that
m ( 2 ) ( A 2 m
L , 1 - - -..2: ) 5 ,t "'- 0,2 .
n=t n + 1 n L., n
n=1
20.
Use the identity
() -2 ( m
e n k - l ) - e')
(kt» )
( 2n ) .
.k+l
21.
All int.egral solutions of ax + by k are given by
x 0;:: 9 + bt, y = h - at, t = 0,
where (g, h) is a particular solution of ax + by = k.
... ,
22.
Prove that
Uk = Vk-l - l'k,
k=O,l,...,
HINT
IIINTS
31
wherc
1
l'k = ,
(a + (k + l)d)... (a + (k + r)d)1'd
, -1,0,1,...
23.
Prove that the st.l'Ollg'r inf-'qu<tlity
11
,
F(xu"rt:"...,.1'"nJn) < L:ri.;
k=l
( . \ 2
] './'I. 1
n <--.. )
k=l
holds by replacing ea.ch Xi by Xi - M for suita.ble AI
(23.0).
AfVb..., xn) in
24. Apply the Cauchy-Schwarz inequality to
r::. 1
\ an -..rn -..rn .
L-." n
n=l
25.
Consider the integers (3m + 2)2, m = 1,2.... .
26.
Use the identity
arct.an ( :'2 )
arcta.n ( n  1 )
arctan (  ) , n = 2.3, . .. .
n+1
27. Suppose that. gn(x) = h(x)k(x), where hex) and k(x) a.re non-
constant polynomials with integral coefficients. Show that h( x) and k( x)
can be taken to be positive for all reaJ x, and that h(p;) :(Pi) 1, i =
1,2,..., n. Deduce that hex) and k(x) are both of degree n, and determine the

32
HINTS
form of both hex) and k(x). Obtain a contradiction by equating appropriate
coefficients in g",( x) and h( x )A( x).
28. Let l)(k), k = 0,1,... , denote the probability that 11 wins when
A has k dollars. Prove the recurrence relation
ap(k+2)-(a+b)p(k+1)+bp(k) O.
29.
If xl:..., X n are the n roots of f(x), the 2n roots of f(x 2 ) are
f:y'Xl, f:y'X2, .... f:ft; .
30. Find a point P such that any two different la.ttice point must be
at different distances from P. Then consider the lattice points sequentially
according to their increasing distanc.es from P.
31.
Determine the generating function
00
L S(n)t n .
n=O
32.
such that
As I(x) is bounded on (O,a) there exists a positive constant K
If(x)1< J(, O<x<a.
Use (32.0) to deduce successively that
{ If(x)1 < Kfn,
x)1 < Kfn 2 ,
etc.
O<x<a,
O<x<a,
II iNTS
33.
33
Consider the derivative of the function
(YI - Y2)(Y3 z)
f(x) = (YI - Y,,)(Y2 - z)
34. Set al 1. Prove the recurrence relation
+ tl a I + . . . + a",-I tt2 + a.,aI ,
an+I aIUn 2 n-
. . Ii ( ) - '\.-00 a x n satbiles
and use it to show that the generatmg functIOn. x - L"",=l "
A(X)2 A(x) - x. Then solve for 11(x).
35.
Use L'H6pital's rule, or use the inequality
t S tan t S t + t\
to estimate the integral J; tan(y sin x) dx.
Osts1,
. el if e is an irrational number,
36. Use a result due to HurwItz, nam y: 0 and GCD(a b) = 1
there are infinitely many rational numbers afb with b> .,
such that
37.
Ie - afbl < 1f(../5b 2 ).
Differentiate (37.0) with respect to x and Y to obtain
(1 + x 2 )f'(x) (1 + y2)f'(y) .

:H
HINTS
38.
Introduce a coordinate svst d .
that max S 4R d . , , -. ern an use sImple inequalities to show
t an mIn.5 = 2R, where R iR the radius of the circle.
39.
Prove that
u ( n An+k ) == X n Y
11.=1 k=1
and
n ( u ,'lnH ) = ,- U Y ,
n=I k=1
where
x -= {O, 2. 4,. .. }, y = {O, 3, (j, .. . } .
40.
Prove that
Pk = { n
qp"
,0<k<n-1
, k: n - ,
,
, n + 1  k  2n ,
and
Pk = (1 kt l Pi ) qpn, k > 2n .
\ .=0
Use these to find a linear equation satisfied by P( x).
41. First prove that the circu ad . .
1 m and n . . b 1111' IUS of a tnangle with side.<; oflen g th
, IS gIven y
lm n
-J( l + m + n)(l + m - nJ(l
Next show that
m + n)( -l + m + n) .
IACI
 bd)(ad+ be)
(ab + cd)
I
HINTS
3.')
FinaUy, apply the above two results to !:"AHC.
I
42. Consider the quadrilateral ABCD as lying in the complex plane.
RepreseIll the vertices A, B, C. f) hy t.he complex numbNs l!, b,c, d respec-
tively_ Prove that. P, Q, R, .9 are represented by the numbers
I
(J.:x i ) la + ib , )'
e2' )lc+d),
rcspediv{.ly. 'l'll"11 rdat" l' - 1- and q

43.
Try a solution of the form
p=xy+X,
(i) (b + ic),
(t;i ) (d+ ilL),
q=y+Y,
where X and Yare polynomials in x, wand z. Substitute in (43.0) and solve
the resulting equations for X and Y.
44. Setn =- f(l)and/ f(i). Provethatlal = 1t31 = 1,la-t31 =.:-12.
Deduce that a 2 + {32 - 0 so that t3 = W, f :l:i. Next from (4<1.0) deduce
that
{ af(z) + af(z)
'iif(z) af(z)
Now solve for fez).
= z+z,
-fiz + fiz .
45. JJet x be a. rational number such that Y = tan 7rX is rational.
Prove that z = 2 cos27rx is a rational root of a monic polynomial with integral
coefficients. Deduce that z = 0, :1:1, :1:2.
46. Let SI,SZ,S3 denote the areas of b.PBC,!:"PCA,!:"PAB'respec-
tively. Prove that
!PAl _ S2+ S3
IPDI - Sl

37
36
HINTS
HINTS
47. Prove that the (i,j)-th entry of N is 1 times the (i,j)-th entry of
AI modulo n,
51. Let Pk denote the k-th prime. Suppose that N > 121 is an integer
with the property (51.0). Let Pn be the largest prime less tha or equ.al to.J!i,
th 1. > 5 nd N < P 2 Use property (51.0) to obtam the mequallty
so a n _ ,a n+l'
N  PIP2 .. . Pw Then use Bertrand's postulate
Pk+1  2Pk, k =:: 1,2,... ,
. h . . 1 . r If.m d IPCI
WIt slml ar expreSSIOns lor IPEI an IPFI'
48. There are (N + 1)" vectors (Yl,Y2,. .., y,,) of intef!;ers satisfying
o  Yj  N, 1  j  n. For each of these vectors the corresponding value of
to obtain
PIP2 . . . Pn-2 < 8
.' 2 Deduce the contradiction n
from the mequahty 1'11-'2'" Pn < Pn+I' .
Check property (51.0) for the integers N 3,4,...,121 dIrectly.
 4.
Li Li(YbY2"" ,Yn) = aiIYl +... +ainYn, 1  i  m,
satisfies -naN  Li  naN, so the vector (L b L2"'" Lm) of integers can
take on at most (2naN + 1)m different values. Choose N appropriately and
apply Diricblet's box principle.
52. Prove that
49. Suppose that f e- x2 dx is an elementary function, so that by
Liouville's result, there is a rational function p(x)/q(x), where p(x) and q(x)
are polynomials witb no common factor, such that
t x 2 + x + 1 d
S =:: 10 x4 + x3 + x2 + x + 1 x
and then use partial fractions to evaluate the integral.
J _x2 d p(x) _",2
e x == -e
q(x)
53. Let IABI -= 2c, IB GI == 2a, IGAI == 2b. S how that
1== J (a-b+c)(a+b-c)
with similar expressions for m and n.
Differentiate both sides to obtain
p'(x)q(x) - p(x)q'(x) 2xp(x)q(x) -= q(x? ,
54.
Evaluate
and deduce that q(x) is a nonconstant polynomial. Let c denote one of the
complex rQots of q(x) and obtain a contradiction by expressing q(x) in the
form q(x) == (x - c)mr(x), with rex) not divisible by (x - c).
H(1,1,0,O), H(0,0,1,1),
H(O,l,l,O), H(1,0,0,1),
using (i) and (iii). Then express H(a,b,c,d) in terms ofthese quantities by
means of (i), (ii), (iii) and (iv).
50. Prove that
n-l (-1)i
X n ==?= i+ l ' n == 1,2,... .
.=0
55.
Set fez) == z"+alz"-l+...+a" and note that for z =1= ° we have
If(z)1 =:: Izn (1+ a; +...+ :: )1

38
HINTS
HINTS
39
Iz"1 1 1+ +."+ an i
z z1t
;;:: Iznl ( 1 _  .. . _ la n l )
Izi Izln
;;:: Iznl (1- 1;1 I,n ) ,
59.
Consider the integers 72k: + 42, k = 0,1,. ., .
60.
Use the identity
56.
Ddinr- intr-gns /; ami I b.I'
k ( k- )
2 k sin k:x cos k x 'L: r t;i n 27':/; .
r=1
2 m 1 l;d, 2 n + 1 = ld ,
and then consider
61.
Express
2 mn = (kd + l)n -, (ld _ l)m .
n: 1 e:)
as the difference of two binomial coefficients.
57. Suppose that f(x) g(x)h(x), where g(x) and hex) are non con-
stant polynomials with integral coefficients chosen so that
62.
Use the identity
Deduce that def?; (g(;1.:») :::; 'In and that g(k,) = :1::1, i = 1,2...., 2m + 1. Let
( +1 (resp. -1) if +1 (resp. -1) occurs at lea.<;t 'In + 1 times among
the values g( k:i) :I:: 1, i = 1,2,..., 2m + 1. Then consider the polynomial
gee) - (.
+ 1 a 2n - 1
2n+i
a 2n + 1 _ 1 '
a> 1.
deg(g(x):::; deg(h(x)).
2" 2"
63.
Prove that
58. Suppose a, b, c, d a.re integers, not all zero, sa.tisfying (58.0). Show
that without loss of generality a, b, c, d may be ta.ken to satisfy
1 ( 2n - 2 ) ( k ) "
a,,=2(-1)n-I;;; n-1 4'
GCD(a,b,c,d) = 1 .
and appeal to Problem 61.
By considering (58.0) modulo 5 prove t.hat
a == b == c == d == 0 (mod 5) .
64 L t C C C denote the columns of Mm. Determine a
. e 1, 2, .. .. .., .1m ..
linear combination of C h C2,"" Cm-I which when added to C m gIves the
column (O,O,...,4>(m)t. Deduce that det U m = 4>('111) det Mrn-i'

40
HINTS
IUNTS
41
65.. . Assume (65.0) holds and use congruences modulo 8 t bt '
contradictIOn. 0 0 am a
for suitable constants a, b, .. . ,f. The case m odd is treated similarly.
66.
Prove that
71.
Note that
1 1 (_l)n-Ixn
a'n d
I + ,x,
o x
alld use this reprt'.sentation of an to de(Juce that
In2
2
In 3 In 4 In 2n
3+1 ...+ 2n
= In 2 ( 1-1 1 + . .. + 1 ) + t I n k
2 n k=l k
2n Ink
L--
k=1 k '
I N 1 I
]; an - l (1: x )2 dx $
1
+2 .
and estiul'l.tl. Lk=l(lnk)fk fur largp 'I/. using I.IH' Euler-Ma.t'!,a1lI'in Rumma.tinl!
formula.
the 67 . b et A b a sequnce of the required type, and let k denote
o num er 0 zeros III A. FIrst prove that k ::: 3. Deduce that A
{ ,O,O,a 3 ,a4,a s ,3},wherel<a3<a 4 <a s < 3 ' fh th N
- - - -' en prove at 1 = 2.
72. Express (v'k+1- Vk)3 in the form Vp (k) + 1 Vp(k), where
p( k) is a cubic polynomial in k.
68. }'rove that
73. Choose al to be any nonzero integer such that a] a (mod n).
Then set b l = b + rn, where T is the product of those primes which divide a]
but do not divide either b or n. Prove that GCD(a},bd = L
g nh g -n == h 2n ,
71.=1,2,...,5.
74. Prove that s(n + 1) == 2s(n) (mod 3), and use this conguence to
show that there are no positive integers n satisfying s(n) s(n + 1).
69.
75. Show that
00 00 1
s== L f L
m,n=l d=1
Prove that every integer N > 2ab is of the form
N==xa+ y b , x>O > 0
- , y - , x + y == 1 (mod 2) ,
and that all integers of this form belong to S.
70.
If m is even, say m == 271., show that
by collecting together those m,n in the sum A L:.n=llf(mn(m + n))
having the same value for GC D( m, 71.). Then evaluate the sum A by proving
that it is equal to the integral
m = (an + b)2 + (cn + d)2 - 5(en + f)2 ,
l lln 2 (1 - x)
dx,
o x

42
HINTS
HINTS
1.3
which can be ('!valuated by means of the transformation J: = 1 _ e--".
ar('! similar if a/b = bfc c/d. Choose positive integers to satisfy this relation
remembering that the triangle inequalities c < a + b, etc must be satisfied.
76. App!y the int,ermediate v<tlu('! theorem to the function T( x) de-
fined to be the tIme taken in minutes by the racer to :run from th . 1. .
'1 I th . . ('! pom .c
Ill! es a ong ('! C01}fS('! to the point x + 2 miles along the course.
82.
(b,c).
Apply the mean vdlue thmrem to f(T-) on th('!intervals (tl,b) a.1l.!
77.
Choose a coordinate syst'rn so that
83.
First show that linIn__;."x. 7W n --' O. Then [pt 1/, . = in
A (-1,0), 0 (0,0), 1) "" (1,0).
Then R = (-a,O) with 0 < a < L Deduce that
n
L k«(Ik - akH)
k=l
n
Lak
k=l
na ll +1.
C = (1 - 2a,0) ,
S = (1 - a,O),
D = f l- 2a,2 V( I.(1 - a») ,
P = 2a 2 - 4a + 1 ,2(1- a)va (1 - a»)
Q 1 2a 2 ,2a va (l- a») ,
a.nd calculate the slopes of PC, PD, QC and QD.
84. Use the fact that the length of the period of the continued fraction
of ..;J5is one, and that D is an odd nonsquare integer > 5, to show that
D = 4c 2 + 1, C ?:: 2. Then determine the continued fraction of HI + ..;J5).
85. lor x, y E G prove that (XY:I--ly-l)2 1.
78.
that U 2
Let I denote the n X n identity matrix. Set [J = 8 + I. Prove
nll. Seek an inverse of S of the form cU - I.
86. Let).. denote one of the eigenvalues of A and let ;£ be a nonzero
eigenvector of A corresponding to)... By applying simple inequalities to an
appropriate row of A;£ = )..;£, deduce that I).. - 11 ::; 1. Then use the fact that
A is real symmetric and the relationship between det A and the eigenvalues
of A.
79.. R:eplace cos(k1r/n) by (w k +w-- k )/2, where w = exp(7ri/n), and
use the bmOIllIal theorem.
80. Let A
[ -1 1 ]
o -2 and show that A3 + 3.4 2 + 2A = O. Then
87. Show that there exists an integer k ?:: 2 such that r = r k . Then
prove that r k -- l is a multiplicative identity for R.
consider (A + 1)3.
81.
Let the sides of the trian g les be a b c and b C d ' I " he t t . 1
' , . ,.,. . wo flang es
88. For 1 ::; k ::; 1 ::; n let 8(k,l) denote the set of subsets of J"
with minsEs S k and maxsES S = I. Evaluate IS(k,l)1 and then compute

44
HINTS
HINTS
45
ESPn w(S) using
is repeated down the left edge of the array from the fourth row down.
L w(S')
4>#-St;;J n
L (l- k)/S(k,l)1
lk<I",
93 L t be any real number such that I X n + l l = IX n + cl, and
. e Xn+l
consider Efjl x; .
89.
Write n in binary notation, say,
n = 2 G1 + 2<>2 + . . . + 2 Gk ,
94. lfwe have f(x) = g(x)h(x) then without loss of generality g(O)
::1:1, 1£(0) ::1:5. Prove that one of the comple..x roots,6 of g{ x) satisfies 1,61 ::; 1,
and then dedl1n! that If(f:i) 1 2': ].
where aI, . . . ,ak are integers such that al > a2 > ... > ak 2': 0, and then use
(1+xf'Q
(1 + x)'"
1 + x 2Q (mod 2) ,
= (1+x)2Ql(1+x)2Q2...(1+x)2Qk.
95.
Set
f(x) = (x al)(X - a2)"'(x - an)'
90. Suppose that Xi, 1::; i ::; n belongs to the Ti-th row and the si-th
column. Show that
Prove that
n
LXi
.=1
n '"
'" 2 '"
== n L.i Ti - n + L.i s; ,
;=1 ;=1
( P(l)' '''' f'("'» ) and ( Pl) '"'' Pn» )
are two solutions of (95.0). Deduce the general. solution of (95.0) from these
two solutions.
and then use the fact that both { Tl,..., Tn} and {81,..., 8n} aTe permuta-
tions of {1,2,.. .,n}.
96. By picking out the terms with n = N in the sum s(N), show that
s(N) = s(N - 1) for N ;?: 3.
91. Let N:r;:r;,N:r;o, No""N oo denote the number of occurrences of XX,
XO, OX, 00 respectively. Relate N",:r;, N",o, No."N"" to a,b,p,q. Prove that
No", = N:r;o, and deduce the value of a - b in terms of p and q.
97.
Prove that
92. Consider the entries of the triangular array modulo 2. Show that
the pattern
L= {I {I dxdY,
10 10 x + Y
and evaluate the double integral using polar coordinates.
1 101
1 000
1 1 1 0
101 0
98. For convenience set p = '!rIll, and let c = cosp, s =: smp. Use
the imaginary part of
(C+iS)l1 = -1,

46
HINTS
to prove that
(Us - 44s: J + 3285)2 == llc 2 (1 48 2 Y,!.
Then show that
tan3p+ 4sin2p
" 2 5
" . r,-:. , , .
=:tvll.
THE SOLUTIONS
l)...duc!' that the + sign holds by confiiderinp; the sign of tile left side.
99.
ists.
Use partial summation aud the fact that limk-+oo(c,,: Ink) ex-
Some pwple think we are W1'01/.Y, but only time will tell: given all
the alternativES, 'I.ve have the solution.
100.
Use the identity
1 x 2 "
 (x + 1)(x 2 4;-,1)(x 4 + 1)...(x2" + 1)
_ 3.'2 1
(x 2 "+' 1) - 1
Lev Davydovich Bronstein Trotsky (1879-1940)
1
-1
1.
product
Let p denote an odd prime and set w = f'.xp(21ri/p). Evaluate the
(1.0) R(p) (W T1 + W T2 + ... + w T (p-'){2)(W n , + w n2 + ... + w n (p-'J{2),
where 1'17, . . , r(p-l)/2 denote the (p - 1)/2 quadratic residues modulo p and
nh"', n(p-l)/2 denote the (p 1)/2 quadratic nonresidues modulo p.
Solution: We set q == (p - 1)/2 and
{ 0 if p == 1 (mod 4),
(1.1) E = 1: if p== 3 (mod 4),
and for k == 0,1,... ,p 1 let
(1.2)
N(k) ==
q
L
1.
itj:1
T.+n,=k (modp)

48
SOLUTIONS
SOLUTIONS
49
If k is a quadratic residue (resp. nonresidue) (mod p) {kri: i = 1,2,...,q}
is a complete system of quadratic residues (resp. nonresidues) (mod p) and
{knj : j 1,2,..., q} is a complete system of quadratic nonresidues (resp.
residues) (mod p). Replacing ri by kri and nj by kni in (1.2), where 1 s: k s:
P 1, we obtain .
W'ri+nj
k=O
i,j=1
Ti+nj=k (roml p)
(1.4)
N(O)
,,} N(k)
k=O
N l () + ]V (I )( (,) + w 2 + . . . + wP I)
N(O) 1\'(1)
t.lJ. (q.. <)/'2, by (1.'1) and (Uj) ,
(1.3) N(k)=N(l), k-1,2,...,p-l.
Next, we Ilotn that
0;;: fq,
that is
i,j=1
Ti=-n 1 (modp)
J'(p )
{ (1 p)/4, if p
(1 t p)/4, if p
1 (mod 4) ,
3 (mod 4) ,
as -1 is a quadratic residue (mod p) for p = 1 (mod 4) and -1 is a quadratic
nonresidue (mod p) for p 3 (mod 4). Now as
as required.
( 1.5)
N(k)
10;;: q2,
2. Let k denote a positive integer. Determine the number N(k) of
triples (x, y, z) ofintegcrs satisfying
{ ixl < k, Ivl s: k, izl s: k,
(2.0) Ix - yl s: k, Iy zl s: k, Iz xl s: k.
we obtain, from (1.3), (1.4), a.nd (1.5),
fq + 2qN(1) q2,
(1.6)
N(I) (q f)/2.
Solution: The required number N(k) oftriples is given by
N(/d
that is
Finally, we have
Ixl$k Iyl$k
Ix-yl$k
Izl$k
Iy-zl$k
Iz-xl$k
B(p) (EW T .) (t wnj )
;,vTi+nj
k k k
0;;: 1,
x=-k y=-k z=-k
x-k$y5,x+k x-k$z$x+k
y-k$z5,y+k

50
SOLUTIONS
SOLUTIONS
51
t1!at is
(2.1)
k
N(k) I: I: I:1,
x=-k y z
-1 -I .r+k
S3 I: I: I: 1 =F,
x=-k y=x z=-k
k x-I k
.";.. L L L 1 F
, ,
,..=1 1/=0 ==:r.-k
k -1 1I+ k
Sf, L L 1 =E,
X=O y=x-k z=.c-k
where the second sum is taken over y = max(-k,x k) to y min (k.x +k),
and the third sum is ta.ken over 2- = max(-k,x k,y k) to z min(k,x+
k,y + k). We now split the sum on th right of (2.1) into six sums SJ,... ,S6,
wher(' x a.nd yare restricted as follows:
O::;:x<;y,
x < 0::;: y,
x::;: y < 0,
0:::; y < x,
y<O::;x,
y<x<O,
in ''''1 :
in fh;
in 8 3 ;
in 8..;
in 8 5 ;
in 8 6 .
/)'6 =
-1
L
.r=-k+l
x-I
L
y==-k
?I+k
I:1
z=-k
1
_::.: -(k
6
1 )k( 41. + 1) .
Thus we have
N(k)
Sl + 8 2 + . . . + ''''6
!(k + l)(k + 2)(4k + 3) + k(k + 1)(2k + 1)
6 3
1
+-(k - 1)k(4k + 1)
6
4k 3 + 6k 2 + 4k + 1
(k + 1)4 k 4 .
Clearly, we have
k k k
51 I: I: 1
x=o II=X z=y-k
k k
I: I:(2k+ 1- y)
x=o 1/=X
1 k
= 2' L(k+ 1- x)(3k+2 - x)
x=o
1 k
2' I:«k + 1)(3k + 2) - (4k + 3)x + x 2 )
x=(J
 (k+1)2(3k+2)- + 2 + + k(k+12k+1» )
!(k + l)(k + 2)(4k + 3).
6
Similarly, with E denoting k(k + 1)(2k + 1)/3, w obtain
3. Let p == 1 (mod 4) be prime. It is known that there exists a unique
integer tV == u:(p) such that
w 2 == --1 (mod p), 0 < 1lJ < pf2.
(For example, '1/)(5) = 2,w(13) = 5.) Prove that there exist integers a,b,c,d
with ad bc = 1 such that
pX2 + 2mXY + (w 2 + 1) y 2 == (aX + by)2 + (eX + dy)2.
p
S2
-1 x+k
I: I:
.r=-k y=o
x+k
I:1
z=y-/.;
=E,
(For example, when p = 5 we have
5X2 +4XY + y 2 == X 2 + (2X + y)2,

52
SOLUTIONS
and when p::::: 13 we have
13X 2 + 10XY + 2y 2 == (3X + y)2 + (2X + y)2.)
Solution: We make use of the following property of the reals: if r is any reaJ
number, and n is a positive integer, then there exists a rationaJ
number h/k such that
(3.1)
IT - il  ken  1)' 1  k  n, GCD(h,k)::::: 1.
Taking T ::::: -w(p)/p and n ::::: [y'P], we see that there are integers a and e
such that
I -w(p) - ! I <, 1  a < vIP.
p a avIP
Setting e ::::: w(p)a+pe, we see from (3.2) that lei < vIP, and so 0 < a 2 +c 2 < 2p.
But c wa (mod p), and so 0,2 + e 2 == 0,2(1 + w 2 ) 0 (mod p), showing
that
(3.2)
(3.3)
p ::::: 0,2 + c 2 .
As p is a prime, we see from (3.3) that GCD(a,e) ::::: 1. Hence, we can choose
integers sand t such that
(3.4)
at-cs==l.
Hence
(as + et - w)( as + ct + w)
::::: (as + et)2 w 2
(0,2 + c 2 )(S2 + t 2 ) - (at _ es)2 _ w2
p(S2 + t 2 ) (1 + w 2 )
- 0 (mod p) ,
so that
(3.5)
as+ct == fw (modp), f::::::H.
SOLUTIONS
53
Hence there is an integer 9 such that
(3.6)
as + et ::::: fw + gp .
Set
(3.7)
b == sag, d t - eg .
Then, by (a.a), (a.4), (a.6), and (3.7), we have
(3.8)
ab + cd == fw, ad - be = 1 .
We now obtain
p(b 2 + d 2 ) ::::: (a 2 + e 2 )(b 2 + d 2 )
(ab + ed)2 + (ad - be)2
::::: w 2 +1,
so that
(3.9)
b 2 +d 2 (w 2 +1)/p.
Then, from (3.3), (3.8), and (3.9), we have
(w 2 + 1) 2
(3.10) (aX + bY)2 + (eX + dy)2::::: pX 2 + 2fwXY + p Y
If f 1 then (3.10) is the required identity. If f == -1, replace b,e,Y by
-b, -e, -Y respectively to obtain the desired result.
4 L t d (n) r ::::: 0 1 2 3 denote the number of positive integral
. e T, , , , , . .
divisors of n which are of the form 4k + T. Let m denote a positIve mteger.
Prove that
(4.0)
m
2.:)d l (n) - d 3 (n)
n=l
00 " [ m ]
?:(-1)3 2j+l .
3=0

54
SOLUTIONS
Solution: We have
m
m
L L (_1)(d-I)/2
"",1 dl"
d odd
L(d1(n) d 3 (n»
n=l
L
L (-1 )(d-I)/2
,1 tJdd 15:.{J.: 771,
L(_1)(d-I)/2
cl (.,tltl
L
1
I <,.k<m/J
L (_1)(d-l)/2 [ 111- ]
dodd d
fe-oj [  ] .
j=O 2J + 1
This completes the proof of (4.0).
5. Prove that the equation
(5.0)
y2 x:, + 23
has no solutions in integers x and y.
Solution: Suppose that (x,y) is a solution of (5.0) in integers. If x
o (mod 2) then (5.0) gives y2 =0 3 (mod 4), which is impossible.
Hence, we must have x 1 (mod 2). If x =0 3 (mud 4) then (5.0) gives
y2 =0 2 (mod 4), which is impossible. Hence, we see that x 1 (mod 4). In
tliis case we have x2 - 3x +9 =0 3 (mod 4), and so there is at least one prime
p 3 (mod 4) dividing x2 3x + 9. Since x2 3x + 9 is a factor of x3 + 27,
we have x 3 + 27 0 (mod p). Thus by (5.0) we have y2 -4 (mod p).
This congruence is insolvable as -4 is not a quadratic residue for any prime
p =0 3 (mod 4), showing that (5.0) ha.'J no solutions in integers x and y.
6. Let f( x, y) ax 2 + 2hxy+ cy2 be a positive-definite quadratic. form.
SOLUTIONS
55
Prove that
(f(Xl, ydf(X2' 1tl))1/2 f(XI - X2, YI - Y2)
(6.0)
2: (ac - ti)(XIY2 X"lytJ2,
for all rea] numbers .1;], X2, Yb Y'l'
Solution: First we lIote that ac - b 2 > 0 as f is positive-definitp. \Ve ue
the identity
(6.1)
(ax + 2bxIYl + cyn(ax + 2bx2V2 + cy) ::::
( ax IX2 + bXIY2 + bX2YI + CY1Y2)2 + (ac ( 2 )(XIV2- XZJ/I?'
Set
E I =f(XbYtJ2':O, Ez f(X2,Y2) 2:0,
F = !ax1x2 + hXIV2 + bX2Y1 + CY1Y21 2: 0,
and then (6.1) becomes
(6.2) EIE2 F 2 + (ac b 2 )( Xl Y2 - X2YI?'
We also have
(6.3) f(xi X2,Y1 V2) E 1 + E 2 :I: 2F.
Hence, using (6.2) and (6.3), we obta,in
1/2 )
(f(xbvtJf(X2,Y2)) f(XI X2,YI - Y2
2': (E1E2)1/2(E1 + E 2 2F)
2: (I<J 1 Ed/ 2 (2(E 1 E2)1/2 - 2F)
2(E 1 J;2) - 2(E 1 1:;2)1/2 F
2F 2 + 2(ac b 2 )(x1Y2 - X2yJ)2
-2F(F 2 + (ac - b 2 )(XIY'l
2F 2 + 2(ac - b'l)(XIY2 X2Yt?
2 ( (ac b 2 )(XIY2
-2F 1 + F2
x2yd)1/2
. 2 ) 1/2
X2Y1)

56
SOLUTIONS
 2F'- + 2(ac b 2 )(XIY2 X2YI)2
_2p2 ( 1 + (ac- b 2 )(XIY2 - X2Yd 2 )
2P2
(a- b 2 )(;/'IY2 - ;l;2yI)2
This cOlllIlletes tIlt; proof of (6.0).
7. Let R, ,.,', T be three rcal numb!'ts, not all thc !\<1lI1e. Give a condi-
lioll which is alisll,,'rl hy tHlP >lIItJ ollly OtH' of rhe thr,".' rripl!';;
{ (R,S,T),
(7.0) ('1', -8 + 21', R - S + 1'),
(R - S + 'l','l.R S,R).
Solution: We let (a,b,c) denote a.ny one of the triples in (7.0) and show
that exactly one of the three triples satisfies
(7.1 )
(i) a 5: b < c or (ii) a  b > c .
\Ve consider six cases.
Case (i): R 5: S < T. Here (a,b,c) (R,S,T) satisfies (7..l)(i) but not
(7.1)(ii), while the other two triples satisfy neither (7,l)(i) nor (ii) as
T < -8 + 21', -8 + 2T > 1l S + l'
and
R - S + T > 21l - S, 2R - S < R .
Case (ii): R < T 5: S. Here (a,b,c) (1', -S + 2T.R - S + 1') satisfies
(!.1 )(ii) bl1t not (7.1 )(i), while the other two triIlles satisfy neither (7.1 )(i) nor
(u) as
R < H, S > T
and
1l - S + l' > 2R - S, 2R S < R .
SOLUTIONS
57
"
t.
i
Case (iii): S < R 5: 'T. Here (a,b,c) == (R - S + T,2R - S,ll) satisfies
(7.1 )(ii) but not (7.1 )(i), while the other two triples satisfy neither (7.1)(i) nor
(ii) a.<;
R > S. S < l'
and
l' < -S+2T, -S+2T> R S+T.
Case (iv): S5:T<R. Here(a,b,(')==(T.-S+2T.R ....'+nsa.li;;lie;;
(7.1 )(i) but not (7,1 )(ii), whilfd,he otlll'r two tl'iples satisfy neither (7.1)(i) nor
 ji) af\
R > S, S 5: T
and
R S + l' < 2R S, 2R - S > R .
Case (v): l' 5: R < S, Here (a,b,c) == (R S + T,2R - 8,R) satisfies
(7.1 )(i) but not (7.1)(ii), while the other two triples satisfy neither (7.1)(i) nor
(ii) as
R < S, S > l'
a.nd
l' > -8 + '21', -s + 21' < R S + T .
Case(v-i): l' < S 5: R. Here (a,b.c):= (R,S,T) satisfies (7.1)(ii) but not
(7.1)(i), while the other two triples satisfy neither (7.1)(i) nor (ii) as
T > -S + 2T, -8 + 21' < ll- S + l'
and
R S + l' < 2R - S, 2R - S  R.
8. Let ax2 + bxy + cy2 a.nd Ax 2 + Bxy + Cy2 be two positive-definite
quadratic forms, which are not proportional. Prove that tho form
(8.0)
(aB - bA)x 2 + 2(aC - cA)xy + (bC - cB)y2
is indefinite.


58
SOLUTIONS
Solution: As ax'}, + bxy + cy2 and Ax2 + Bx!) + Cy2 are positive-definite
we have
a > 0, c > 0, b 2 4ar. < 0,
A > 0, C > 0, B2 - <lAC < 0 .
Tb show that tIll' form
{aB bil)J: 2 + 2iaC  cA):q, + (bC - rBhl
is inddinite we HIIISt. "how that its disnimimmt
D = 4{aC - CA)2 - 4(an - bA)(bC - cfl)
i;; po!;itive. We first show that D ? O. This follows as
0,2 D (2a(aC' cA) b(aB - bA))2 - (b 2 4ar.)(aR _ bA)2 .
Moreover, n > 0 unless
aB bA = aC cA = 0
in wMch case
a b c
A B r;
This does not occur as ax 2 + b;cy + cy2 and Ax2 + Bxy + Cy2 aT(' not propor-
tiona.l.
9. Evaluate the limit
(9.0)
n n 2k
lim - '" - .
n_= 2 n L.J k
k=1
L
Solution:
We show that L = 2. For n ? 3 we have
n 2 k
"'-
2 n L.J k
k=1
n n-l 2n-k
'}n L n - k
- k=O
SOLUTIONS
,,-nd so
I nn 2 k
"n Lk
 k=l
on-II I
L 2k
k=O
59
n-I J n
== L 2 k 12. - k
k=O
r.-l 1 k )
L 2k(1+k
k=O
n-l 1 on-I k
L 2k + L 2k(n - k )'
k=O k=l
== \ on-l k I
t; 2 k (n - k)
n-l k
L 2k{12. - k)
k=1
n-l k
1 '"
::; 2(11 -1) +  (k 2 - k)(n - k)
1 n-I 1
2(n 1) +(k-t)(n-k)
1 1 n-I 1 1
, +-L(-+-)
2(n - 1) '11 - 1 k=2 k - 1 n - k
1 2 n-2 1
+-L
2( n - t) n - 1 r=1 r
1 2
< + Inn.
2(n - 1) n 1
....--L +  In n --+ 0 and so
As n --+ +00, 2(n-l) (n-l)
n 2 k
L== lim L-
n_co 2 n k
k=1
n-l 1
== lim L 2 k
n-+X"
k=O
00 1
L 2 k
k=O
2.

60
10.
(10.0)
SOLUTIONS
Prove that thcre does not exist a constant c  1 such that
for aU pt)$itiv, integers nand m satisfying n  m.
nCt/J(n)  mCd>(m),
Solution: SlJ pposc then exisl s a constant c  1 such that ( 10 0 ) I ld fi
all T't .losor
. Po] lVe m ,e/1:el'S m and -n satisfying n > m I 1. h .
'-1;'111 'f' > ,.... Thcn, w" hav': - . Je pea pflme
3 p+1
-  ( as p > 1c  4)
4 2(p- 1)
> tj>(p+ 1) (as tj>(p + 1)  (p+ 1)/2, tj>(p) = p - 1)
tj>(p)
 (-L)" (by (10.0»
.p+ 1
(1 _ 1
p+
 1 c (using XC - 1  c(:c - 1), x > 0)
p+]
> 1- ::.
p
3
> - (as p > :1c),
.4
which is impossible, and no such c exists.
.1.1. , Let D be a squarefree integer greater than 1 for which there exist
posItIve mtegers A}, A2' B}, B 2 such that .
(11.0)
{ D = A + B = A + B,
(AI,B I ) =f. (A 2 ,B2)'
J>rove that neither
2D(D + AIA2 + B I B2)
SOLUTIONS
,r"
"I..
nor
2D(D + AIA2 - B I B 2 )
is the square of an integer.
Solution: Suppose that 2D(D + AIA2 + fB 1 1l2) X 2 , where X is an
integer anti f =- :i: 1. We consider two caSE'S aCt:orrlinl!; as {) is ....1<1
or even.
If P is odd, as it is squarefr<'c, 2D divides X, say X = 'lDl.', when' (J is
a.n integer. and so
(11.1)
D + AIAz + I:;B l 1l 2 = 2DU 2 .
Next we have
2D(D
AIA2 - I:;B I B 2 )
= 2D (D 2 - (AIA2 + fB I B 2 ?)
D + AIA2 + fB I B 2
2D(AIB2 - fA2Bd 2
=:
D + AIA2 + fB I B 2 '
that is
(11.2)
2D(D AIA2 - I:;B 1 B 2 ) = ( AIB2  fA 2 B 1 r .
Since the left side of (11.2) is an integer and the right side is the square of
a rational number, the right side of (11.2) must in fact be the square of all
integer. Hence, there is an integer Z such that
(11.3)
2D(D - AIA2 I:;B 1 B 2 )
Z2
,
(11.4 )
AIB2 - fA2BI U Z .
From (11.3), as above, we see that 2D divides Z, so there exists V such that
Z 2DV. Then (11.3) and (11.4) become
(11.5)
D - AIA2 - fB 1 B 2 = 2DV 2 ,

62
SOLUTIONS
(11.6)
AlB2 f..4 2 B 1 = 2DIJV .
Adding (11.1) a.nd (11.5) we obtain 2D =: 2DU 2 +2DF2, so that [[2+ V2 = 1,
giving
(11.7)
(U,v) (:1:1,0) or (0,:1:1).
Now from (11.1), (11.5) and (11.6), we have
{ 11/12 + (BIHz == ?(U. 2 -- V 2 ),
-f H 1 .4 2 + Al U 2 = 2DIJ V .
Solving thesc equations for A 2 and R 2 gives
(11.8) Ih =: (U 2 - VZ)Al 2dlFB 1 , B 2 =: ZUV A 1 + f(lJ2 V2)Bl'
Using the valus for (ll, IT) given in (11.7), we obtain from (11.8) (A:;;,B 2 ) =
:1:( AI, fBd, which is clearly impossible .,.; Ah ..1.2, B1, B 2 ar€ positive and
(AbBd f (A 2 ,B 2 ).
The case when D is even can be treated similarly.
12. Let Q and R denote the fields of rational and real numbers
respectively. Let K and L be t.he smallest subfields of R which wntain both
Q and the real numbers
/ 1985 + 31 V1985 and / 3970 + 64 V1985 ,
respectively. Prove that K =: L.
Solution: We set
(12.1)
{ 0+ =: .j1985 + 31.Ji985  58.018,
I.L /19 85 - 31 v'1985  24.573,
(12.2)
{ ,£'J+ /.3970 + 64.Ji985  82.591,
=: )3 970 - 61 v'1985  33.445 .
SOLUTIONS
63
[1. is easy to check that
(12.3)
32 V1985 , fhfL =: 62Vi 98& ,
Oi+I.L
( 12.4)
{ (0:+ + I.L? 3970 + 64 -J '
(a+ - I.L)2 =: 397() - My!1985 ,
from whidl Wf' obtain
>
(1Z.5) (1'++(L I'J+_ 0t-(.L=P_
W . t ' Q ( 'II 'II ) for the smallest subficld of R contaillin both Q <1[111
flIng tl,...'tn
the real numbers 11,. . . ,1n, we have
Q(Oi+) Q(o+,at)
Q( Oi+, V1985) (by (12.1»)
Q(O+,Oi_) (by (12.3))
'2 Q(o+ + Oi_)
:::= Q(/h) (by (12.5))
= Q(fh,{3)
QU:J+, V1985 ) (by (12.2»)
= Q(!3+,{3-) (by (12.3))
'2 Q(/h + (L)
Q(o+), (by (12.5))
so that K Q(o+) = Q(,g+) L.
13. Let k a.nd l be positive integers such tha.t
GCD(k,5) = GCD(l,5) GCD(k,l):::= 1
and
II
_k 2 + 3kl- z2 = F 2 , where GC D(F,5) 1.
Prove that the pair of equations
{ k = x 2 + y2,
(13.0) l = x 2 + 2xy + 2 y 2,
'
....4

64
SOLUTIONS
has exactly two solutions in integers :1: aud y.
Solution: We have
1'"2 = 4k 2 + 8kl + 41 2 == 4(k + l)2 (mod 5)
so that F :f:2(k + I) ( d 5 ) R 1
1110 .'. Pp adTlI1; F by - 1", if 1)1'((':-:8ar , \' , we J ' 1 " ,, '
RUpp08e Lllat ' 
(l:1.1 )
p  '2(h + l) (mod '»
Then we have
(13.'2)
{ 4k - 1 - 2P
--3k + 21 - F
k + 1 + 2F
o (mod 5),
o (mod 5),
o (mod 5),
and we may define integers R, H, T by
(13.3)
{ 5R =
58 =
5T =
4k - 1 - 2}
-3k + 2l F,
k + 1 + 2F .
Purther, we llave
25(RT - 8 2 ) = (4k - 1- 2F)(k + 1 + 2F) (-3k + 2l F)2
= -5k 2 + 15kl 51 2 _ 5F 2
0,
so that
(13.'1)
RT = !p .
We now treat t1lree cases:
(i) R = S 0, (ii) R  0, S = 0, (Hi) S  0 .
Case (i): R = S = O. From (13.3) we have 'lk - 1 - 2F = 0 . d -3k
21 - F = 0, so that k = F 1 - 2F B k I ' . . .' an ,+
,- . ut , are positIve COprIme integers, so
SOLUTIONS
65
F = ], k = J, 1 2. In this case (13.0) has two solutions (x, y) ::1:(0,1).
Case (ii): R  O,S O. From (13.4) we have T = 0, and so from (13.3)
we obtain
{ -3k + 2l - F  0 "
k + 1+ 21" = O.
so that k = 1 = -F. As k, I are positive coprime integers we have P -1, k =
1= L In this rase (U.O) ha" two "oll1l.ions (:;::,y) ::1:(1.0).
Case{iii): S  O. From (13.4) we ha.ve RT> O. If R < 0 then P < () and
Wt' navf' k = H + T < 0, contradktillJ:'; k 2': 1. Ilenc!' U and Tar" l'0sitiw
integers. !\i ext, ohberve tha,t
(-1k
2F)(4k 1 + 2F) (1k .._lj2 - 41"2 = 5(l- 2k)2 ,
so that
(13,5)
R(4k 1 + 2F) (1- 2k)2 .
Clearly, we have 'lk 1 + 2F  0, otherwise 5R -:c -4F and so 5 I P, con-
t.radicting GCn(F,5) 1. Hence we may define nonnegative integers tL,b,c
by
(13.6)
2 a II R, 2 b II 4k - 1 + 2} 2 c III - 2k .
We have from (.13.5) and (13.6)
(13.7) a + b 2c
and
(13.8)
( l- 2k ) 2
2<.' '
2 a
where
R
2 a '
are odd positive integers. Suppose that
2 C
Gcn ( R 4k-l + 2F ) 1
2 a ' 2 b >. .

66
SOLUTIONS
Then there is an odd prime p which divides R/2° and ('1k -1 + 2F)/2b, and
thus p divides "k 1 - 2f 4l' -1 + 2F, and 1 - 2k, giviIlg successively
p ISk 2l, pi 4k l, p 12k, pi k, p' 1 ,
cont,ra.di<-.tiIlg fie D( /':, l) 1. lIence we have
(T3}))
(in lJ ( !!. 'i + }= ) = I .
.2°' 2 b ,
Fm!l1 (I :U) a1l(1 (l;i.9) we 8<>e t 11'1,(.
(13.10)
.!i =X2 ,
2 a
for some integer X. '!\ext we show that a is even. This is clear if a = 0 so we
may SUppose that a  1. Thus 21 R and so 1 is even. As GCDU",l) = 1 we
have k odd. Then, taking -k 2 + 3kl 1 2 = F2 successively modulo 2,4 and 8,
we get
(13.11 )
F
(mod 2) ,
(13.12)
1==2 (mod 4) ,
(1:!.13)
1 == 2k (mod 8) ,
Thus we hav(' 4/' 1:f: 2F 0 (mod .1) and so a  2,b  2. Also we have
2 mU, (a,b) I (4k -l + 2F) - (4k - 1 - 2F) 4F,
and so as F is odd we have min(a, b) :::; 2. If a :::; b then we have a:::; 2, which
implies that a 2. If b < a th<m b :::; 2, which implies that b 2, a = 2c _ 2.
In both cases a is even as asserted.
Setting a 2d, Xo = 2 d X, we have R = x. Then from (J:J.4) We deduce
that '1' yiS, S :l:xoYo. Changing the sign of Xo if I1ccessary we may
suppose that S = XoYo. Thus we obtain x + YiS :::: R + '1' :::: (5R + 5'1')/5 ::::
(4k-1- 2F+ l: + 1 + 2F)/5 = l' and x5 +2xoYo + 2Y5:::: R+2S +2'1':::: I, so
that (xo, Yo) is a solution of (13.0).
J:
,
!
SOLUTIONS
67
I 1. , f ( 130 ) Then using (13.0) we have
:\ow let (x,y) be any so u IOn 0 ,. .
F 2 =-k 2 +3kl l2 (x 2 +xy_y2?,
th - t (WI ' t! 1 F chosen t.o sath-.fy (13.1»
.'-\0 d. j
x2 + xy y2 = :1: F .
( 13.14)
Solving (13.0) and (1;\.14) j{)'r :2, xy. y2, we get
{ £);,.2 .lh 1-1 2F,
5xy -3k + 21 :f: t',
(l:tID) 5y2 = k + 1 =F 2F .
As
F == 2(k+ l) tc 0 (mod 5)
the lower signs must hold in (13,15), and so
{ x2 :::: (4k-l-2F)/5,
xy = (-3k + 21 P)/5,
(13.16) y2 :::: (k + 1 + 2F)/5 .
.. " lution of (13.0) we must have that (13.16) holds
Since tillS IS true lor any so . h 1.
with x, y replaced by xo, Yo respectively. ThIs means 1. a
x 2 x, xy XoYo, y2 Y5,
giving
(X,y) (xo, Yo), or (-xo,-Yo) ,
. h t (13 0 ) has exactly two integral solutions.
and provmg 1. a '
14.
(14.0)
Let T and 8 be non-zero integers. Prove that the equation
2 2 ) 2 1
(r 2 - S2)x 2 - 4T.XY -(r - s y
has no solutions in integers x and y.

68
SOLUTIONS
Solution:
We SUppose that x and yare integers satisfying (14.0). Factoring
the left side of (14.0), we obtain
(14.1)
«r - s)x (, + s)y)«r + 8)X + (, s)y) == 1.
As each factor on tIle left side of (14.1) is an integer, we s that
(14.2)
{ (r - s)x - (, + s)y == f,
(,ts)x+(r-s)y f,
where f =:; :1:1. Solving (H.2) for x and y, we ohtitin
(14.3)
,f -Sf
x-_ y_
-r 2 +s 2 ' - +
Hence we have (x 2 + y2)(r 2 + 8 2 ) == 1, so that ,2 + s2 1, that is
("s) == (:1:1,0) or (0,:1:1),
which is impassible as rand s are botll non-zero, thus showing that (14.0)
has no integral solutions.
15. Evaluate the integral
I L 1 1nXln(1_ x) dx .
(15.0)
Solution: The function In x In(l- x) is continuous for ° < x < 1, but is not
defined at x == ° and x == 1, so that
(15.1)
1== lim
£-+0+
S-+o+
1 1 - S
 Inxln(l-x)dx.
For x satisfying
(15.2)
O<f.:$x.:$1-6<1,
SOLUTIONS
'\ ...
f
J'
.II;
\\:
.  r
, '
,.
69
11,IId n a positive integer, we have
00 xl.
-In(l x) = :L k
1.=1
n 1. 00 xk-(n+1)
:L  + x n +1 :L
k=1 k k=n+l
"nd so
Iln(l
n xk!
x)+ k'
1:=1
00
xn+l
11.+1+
k=O
n+l '"
"xk
.:$ L.-
n+ 11.=0
xn+l
(n + 1)( 1 - x) .
=
Thus we have
111-S Inx(ln(l
(15.3)
n xl. ) I
x) + {; k dx
-I-S xn+l
1 1 (-lnx)- ) dx.
.:$ (n+ f (l-x
Now, for y  1, we have
l Ydt< 1 11dt y 1.
(15.4) O:::;lny= 1 T- 1
Taking y == l/x in (15.4), we have
( 1 ) <1_1=1 x
(15.5) 0 :::; In x = In ;; - x x
. ( 5 5 ) ' ( 153) we deduce
Using the inequality 1. I l n 1 1 - S ' ..;;-. 1 1 1 - S xklnxdx l
 Inxln(l- x)dx +  k 
1 1 1 - S
:::; - xndx
n + 1 f
 (1 xndx
< n + 1 10
1
( n + 1)2 '

70
SOLUTIONS
SOLUTIONS
71
a.nd letting n ---> 00, we obtain
We llext show thal
(15.6)
j l-6
, In x In(l x)dx
'X> 1 1-6
k j xklna:dx.
k=l f
(15.10)
lim (InE:)A(E:) := 0,
(';-+0+
As
( a:k+lln:r. xk+l
d.r ' + 1 (/.: +
:r. k In x,
(15.11)
lim (In(l- o»A(I- 0):= 0,
6-->0+
hy the flludam('ntal t1worem of "akulu.-;, W'. ha,\'f'
(15.12)
limB(c)=O,
(---)00-1
1 1-'"
, xk In xdx
( 1- 0)k+1
(k+1)2
Ck+t
(15.13)
11"2
lim B(l - 0) = 2 - _ 6 '
6-->0+
so that by (15.6)
so that (15.1) and (15.7) give
j I-6
f Inxln(l- x)dx
00
InE:L
k=l
"" c k + 1
- E k(k+ 1)2
00 (1 0)k+1 00 (l 0)k+1
0) k=1 k(k + 1) + t; k( + 1)2 '
(15.14)
11"2
[=2--,
6
In(1
as asserted in the HINTS.
Before proving (15.10)-(15.13) we show that
that is
(15.15)
lim (Inc)ln(l- c):= O.
(':-+0+
(15.7) 11-61nxln(l_ x)dx
(InE:).l1(c) B(f) (In(l o))A(l-o)+B(1
where, for 0 < y < 1, .t1(y) and B(y) are defined by
For 0 < I' < 1 we have
In(1 - €)
1'2  {
f+-+-+'"
2 3
> 1',
< C+f2++...
f
1- I' '
0) ,
(15.8)
A,(y)
0.0 k+J
t; k(,+ 1) '
so that
-€In c < (In c)ln(l- 1') <
dne
1 I'
(15.9)
B(y)
,'.0 yk+]
E k(k + 1)2 '
from which (15.15) follows, as
(15.16)
lim €ln I' := 0 .
,-->0+

72
SOLUTIONS
SOLUTIONS
73
Now for 0 < t < 1 we have
A(t)
O k co k+1
(L L t
k=1 k k=1 + 1
-dull f) + 111(1 to) + f
(1- f)ln(1-/) + <:,
plUving (15.13), and completing the proof of (15.H),
16. Solve the recurrence relation
so thai
(16.0 )
t ( n ) 0,( k) = !.:, 11 1, 2, .. ..
k=l k 11. + 1
Jim (In(),,1(c) O.
(. -0+
'nis proves (lr,-,o).
Kext we have, by Abel's theoreID,
lim A ( 1 6 )
6-+0+
. = (1 '- bY+! 00 1
hm L - "
6-+0+ k=1 k(k + 1) - f;;;;, k(k + 1)
1 ,
Solution: W('m,1kt'lh"illdlldiwhypot!i""i:;thal a(,I) ( 1)"H/(1I1 I)
for all positive integers n satisfying 1 :$; 11 :$; m. This hypothesi:>
is true for 'In 1 as a(l) 1/2. Now, by (16.0) and the inductive hypothesis,
we have
( ) _ m + 1  ( m + 1 ) (-I)k+ 1
am+l - -L...
m + 2 k=1 k k + 1
Thus we must show that
so that
61!..W+ (In(1- 6)A(1 - 8) = In 1 0 .
This proves (15.11). Also we have
IB(f)1 :$; f L
k=1
f ( m + 1 ) (-1)k+1 == m+ 1- (_l)m
1=1 k k + 1 m + 2
00
so that
or equivalently
lim B(t) == 0,
(-+o+
proving (15.12). Finally, by Abel's theorem, we have
. 00 / 1 _ C ) kH
lim B(l 8) == Inn" \ "
6....0+ 6-+0+ f;;;;, k( k + 1)2
">0 1
t; k(k + 1)2
'x- 1 I
.t;( k(k + 1) - (k + 1)2 )
( 11'2
= 1- 6-1)
11'2
= 2-
6 '
 ( m+1 ) m+ 1
k=1 k k + 1 = m + 2
By the binomial theorem, we have for any real number x
(16.1 )
m+l ( m + 1 ) k
O+x)m+1 L k X.
k=(J
Integrating (16.1) with respect to x, we obtain
(1 + x )m+2 m+1 ( m + 1 ) xk+1 _ -2-
(16.2) m + 2 t;) k k + 1 + m + 2 .
:raking x = -1 in (16,2) we have
I: ( m+ 1 )
k=O k
1
=- m+2 '

74
SOLUTIONS
SOLUTIONS
75
and so
This estahlishes the existence of a pair of integers as required.
( m + 1 ) (_I)kH = 1- 1 =
k==1 k k + 1 m + 2
m 1
m+2
18.
Let
as required. The result now follow!> by the principle of mathematical induc-
tion.
111
Un"" 4n+l + 4n+a - 2n+ 2' n=O,l,...
17.
Let nand k be positive integers. Let p he a prime such that
Do('s the infinite s('ris I:=() 11." conv('rgp, and if so, what is it>; sum?
p> (n 2 + n + kj2 + k .
Solution: Let s( N) I:==() an, N = 0,1,. . .
We have
Prove that the sequence
(17.0)
n2, n 2 + 1, n 2 + 2, . . ., n 2 + 1 ,
s(N)
Nil 1
E Cn+l + 4n+ 3 2n+2 )
N ( 1 1 1 1 1 1 )
.- n=() -i n + 1 - .1n + 2 + 4n + 3 - '1n + 4 + 4n + 2 -- 4n +4
4NH ( -1 )m-l 1 2N+2
L m +2 L
m=1 m=1
where 1
that
(n 2 + n + k)2 n 2 + k, contains a pair of integers (m, m + k) such
( ; ) = ( m; k ) 1.
m
Solution: As nand k are positive integers and p > (n 2 + n + kp + k,
2 none of the integers of the sequence (17.0) is divisible by p. If
( n:k ) 1 we can take (rn,m+k) (n2,n2+k).If(J 2+k ) 1 we can
take (m, m + k) = «n + 1)2, (n + 1)2 + k). Finally, if P
( n 2 : k ) ( n + 2 + k ) -1,
we can take (m,m + k) = «n 2 + n+ k)2,(n 2 + n + k)2 + k), as
( n 2 + n; k? + k ) ( n 2 + k)«: + 1)2 + k» )
( n2:k )( n+2+k )
(-1)(-1) 1.
Letting N . (X; we 11ave
lim s(N)
N-+-;'JC)
00 (_l)m-l
L +
m=1 m
3 00 (_I)m-l
2 L m
",=1
3
2 1n2 ,
1 00
2 L
1n=1
m
so that I:=() an converges with sum  In 2.
19. Let 11.},... ,G.", be m ( 2) real numbers. Set
An 11.1 + 11.2 + ... + an, n = 1,2,..., m .

76
SOLUTIONS
SOLUTIONS
77
Prove that
(19.0)
Solution:
and so
(19.1)
But as
we have
that is
(19.2)
Using (19.2) iD (19.1) we obtain
f ( n r  12 f a .
n2 n=l
For 11. 1,2,...,m we have
tha t i
( n r (a n + 11. anf
2 ( An ) 2
< 2a +2 --an
n 11.
2 ( An ) 2 An
4a n +2 - -4a n -,
11. 11.
(19.:-1 )
f( n r 
n=l
m A 2 m A2
+2;( nn ) - 2E n(nl)
m A2
2" n
!;t 11. 2 (11.+ 1) ,
m
4La
n=l
m
4La +
n=l
£:(1
n=l
2
( n r4f>.
n=l
11.+
The inequality (19.0) now follows from (19.3) by noting that 1
n =: 1, and 1 2': l for n 2': 2.
 ( n r 4a + 2 ( n r -4an n .
= 0 when
20.
Evaluate the sum
n (n)
S = L ( 2nl )
k=O k
-2anAn = -(A
A;_1) - a  -(A - A_l)
for all positive integers 11..
Solution: We have
-2
n=l
11.
m ( A2 _ A2 )
n n-l

n=l
m-l A;
-L
n=l n(n+ 1)
A2

m
so that
m A m A2
-2 L an < - L n .
n=l n - n=I11.(n+l)
()
ek')
(k:l)
(k 2 ;1 )
11.! (211.- k)! 11.! (211.- k -I)!
(11.-k)!211.! (n-k-l)!2n!
n!(2n-l-k)! ( 2n-k) _ (n-k» )
(11. - k)! (211. - I)! 211. 211.
1 (n)
= -
2 e n ;l) ,
S = 2 t ( G) _ (k:1) )
k=o ek') (kl)

78
SOLUTIONS
SOL UTIONS
79
= 2 ( ()
e;)
2.
(nl) )
(n;\)
values of t satisfying (21.3). Hence we have
(21.5)
N(k) = []
[ 7] -
>'(b,g) + 1 ,
and so
21. Iet II. and h be coprime positive int...!!;ers. For k a positiv(' intep;er,
Id IV (k) denote the 111l1llber of integral solution" \.0 the equation
IN(k)
h _ [L I < 1 + 1 + 1 + 1 4,
a b-
giving, by (21.1),
(;ll.n)
It./: + by:::: h, ,r  0, !I?!J .
( ;ll.{j)
I N(k) I I ,4
---- -- -. < -
k ab'-'k"
1..
lim N(k) .
k--++oo k
Letting k -+ +f)Q in (21.6), we obtain L = l/ab.
Evaluate Ute limit
22. Let a, d and r be positive integers. For k = 0,1, ... set
Solution: As a and b are coprime there are integers g and h SUell tbat
(22.0)
1
Uk uk(a,d,r) = (a + kd)(a + (k + l)d)... (a + (k + r)d) .
(21.1) ag+bh=k.
Evaluate the sum
n
S = L Uk ,
k=O
Then all solutions of ax ,I, by k are given by
(21.2)
where 1/. is a positive integer.
x 9 + bt, y = h - at, t 0, :1:1, :1:2, .. .
Thus the solutions of (21.0) are given by (21.2) for tbose integral values of t
satisfying
Solution:
Fork -1,0,1,... we set
(21.3)
h g
- > t > -- .
11.- - b
(22.1 )
Vk = vk(a,d,r)
1
(a t (k + 1 )d) . . . (a + (k + r )d)rd '
so tbat
Set
( 21 4 ) >. ( b ) = { 0, if b divides 9 ,
. ,g 1, if b does not divide 9 ,
Tben there are
[] - [ 7 ] - A(b,g) + 1
Vk - Vk+l
1 ( 1
= (at(k+2)d)...(a+(ktr)dird (a+(k+1)d)
- (a + (k + r + 1 )d» )
1
= (at(k+1)d)(a+(k+2)d)"'(a+(k+r)d)(a+(k+r+1)d) '

80
SOLUTIONS
SOLUTIONS
81
that is Vk - VkH Uk+l' Hence we bave
n n-] 'n-l
S = L Uk = L Uk+I = L (VI, - Vk+l) LI - 'I'" ,
k=O k=--.] k=-]
Solution: By the Cauchy-Scbwarz incquality we have
that is
-' 1 ( 1 1 )
.. 'I'd a(a+,/)...(tI+('I'-nd) - (a+(n+l)d)--.(+c;+ r)d) .
(24.1 )
C" ) 2
.a"
1ft 1 ) 2 1ft m 1
(a"vn y'n S na  .
Xpxt, we have
23. Let xJ,... ,x" be n (> 1) rea.l numbers. Set
Xij=Xi Xj (ISi<jsn).
Let F be a real-valued function of the n(n - 1)/2 variables :rij such tbat the
inequality
 1 < 1+ {1ft d,x <1+ {"'!!!.
L.. n - J] X - JI..jX
n:;;:;.::l
1 + (2.jm - 2) 2..jffi - 1 < 2,fm .
We obtain (24.0) by using the lattN' inequality in (24.1).
(23.0)
n
F(Xll,XI2,...,X n _ln) S Lxi
k=l
25. Prove that therr exist inJinitcly many positive integers which are
not expressible in the form n 2 + p, where n is a positive integer and p is a
prime.
holds for all a], . . ., Xn.
Prove that equality cannot hold in (23.0) if Lk=l Xk -10.
Solution:
Set Ai = (Xl + ... + x,,)/n, and replace each :r.i by Xi - AI in
(23.0). Then (23.0) gives the stronger inequality
.!. ( t Xk ) 2 .
n k=]
Solution: We show that the integers (3m + 2)2 , m 1,2,..., cannot be
expressed in the form n 2 + p, where n ;::>: 1 and p is a prime. For
suppose that
" n
F(xJ1, XI2,... ,X n -] n) S L(:r.k - M)2 L x
k=I k=]
lIellce if Xl, . . . ,.'l'n are chosen so that Lk=l x k -I 0, equality cannot hold in
(23.0).
{3m + 2?
n 2 + p,
where n ;::>: 1 and p is a prime, then
(25.1 )
p (3m+2 n)(3m+2+n).
Since p is a prime and 0 < 3m + 2 - n < 3m + 2 + n, we must have
24. Let a],... ,am be m (;::>: 1) real numbers which are such that
L::'=I an -I O. Prove the inequality
(25.2)
3m + 2 n = 1,
3m + 2 + n p .
(24.0)
( fna ) / ( fa n ) 2 > 2 m '
,,=1 n=1 V
Solving (25.2) for Tn and n we get
m (p - 3)/6,
n (p 1)/2,

82
SOLUTIONS
SOLUTIONS
83
so that p 3(2m+ 1). As P is prime. we must have rn 0, which contradicts
m1.
26.
Evaluate the infinite series
27. Lei 111",.,]in d('nol. n
the polynomial of degree n givcn by
1) distinct inhJ';ers and let In(:r:) be
s f arctan ( 22 ) .
n=1 n
/,,(:1:) = (x - PI )(:r: 1'2 L.. (x - p,,).
Prov{' that the polynomial
Solution: For 11.  1 We have
!/n(:r:)::: U:"'t:r:»2 + I
arctan ( ! ) - arctan (  )
n 11.+2
arctan (  - nr )
1 + n(n+2)
arctan ( 11.: 1)2 ) ,
cannot be exprcssed as the product of two non-.consta,nt polynomials with
integral coefficients.
Solution: Suppose that Yn(;r;) <:an be expressed as the product of two non-
c.onstant polynomials with integral coeHicients, say
80 that for N  2 we have
t arctan ( 2 2 )
n=2 n
(27.1 )
9n(X) h(x)k(x) .
3=1 arctan ( 2 ,, )
n=1 . n + 1)-
I:I ( arctan ( ! ) - arctan (  ))
n=1 n 11. + 2
::: arctan(l) + arctan () - arctan (  )
arctan (  ) .
N+l
Neither hex) nor k(x) has a rea.l root as 9,,(X) > 0 for aJlrcaJ. x. TIms,
lLeither hex) nor k(x) <:an change sign as x takes on all real values, and we
may suppose that
(27.2)
h(x»O, k(x»O, forallrealx.
tetting N -7 00 we get
E arctan ( :2 ) arctan( 1 ) + arctan G) =  + arctan G)
Since gn(Pi) 1, i 1,2,...,11., we have h(Pi)::: k(p.) 1, i::: 1,2,...,11..
If the degree of either h( x) or h(:f.) were less than n, then the polynomial
would have to be identically 1, which is not the case as h( x) and k( x) are
nOll-constant polynomials. Hence both h( x) a.nd h( x) have degree 11., and
(27.3)
{ hex) 1 +a(x - pd'''(x Pn),
k(x)=l+b(x pIJ...(x Pn),
S :::  + arcta.n(2) + arctan () ::: 3 4 7£' .
for integers a and b. Thus we have
(27.4) (x - pd(:/': - P'l)2 ...(:r: - Pn)2 + 1
=1+(a+b)(x pd...(:r:-Pn)+ab(x-Pl)2...(x- p,,)2.
and so

84
SOLUTIONS
Equating coefficients of x 2n and x" in (27.4) we obtain
(27.5)
{ ab 1,
a+b =0.
Thus we have a contradiction as no integers satisfy (27.5).
28. Two people, A and B, playa game in which the probability that
A. wins is p, th( probability that B wills is q, and tilt.' probability of a draw is
r. At the beginning, A has m dollars and B has n dollars. At the end of each
game the winner takes a dollar from the loser. If A and B agree to play until
one of them loses all his/her money, what is the probabilty of A winning all
the money?
Solution: Let p(k), k = 0,1,. .. , denote the probability that A wins when
he/she has k dollars. Clearly, we have
(28.1)
p(O) = 0, p(rn + n) 1.
We want to determine p(m). Consider A's chances of winning when he/she
has k + 1 dollars. If A wins the next game, A's probability of ultimately
winning is ap(k + 2). If A loses the next game however, A's probability of
ultimately winning is bp(k), while if the game is drawn, A's probability of
ultimately winning is cp(k + 1). Hence we have
p(k + 1) = ap(k + 2) + bp(k) + cp(k + 1) .
As a + b + c = 1 we deduce that
ap(k+2) (a+b)p(k+l)+bp(k)=O.
Soving this difference equation, we obtain
P( k ) { A + Bk , if a = b ,
. = A + B(b/a)k ,if a l' b ,
SOLUTIONS
85
where A and B are constants to be determined. Using (28.1) we obtain
{ A= 0,
A= -B
B=I/(rn+n) ,ifa:::b,
1/(1 (b/ar+ n ) , if a =F b,
so that
{ m/(m+n) ,if a b,
p(m) = (1 _ (bitt)"'') I (1- (blar n + n ) ,if a =F b.
29 Let f(x) be a. monic polynomial of degree n  1 with complex co-
efficients. Let xl:'" ,X n denote the n complex roots of f(x). The discriminant
D(f) of the polynomial f( x) is the complex number
(29.0)
D(f) II (x. - Xj)2.
l:<;;i<j:<;;n
Express the discriminant of f(x 2 ) in terms of D(f) .
Solution: As Xl,' . . , X" a.re the n roots of f( x), the 2n roots of f( x 2 ) are
Y1
,fiil, Y2 ..jX2,... ,v" =,,;x;; ,Yn+l = -.;xi, ... ,Y2n = -,,;x;;.
Hence, the discriminant of f(3: 2 ) is
II (Yi - Yj)2
1 :<;;i<j:S2n
II (Yi - Yj? II (Yi - Yj)2
l:<;;i<j:<;;n l:<;;i:<;;n<i:5;2"
II (Yi - y;)2
n«<;:<;;2n
(.[Xi - ,jXj)2 II (VXi + .Jx;-n?
1$i$n<;2n
II
l:<;;i<i.:<;;n
II (- .JXi-n + .JX;-n ?
n< i<j :$2n
II (.[Xi -,;x;l II (VXi + ..jXj)2
19<;$" :

86
SOLUTIONS
II ( Vii
I9<.iSn
II ( -Vii + vfXj)2
l$i<iSn
vfXj)4 II (.JXi + vijl
ISi<iSn
II (2y'iJ2
lSiSn
n
n (a:, .l:J)4 2 2n n Xi
I:$i<.iSn i=.l
2211.( -l)nJ(O)(l)(J)) .
, O. ,.Prove tht for each positive integer n there exists a drcle in the
xy-p ane wInch contams exactly n lattice points.
Solution: L t P b h t;>
I e. e.t e point (v 2 ,1/3). First, we show that two different
attIce pomts R - ( X ) d 8 (
d' t f ' - 1, Yl an X2, Y2) must be at different
15 a ld llc i es rom P. For if Rand 8 were at qual distances from P then we
wou lave '
(Xl - V2? + (VI - )2 == (X2 - V2)2 + (Y2 _ )2 ,
so that
(30.1 )
2(X2-:l:I ) V2 2 2 2 2 2
x2 + Y2- Xl YI + 3(Yl 112).
As V2 is irrational, from (30.1) we see that x - x - 0 . d h 2 2
(Yl - Y2) = 0, that is 1 '2 - ,an ellce Y2 - Yl +
(Y2 Vl)(Y2 + Yl 2/3) O.
Since Y1 and Y2 are integers, 'we have Y2 + Y - 2/3 t= 0 . d
contrary to the fact that 1l and 8 are a:;slJme distinct.. ' an so Yl = Y2,
SOLUTIOl.'\S
87
Now let n be an arbitrary natural number. Let C be a circle with centre
P and radius large enough so that C contains more than n lattice points.
Clearly C contains a finite numher m (> n) of lattice points. As the distances
from P to the lattice points are all different, we may arrange the lattin' points
inside C in a sequence PI, 1'2, . .. 'Pm' according to their increasing: distances
from P. Clearly, the circle C n with centre P, passing through Pn+1' t:ontains
f>xae1.Jy u. Ja,ttifE-' points.
31. 1,('1, f1 1)(' a /!,ivf'1I 1I{)1J-IH',a.l.iv(' int.I'W'f. Ikt"l'lilinp I.h.' numlJf'f
8(n) of solutioHs of the equation
(31.0)
x+2v+2z=n
in non-negative integers x,y,z.
Solution: We have for It I < 1
0(,
L 8(n)t n (1 + t + t 2 +... )(1 + t 2 + t 4 +... f'
1'1=0
=
1
(1 - t)3(1 + t)2
3/16 1/4 1/4 3/16 1/8
It + (] t)2 + (1 - t)3 + 1+t + (1 + t)2
3 ov 1 ov
= 16 L t n + 4 L(n+ l)t n
n=O n=O
+! I: (n + l)(n+ 2) tn + 3 I:(_1)ntn
4 n=O 2 16 n=O
1 (X)
+8' L( -It(n + l)t n
n=O
1 ""
16 L(3 + 4(n + 1) + 2(n + l)(n + 2) + 3( -It
n=O

88
SDLUTIONS
+21-1)n(n+ l))t n ,
giving
S(n)  {
n(n+ 6)
-+1
, if n is even,
,Cn+ 1)(n+3)
, if n is odd.
R
32. Let n be a fixed integer>') D .
are bounded for 0 < ' . - -. etermme all functioThs f(x). which
x < a, and WhiCb satisfy the functional e(!Iation .
(32.0)
f(x)==: :2 (f(;) +f( x:a ) +...+f C+(:-l)a )).
Solution: L t f( ) b
e ;c e.a bounded function which satisfies (32.0 ) for 0
. a. As f( x) IS bounded on ( 0 a ) th ' . . ' . < x <
Ii. such that . ere eXIsts a pOSItIve constant
(32.1 )
If(x)1 < K,
O<x<a.
l'br s - 0 1
- , ,..., n - 1 we have
0< :r.
so,
< a,
ifO<x<a,
n
so that by (32.1) we obtain
I 'f ( X + n sa ) I < K ,
o .s; s .s; n - 1, 0 < x < a .
Then, for 0 < x < a, we have from (32.0),
1
Jf(x)1 < 2([( + K +... + K )
n ,
SOLUTIONS
89
that is 1ft x)\ < J{ In. Repeatinj!; the argument with the bound J( replaced by
Kin, we obtain
If(x)1 < Kln 2 ,
O<x<a.
Continuinl!: in this way Wf' get
(32.2)
'f(x)1 < Kin',
O<x<a,
for I = 0,1,.,. , ;Hd letting 1 -,. 00 in (3:.!.2) !!;ives !(;c) 0 for () <: :r < a.
33. Let I denotl the dosed intervdJ [u,b], a < b. Two functions
f( x), g( x) are said to be complctely different on I if f( x) # !I( x) for all x in L
Let q(x) and 1'(X) be functions defined on I such that the diffrential (.quation
dy
- = y2 + q(:r)y + t'(x)
dx
has tbrel} solutions VI(X), V2(X), V3(X) which are pairwise completely different
on I. [f z(x) is a fourtb solution such tbat the pairs of functions z(x), y,(x)
a.re completely different for i 1,2,3, prove that there exists a constant
[( (# 0,1) such tlta.t
(33.0)
 _ YI ([( Y2 - Y:3) + (1 J()Y2V3
If( - l)YI + (Y2 - KY3)
Solution:
Y4 are pairwise completely different 011 I, the
As VI. Y2, Y3, Z
function
(33.1)
f(x)
(YI - Y2 )(Y3 Y4)
(YI - Y:d(Y2 - Y4)
is well-defined on I. Also, as Y1:Y2,Y3,Y4 are differentiable functions on J,
f( x) is differentiable there alld its derivative is given by
l(x) = g(x)
(VI - Y3)2(Y2 V4)2'

90
SOLUTIONS
where
g(x)
(Y - Y)(YI- Y3)(Y2 - Y4)(Y3 - Y4)
(YI -- Y2)(Y .- !IJ)(Y J/4)(YJ .'1.d
(YI - Y2)(YI Y3)(Y -- Y;I)(Y:J - Y4)
+ (!II - YZ)(YI 1/:I)(Y2 !l4)(Y(, Y'l)
,\"
:If.
Z
Y.., + IIY. r, -'
I, :J,:1. 1 ,
we have
g(x) ((YI + Y2 + q) (YI + !/3 + q) - (Y2 + 1/4 + q) + (Y3 + Y4 + q))
(YI - Y2 )(UI Y3 )(;112 - Y4 )(113 11,1\,
that is g( x) = 0, and so /'( J:) := 0, showing thltt I( x) = J( on I for some
COIlstant ](. Finally, (33.0) is obtain(.-'d by solving; (33.1) for z Y4. J( i= 0,1
as ;; i= Y3,!l1 respectively.
34. It. an, 1). 2, ;, , .., denote the number of ways the I Ho duct
b2'" b n can be bracketed so that only two of the bi are multiplied together
at anyone time. For example, a2 = 1 since b I b 2 can only be bracketed
as (bJb z ), whereas a3 = 2 as b l b 2 b 3 can be bracketed in two ways, nanlely,
(bI(bzb:,» and «b J b 2 )b,,). Obtain a formula for an.
Solution: \Ve set al =- 1. The number of ways of bracketing b l b 2 .. . b n +1 is
n
I:: N(l,i) N(i + 1, n + 1),
i=I
where N (i, j) denotes the number of ways of b1'lcketiIlg bibi+1 . . . bi' if i < j,
and N(i,j) l,ifi=j. Then
(34.1)
a n +l = alan + a2 a n_J + '" + a n -Ia2 + anal, n =- 1,2,... .
SOLUTIONS
91
Set
( 34.2)
00
A(x) = I::anx n .
n=l
From (34.1) and (3-1.2) we obtain
A(x)2 = (u,.i) (a)xj)
('::-.0 '.1
I::L:
n=l I,J;;;;1
'+3=1'1-+1
(tiU,)X'+)
'" a'a Xi+i
u ')
i,i=l
00 ) n+1
I::(ala n + aZUn-1 + ... + anal x
n=l
00
I:: a n +1 Xn + 1 = A(x) - x ,
n=l
that is
(34.3) A(xi - A(x) + x = O.
Solving the quadratif equation (34.3) for A(x), we obtain
A(x) (I::1: Vl-4x )/2.
As A(O)
(34.4)
o we must have
A.(x) = (1 V1- 4x) /2 .
Hy the binomial theorem we have
(34.5) VI - 4x :=  e2)< -4Y'x n ,
so that, from (34.4) and (34.5), we obtain
(34.6) A(x) -  e2)(-4Y'Xn .

Equating coefficients of x n (n  2) in (34.6), we obtain
an = -e2)(-1r22n
(_1)n-I22n-I e2)
(_l)'H22n-I( _1)n- I 1.3.5... (2n 3)
2 n n!
{ get) =tt.an 2 t-3tant+3t,
g'( t) = sm} (2t - sin 2t) .
cos t
Hence g'(t) > , 0 <.t  1, .which implies that get) > g(O) = 0, 0 < t  1. We
deduce that f IS an mcreasmg function on 0 < t  1, so that
92
thal is
35.
(35.0)
Solution:
(35.1 )
We set
and deduce that
where
SOLUTIONS
SOLUTIONS
93
that is
1 tan t - t
'3  t 3  tan(l) - 1, 0 < t  1.
Since tan(l) < tan(-n-j3) = .J3 < 2 , we have
tan t - t
o < t 3  1, 0 < t  1 ,
which completes the proof of (35.1).
For 0  x  7r and 0 < y  1 we have 0  sin x  1 and so
1.3.5... (2n - 3) I
an = , 2 n -, n  2 .
n.
(35.2)
o  ysinx  1 .
Hence, by (35.1) and (35.2), we have
Evaluate the limit
ysinx  tan(ysinx)  ysinx + (ysinx)3,
. I i "
L = 11m - tan(ysinx) dx .
".....0 y 0
so that
tan(ysinx)- ysinx 2. 3
O ysmx.
y
(35.3)
We begin by showing that
ttantt+t3, Ot1.
Integrating (35.3) over 0  x  7r, we obtain
(35.4)
O! r(tan(ysinx)-ysinx)dxy2 rsin3xdx.
y 10 10
f(t) = (tant - t)/t 3 , 0 < t  1,
Letting y --+ 0+ in (35.4) we deduce that
lim ! 1 " (t an( y sin x) - y sin x) dx = 0 ,
".....0+ Y 0
f'(t) = g(t)/t4, 0 < t  1 ,
and thus
lim ! 1 " tan(ysinx) dx = r sin x dx ,
".....0+ Y 0 10
that is
f(O+)  f(t)  f(l), 0 < t  1 ,
(35.5)
lim ! r tan(ysinx) dx = 2.
".....0+ y 10

U4
SOLU'I'IONS
Replacing y by -y in (35.5), we see that
(35.6)
1 1 "
lim - tan(y sin x) dx
y-+o- Y 0
2,
also. lknce, from (35.5) and (35.6), we find that L = 2.
36. Let t be a rea.l !lumber with I) <: t <: 1. Prove tlmt there a.re
infinitely lIla.ny integers n for which
(36.0 )
cosn  1 c.
Solution: According to a theorem of Hurwitz (1891): if e is an irrational
number, there are infinitely many rational numbeIs afb with b >
o and GCD(a,b) = 1 such that
Ie I <  b2 .
As w: is irrational, Hurwitz's thcorem implies that there are infinitely many
rational numbers nfk with k > 0 aud GCD(n,k) = 1 such that
12w:-1<
k 2 '
or equivalently
(36.1)
12w:k - nl < 1f(vI5k).
Let 0 < E < 1. vVe consider those integers nand k satisfying (36.1) for which
k > 1fh/5£). There are dearly an infinite number of such positive integers
k, and for each such k there is an integer n such that 12w:k - nl < £. For such
pairs (n, k) we have
1 cosn :$ 11- coonl
SOLU'I'IONS
95
21 sin ( br + %) }in (k1r  ) I
< 2\sin(k'ir %)\
1
< 21br - I
'2/.:1£' - 11\
< l'l'
"howiu!-!, I hal l)fJ.(J) hold" !'or inliuildy many inl('('r 11.
7 De termine aU the functions f, which are everyw here di1Terentiablp
3 .
and satisfy
(37.0)
f(x)+f(y)
f (  . +Y )
1 - xy
for aU real x and y with xy =I 1.
Solution:
Let f(x) satisfy (37.0). Differentating (37.0) partially with re-
sped to each of x and y, we obtam
1 + y2 f' ( X + y )
rex) (J - xyp 1 - xy
(37.1)
and
1+X 2 , ( X+Y )
(::W.2) f'(y) = (1 xy)2 f 1 xy .
. 37 1 ) d ( 3 2 ) we deduce tha.t
Eliminating common terms IT! (, . an . I. ,
2 " )
(37.3) (1 + x 2 )f'(x) = (1 + y )f(y .
As the left side of (37.3) delJends only on x and the right side only on y, each
side of (37.3) must be equal to a constant c. Thus we have
c
f'(X) = 1+
ill

96
SOLUTIONS
and so
f(x) c arctan x + d,
for sOllie constant d. However, taking y 0 in (37.0), we obtain f(:1;) + frO) =
f{.I;).1'O that f(O) = 0 and d O. Clearly f(x) carctanx satisfie" (37.0), ,
and so aU solutions of (a7.0) are given by i
f(:r) = (' antan. ,
\v1lcr(' c is a. const;mt,
38. A point Xis chosen inside or on a circle. Two perpendicular
chords AC and ED of the circle are drawn through X, (In the case when X
is on the circle, the degenerate case, when one chord is a diameter and the
other is reduced to a point, is allowed.) find the greatest and least values
which the sum H = 1.1C1 + IBDI can take for aU possible choices ofthe point
X.
Solution: We can choose an (x,y)-coordinate system in the plane so that
the centre of the circle is at the origin, BD is parallel to the
x-axis, AC is parallel to the y-axis, B lies to the left of D, and /1. lies above
C. Let X denote a point (1',8) such that
(38.1 )
1'2 + B 2 :::: 11 2 ,
where R is the radius of the cirde. Then the coordinates of the points
A.,B,C,D are
(r, VR2 - 1'2 ), (- V R2 - s2,B),
(1', - VR2 - 1'2 ),
( VR2 - s2 , ,9)
respectively. Thus we have
IA.CI = 2 VR2 - 1'2 , IBDI = 2 VR2 - s2 ,
and 1'0
S(r,s) IACI + IEDI = 2( VR2 - 1'2 + VR2 - B2) .
SOLUTIONS
97
We wish to find the ma,ximum and minimum values of H(r,s) subject to the
constraint (38.1). '. > . .
First we determine the maximum value of S(r,s). Clearly, w(. h,1V(
anti l;}riR prove!: tllat,
VRZ - r Z + Jii z .9 2 :::: '2R ,
max 8(r.s) 8(O,O)=:W
1 .. A:-s? '5: J {l
. I f S ( T S ) We haNe
Finally, we deterrmm\ the mimmum va ue 0',' , .
2HZ __ (1'2 + s2) + 2ff---.;:i ,jil2' .9 2
2: 2R 2 - (r 2 + s2)
2. 2R Z _ (1'2 + s'2) + (1'2 + S2) R Z
R 2 ,
(JR 1'2 + JR2 _ B2 )2
so that
This proves that
V R2 - r 2 + Jn 2 -" 8 2 2. R .
min SIr,s) = S(:!:l-l.O) = 8(0,:l:R)
,,2 + s2:SR2
2R.
39.
For n 1,2,... define the set A" by
An = {
Is it trlle that
{ 4 6 8 } if n 0 (mod 2) ,
0,2, , , ,... ,
{ 6 3( 1 )/ 2 } if n == 1 (mod 2) .
0,3, ,..., n-' ,
u ( n An+k )
n=1 k=1
n ( G r\Mk ) '!
n=1 k=l

98
SOLUTIONS
SOLUTIONS
99
Solution:
We set X = {O,2,4,6....} and Y = { o 3 6 9 }
have ' , , ,... . Clearly, we
for all n = 1,2,. . . , so that
CXJ
II I ell:.! c AT, c ... c U t12n+l = )'
n=O
D1 C91 An+k) = X U Y
Hence, we see that
and
,Ql ([51 A"+J,) # !.:\ (91 An+k) ,
ih  A4  A(; = . . . = X .
lIPIIC(', WI' have for n = L 2. __.
;),8 2 Iwloll/,:s to .Y l J Y hut. dill's IIOt. belong to X nY,
'X>
n ;\'n+ h
h=J
..x,
n An+h n
k=J
n+k=O (mod 2)
XnB n ,
0"
n A nH
k=l
n+k=l (mod 2)
40. A sequence of repeated independent trials is pprformed. Each
trial has probability p of being successful and probability q = 1 - p of failing.
The trials a,re continued until an uninterrupted sequm]Cp of n successes is
obtained. The variable X denotes the number of trials required to achieve
this goal. If Pk = Prob(X = k), determine the probability generating function
P( x) defined by
where
Bn = {
AnH, if n == 0
An+2, if n == 1
(mod 2) ,
(mod 2) ,
and so
(40.0)
CXJ
P(x) = LPk Xk .
k=O
u ( n AnH )
n=l k=J
CXJ
U (X n En)
n=l
Xn CQ En)
X n C91 A 2nH )
XnY.
Solution: Clearly, we have
PF{ D
pn
qpn
, k = 0,1,..., n - 1 ,
,k = n,
, k = (n + 1), (n + 2),..., 2n .
For k > 2n we have
On the other hand, we have
Pk = Prob(A) Prob(E) Prob(C) ,
CXJ
U AnH
k=l
CXJ
U .4. nH U
k=l
n+k=O (mod 2)
XUY
CXJ
U A nH
k=l
n+k=l (mod 2)
where A, E, C represent events as follows:
(A) no n consecutive successes in the first k - n - 1 trials;
(E) (k - n) th trial is a failure;
(C) n successes in last n trials.

100
SOLUTIONS
Then Pk (1 - rob(D))qpn, where D represents the event of 1
run of n consecutive successes in the first k n 1 t . I h a.t east one
. - - rIa s, t at is
Pk (1- klp.) qpn, k> 2n.
Hence we have
PC:!') = 1''':!''' + q1'''(x''+1 + '" + J.2n ) + ql n " k- n -l
L, (1 -. L p,);/:l-,
k=21<+1 '=0
and 1>0
!(x)
pnxn
l+q(x+'''+xn)+ q  xk-n_ q -{ k-J
L, L, L, Pi xk - n
k=2n+ 1 k=2n+ 1 i=O
1 (x - x"+1 ) xn+l 00 n+1
+1] (l-x)+q (l-x) -qx n + 1 LLP;x l
1=0 ;=0
(1 - x + qx) co n+l
(1 - x) - qx n +1 L Lp;x l
/=0 ;=n
== (1 - x + qx) 00 1
(1 x) - qx n +1 L LPn+r XI
1=01'=0 .
(1 - :c + qx) 00 1
- qxn+l,", '"' 1'+s
(1 x) L, L, Pn+,'X
1=0 T.S=O
r+s::::l
(1 - x + qx) 00 'XJ
== - qx n + 1 '"' '"' r+s
(1 - x) L, L, Pn+,'x
,'=0 s=o
(1 (1  : r) - qx ( f: pn+1'xn+r ) ( £>s )
1'_0 8=0
P(x) (1- x + qx) P(x)
pnxn (I-x) qX (l_x) '
P(x) = (1 px)pnx" .
1 - X + qpnxn+l .
that is
so that
SOLUTIONS
101
41. A, B, C, D are four points lying on a circ1e such that AB CD ifi a
convex qmldrilatera.l. Detel'lnine a. formula for the radius of the circle in term"
of a 1.4BI, b == IBCI, c "- ICDI and d ID.41.
Solution: We first prolle the following result:
Th#> radius of the circumcircle of a !':::.L.,'l-.fN is givt'n by
(.n.1)
lmn
In+m+n)(i+m-n)(l m+n)(-l+m+n) '
R
where
IMNI, m INLI, It ILM!.
Let C denote the circumcentre of !':::.LM ]\1, so that ILCI 1M CI IN CI == R.
Set
a LMCN, f3 == LNCL, 1== LLCAf,
so that 0'+ f3 + I 211'. By the sine law applied to .6.MCN we have
R
sinO' sin«lI' - 0')/2)
so that
sino .
l = R-. ( /2) = 2Rsm(0'/2).
cas a
SimiJarly, we have
m = 2Rsin(fJ/2), n = 2Rsinh/2).
Thus we obtain
n
2R
== sin(I/2)
sin (11' ( a  f3l )

102
SOLUTIOKS
' ( a 3 )
SIll -- + 
2 2
sin(oJ2) cosCi:}/'!.) + cos(a/2) siu( 8/2)
1 /--;;;2 m r [2
2R V l 1R i + 2R Vi  11(.2
;md so
1/
r-
IV I
. 1 --....
m. l I l2
- + 1f1 V I ---
lFl . .j Hl '
Squa.riul!> bof.llliid,'., WI, obtain
2 2 ( m2 ) . ( z2 ) 0 ( ) (
')). ==1 .1- 4R2 +m 1- . 1R i . +21mV I- 4H i ]
i;) ,
an d so
I --
[2 m 2
2lm ( 1 - - ) ( 1 - - )
4R2 4R2.
(n2 l2
l 2 m?
1'11. 2 )+ -,
2R2 .
S(!uaring again we find that
'1l 2 m 2 (1
[2 ) ( m2 )
4R3 1 - '1R i
= (n 2
1 2 2 2 [4m 4
m) +-.
4R4
2 2
.11/1. ( 2 2 2
T . R2 n - 1 '" Tn ) ,
giving, after some simplification
(n:! - z2 - m 2 )2 _ 4z2m 2 ==
l 2 m. 2 n 2
whidt esl,ablishes (41.1).
Returning to the original problem, we set .1: -= lAC!, and 0 '= LABC, so
t,hat LC 1>.11 7r - O. B,v the cosine law in D..4BC and D.4CD, We have
( 41.2)
:1: 2 == a 2 + b 2 - 2ab cos 0
SOLUTIONS
103
a.nd
t 41.3)
.r. 2 c 2 + d 2 - 2cdcos(7r - (I) c:! + IP + 2c(lcosO,
BliIlinatinv; :1'2 fwm (.Jl.2) and (.:lI.;J), \\'(, obtain
It ;- /)2 .... e 2 tLl
cosO
2( ab + ('t1)
liin!( this t:xpreshioll for ('os () in (.J 1.2), we 1';t't
,1:
(/1'£ +11 . (''l
(/,:'... Ii ab
(ab + e.[)
( ar + b dJ!a d + bc)
(ab + cd)
d l )
so 1.11 at
I «(lr + bd)( ad + b;;)
X:..o; V (ab+cd) .
The ra.dius T of the drcle passing through i1,B,C,D is the circllmradius of
D.ABC, and so by (,iLL) is gh'en by
j' --
nln:
J; )( a .. - b + .1:)( -(I + b + x)
J{o.+b+x)(a+b
abx
,if
N ext we ha.V(
I
'j
,

'I'
I'
, '
,i
(0 +b)2 - Xl
.. 2 ( ac+bd)(o,d+b.:)
(a + b).- (ab + cd)
ab«o+W-(c d? )
(ab + cd)
+b c+ c-d2
(ab+
and similarly
l
To
2 ab(-a+b+c+d)(a-b+c+d)
(a - b) == (ub + cd) ,

104
SOLUTIONS
SOLUTION S
105
so that
so that g 8 = -i(p - r), provinl!; that IF RI = IQSI a.nd P R .1 Q8.
r=
(ab + cd)( ac + bd)( ad t- 1)(:)
( -(1, + b + c t d)( a b + c + d)( a + b c + d)( i! -I- b + c . d)
43. Det"I:Jllinp polynomia.ls (I( J:':I/, .:, 11.» and q( x, y, ;;, II') with real
('"pfficipnt snch tha.I.
( <I:{'O)
(:ry + ::: + 11')2 (;1:2.. 2::)(rP - 2w)
(pI.D,If..:.W)) . (.1'2 2;:')(q(;I',If,,.w)f,
42. L<:t ..l1U' I) he a COllvex quadriJatpral. Lei, l' 1)1' tl1P pnint ullhide
ABeD such that kiPI iJ'B, ani! LAPH ... !lilt'. The point q,u,s are
,.,jmilarly (!i.fined. Provp 1.lt:\1 tlw lhlPs pn alld q.i.,' arp or I'qlla.l IPHglh alld
perp<:lIdicular.
Solution: We cOTisider the quadrila.teral ABCD to be in the complex plane
and denote the vertices A,B,C,f) hy the comlJlex nmnh"rs a,
b, c, b. Then the midpoints H,I(,L,J-l of the sides /lB,BC, CD, TJA (re
mpresentpd by (a + b)/2,(b + c)/2,(c + d)/2,(d + a)/2. Let p represent Ihe
point r. As IPHI = IBHI a.nd PH .1 BH we ha.ve
Solution: We sed, a solution of (4:3.0) of the form
{ 1-'V,y,z,'lI.') = ;ty + X ,
q(:l;,y,.:,w) = y+l',
(:U)
where X and Yare polynomials in X,lI', amI z. Substituting (4:U) ill (43.0)
and simplifying, we obtain
( a t b )
p- -
2
i (b (al))
('13.2)
((.:; ... tv)2 I- 2x 2 't1:) + 2x(::; + W):IJ
(X2 (X2 _ 2Z)y2) + 2 (xX (:Z;2  22)\'') y,
so that
( 1 - i )
p = -2- (at ib) ,
which givps
Similarly, we find that
( 4:3.3)
{ X:: - (x 2 2Z)1'2
:c\" (x,2 - 2z)Y
(z 1.11)2 + 2J;2 W ,
x(:; 1 w) .
{ q ["T' fie),
r = i (t: lid) ,
s = 1 2 i (rl+ in).
From the second equation in (4:3.3) we have
x = (:1: 2 - 2z)Y + x(z + <1')) /J: ,
and, using this in the first equation ill (43.::1), we obtain after simplification
From this we obtain
zy2_X(Z t.())y 1.:r. 2 w=0.
q
(L2'i) «a - c) + i(b - d)) ,
oS = .( \ji )«b d)+i(c-n))
= -2 (t2' ) «a c) I i(b -- d») ,
Solving for Y we fiud l.hat Y = :1;1.11/::' or Y .1:. Dis,a,rdinf!: the first sob1l.ion
;,s we are seeking pnlynorniah X a.nd Y. we have
p-r
x = :r. 2 ;;; + W, Y ;r. ,

106
ane! so we may take
p(x,y,;::,w)
X y + x 2 - z + .w , <l e x )
\I' ,y,z,W.
SOLUTIONS
x +y.
44. Let C "(u{)te the field of t'OlllpJ"X II 11111 her,... Ld, f : C  C 1)(' a.
f11l1C1iofi satis{\'inp;
(-1-1.0)
{ f(0) :-:; 0 ,
If(z) f(w)1 Iz 1/)1,
for aU z in C and 111 = 0, 1, ,i. Pmv. tlla.t
f(z) f(1)z or f(1);) ,
where If 0)1 = 1.
Solution: Fl'Om (,H.O) WI' haw
(44,11
lJ(z)1 = I'; ,
(44.il
If(z) -. od = Iz 11,
( 4,1.3)
If(z) -iij = Iz - il ,
which hold for all z in C, and whore
(44.4)
a f(l), i3 f( i)
Taking z 1,i in (44.1) and z = i in (41.2), we obtain
(/14.5 )
lal 181 1, 10 - /11 = h .
SOLUTIONS
107
Hence, we ha,ve
({2 +;P _ a 2 (J/1 + oCi;,P
a(J( all + afl)
- a(1 (na + ,r{fJ - (c.r - ,fJ)(a - i:j))
(r[1 (10'1 2 + !!W -In !W)
u;}(1 + 1 ..2)
(I ,
so that
(4-'1.6)
{1 = fa, ( :l:i.
Next, squaring (44.2) and appealing to (44.1) and (-14.5), we obtain
(44.7)
af(z) + O' f(z ) z + if,
for all z in C. Similarly, quaring (44.3) and a.nd appealing to (44.1), (44.5)
and (44.()), we obtain
(44.8)
Cif(z) nf(z) = -fi.? + ci'f .
Adding (4,1.7) a.nd (4'1.8), we deduce that
20.f(z) (1 - d)z + (1 + (i): ,
that is, as c = :l:i, Cif(z) = z or z. Hence we have
f( :;) = .f(l)z or i(l)z,
where If(l)! = 1, and it, easy to check that, both of these satisfy (41.0).
45.
(45.0)
If x and yare ra.tional numbers sueh that,
tan'/!':t. = y,
prove that x = k/4 for some integer k not congruent to 2 (mod 4).
'lIB I

108
SOLUTIONS
Solution:
As x and yare rational numbers there are inte g ers P q . l' s such
that ' , ,
{ x=P/q, y=r/s, q>O, 8>0,
GCD(p.q) = GCf)(r,s) ].
The equation (45.0) becomes
(,J!).l)
l'
tan
q S
\Ve have, appealing to Dc:Moivre's theonm.
( s + r ) q
s - zr
( 1 + i1'/S ) q
1-ir/s
( 1 + ita:n(7r p / q » ) 1
1 itan(7rp/q)
( cos(7r p / q ) +  sn(7rp/q» ) q
cos(7rp/q) zsm(7rp/q)
cos( 7'ip) + i sin ( 7rp)
cos( 7rp) - i sine 7rp)
( m 1)P + i.O
( -1)p - i.O
:;:: 1,
so that, appealing to the binomial theorem, we have
(s+ir)q
(li - ir)q
«s +i1') 2ir)q
t ( k ) (S + ir)q-k( -2i1')k .
k=O
Hence, we have
( -2ir)q
-  (  ) (s + ir)q-k
k=l k
= -(8+ir) ( k ) (S+i1')q-k-l,
k=l
SOLUTIONS
109
that is
( 45.2)
ij
'i}
,
(-2ir)q = (s + ir)(x + iy),
for some integers x and y, Taking the modulus of both sides of (45.2), we
obtain
2 2q 1' 2q (s2 + r2)(,r,2 + y2) .
Let 7' be !In odd prime dividing 8 2 + 1''l., Then p divides 2 2q r 2q a,nd so 1) divides
1'. Thus p divides s2 (8 2 + 1'2) - 1'2, tha.t is, p diddps s. This fOlItra.(lids
GCD(r,s) = 1. Thus 8 2 + 1,2 has LlO odd prime divisors a,nd so Hlust be a.
power of 2, say
.,
'ti
I
I
I
of
S2 + r 2 = 2 1 , I 2: ° .
Further, if I 2: 2, then sand l' are both even, which is impossible. and so 1 = °
or 1. As s > ° we must have
(r,s) == (0,1) or (:1:1,1).
The first possibility gives x = k/4, where k = 4p, while the second possibility
gives x = k/4, where k 1 (mod 2), thlL<; completing the proof.
snch that
Second solution: (due to R. Dreyer) "Ve m.ake use of the fact that there
are integers c(n,r), n = 1,2,... ; l' = 0,1,...,[n/2],
(45.3)
[n/2]
2cosnO L c(n,1')(2cosOt- 2r
,'=0
II
for a.ny rea1 number (I. The integers c(n,1') are given recursively by
c(l,O) = 1, c(2,0) = 1, c(2, 1) -2,
..
and for n 2:. 3
I
,I!
iN,
I
{ c(n,O)
c( n, r)
c(n,n/2)
= 1,
=c(n-1,r)-c(n 2,1'-1), 1:S:r:S(n-1)/2,
= ( -1)n/22, n even.

110
SOLUTIONS
Now, as x is rational, we may write x = plq, where GCD(p,q) 1 and q > O.
Furth"r, as y tan 11'X is rational, so is the quantity
- 2 .) __9 (l-t<l1l211';/;) ,(I_y2)
k - cOS11'X -..  - , '-= 2 .--
(1+tan 2 r.x) (1+1;2)'
Al'pealin to (45..3), with 1J. c- fj and 8 = 211'X = '211'7Jlq, W0 S{'(' that :: is a,
rational 1'0111. or UII' monj{' illtPgnl1 polynomial
'nIl]
/(:1:) = 2..:: e(n,'I'J:t:""2r 1..
,,,,0
Hence, z must be an integer. But Izi = 21 ws 211'xi :S 2 so that z = 0, :1:1, or
:1:2, that is
cos('111'plq) = 0, :1:1/'2,:1:1 ,
giving
211'p
q
.11' . 11'
(2l + 1) 2 ' (31 :I: 1):1" l11',
for some integer l. Thus, WE have
x
11
q
2l + 1
4 '
1
or
2
3l :f: 1
6
Only the first possibility, a,nd the third possibility with 1 even, have y tan 1IX
ration<l,l, and hence :r. = k/4. where k Ii; not wngruent to 2 (mod ,f).
46. I,et P be a point inside the triangle ABC. Let AP meet BC at
D, HI' meet CA. at'.B, and CP meet All at F. prove that
('16,0)
IPAIIPBI IPBII1'CI IPClIPill
IPnTlpi[ + iPE[ IPF I + IPF I IPDI 2: 12.
Solution: Let S,Sl,!:h,S3 flenote the areas of 6..4.8C, 6.PBC', 6.PCA,
6.P AH respectively, so that S S} + S2 + 83- Since 6.,1HC and
,
"1111
J
iir-
"
11
:.
/
'ilB
SOLUTIONS
111
6.PBC share the side BC, we have
IADI S
II'D\ Sj '
"i.
.J
tlU
1m
so that
Will I ADI-I/'DI "Wi
1/'1>1 IpO[ /-'VI
8 ,f,' ... ::J\ Sl, + S';)
1..,.-.....---...-
8 1 ... 8 1 ... ,1.;1
;'
Similarly, w(' haY<'
IPH.I_:,-'! + 8 1
IP /':1 - .'h
)pGJ
jPFI
8 1 + ,','2
S3
Hellc(', we ha,ve
JPAIIFBI IJ'RIIPCJ IPClIP"11
I PJJ, I PEI + j' PR I H)FI + WI'I jl> DI
(.)2 + S:!)(S3 -I HI) (5':, + 8 1 )(8 1 + 8 2 ) (HI + .'h)(S + '2:: t
+---..., + c<c
5\,'h ,'i2.'h 0:,.71
( Sa ,)'3 ' I , 8 ) ( .1.;1 -I ,'it + l + Sf )
-. -;- + -;- + + ,---;- + --;- .. .
S1 5'1. 5}5 2 8 2 S3 8z8:1
( .')2 ."/2 .. Si )
+ . 5 -;- -t (-;- +1 + '] '. S '
. :I "1 . ". 1
( ' 3 + 5 ,., > ) ' + ( ' 5 3 + :2. ) + ( I + '2 ) '
.5 1 . 3 2:! 2' 1
( Sf s s )
+3 + 8 2 1; , + S 3 8 , + 8}S2
> 2 + 2 + 2 + :) + 3 12 ,
by t.he arithmdic-geom('tric mean inequality, which completes the proof of
( 46.0).

I
47.
L<'t I and II. be. positive integers sllch that
1<;I<n, GCD(l,n) = 1.

112
SOLUTIONS
Define the intefJ:er k uniquely hy
1<.k<n,kl -l(modn).
Let AI hI' the /... x 1 ma.trix whose (i.,JJth entry is
( i-l)1 + j .
LP1. IV ht lilt' k X 1 ma.trix t()J'rnt.'l! by ta[{ing tlw (,O[I11I1T1 of :11 inl'('\'ers(' order
;111<1 wril illg lhe ,'ntl'i,s as th,' rows of y, \i\.'hal is the I't.laliolJsIJip IIdween
Ihe (i.jHIJ {'ull'v or AI ami '.]1<' (i.jl-th (,Idry 01'\' !IIodul" /I':'
Solution: If.4 [aij: a.nd B [bij: a,re two J.: X 1 m;;,trices, we write ,1
B (mod n) if aij bi i (mod n), i:= 1,2,.,. ,l:j j = 1,2,.., ,1.
As Jd -I (mod n) we have Inodl] 10 'It
I 2 1 2 1 1 I ]
1 + 1 1+2 'll 2 2l- I 2l
M .- 21 + I 21 + 2 :Jl 2 3l 1 :
( k - l)l + 1 (k 1)1 + 2 J.:l- 2 kl- 1
(ld - (I.; 1))1 (kl (2k l))l (2l.: + l)l (f..' + 1)1 1
(kl (I..: 2))1 (1d (2k 2))1 (2/" + 2)l (k + 2)1 21
(/;;/ - (k 3))1 (/d - (2/..: - 3»)l (2k + 3)l (k + ;3)l 3l
(kl)1 (ld -1.:)1 :3kl 2l.:1 kl
from which iI is deal' that Ow (i,j)-th ent.ry of N is 1 times the (i,.i)-th entry
of Af modulo n.
48. Let 11/, and It be int.egers l'\11ch that. 1 :S 'II/. < n. Let t1.ii, i
1,2, . . . ,m; j 1,2,. . . ,11, be rnn integers which are not all zero, and set
a= max la,'I.
l<i<,." J
15.j5.1l
SOLL'TIONS
113
Prove that the :>ystem of equations
.4
If
(-18.0)
{ aU Xl + aU"'2 + . . . + atn X "
:".,' t n",,+ .. . +0,,,,,,
Uml:l:l + a m 2xZ + ... + a,."".x n
o .
=0,
O.
'
has a. oluti{)11 in illt./'gers ;1: I, ,1;2," , ,:/:n, not all zero, sa.tbfying
J
I lit '1
I£,I:S. ;1.2I1IL)" '" i, I'::: j::; I/.
. l. .
Solution: We fi(,1.
i'
N (27H)n:'..";:;;]
tiO that
i
,
N > (2na)"c:',;; 1, which implies (N + l),,-m > (1na)"'" .
Hen....., we ha.ve
(N+1)" > (2na)'''(N+l)'n
(2naN + 2na)'" ,
\. that is, as a 2: 1,
( 18.1)
(N + 1)" > (2naN + l)m .
Set
,f.
!
Li = l'i(Yt,Y2,... ,y,,) a;lYI + U,2Y2 +.., + ainJ/n,
for 1 ::; i :S rn. If (Yb 11'1.,. .., Yn) is a vc'ctor of integers satisfying 0 :::: Y! ; N,
1 :'S; j::; 11, the corresponding value of Li Li(Yl, Yz" .. ,Yn), 1 ::; 1 ..::: Tn,
sa,tbfies
;'1
'Ii
naN :S /-i ::; naN, 1:S i :S Tn ,

114
SOLUTIONS
SOLUTIONS
115
and fiO t.ht' vector (1. 1 , 1. 2 , . . . . L n ,) of integers can t.ake on at most.
('!.naN + l)m differellt. values. As t.here are (N + 1)'" choices of the vector
(:t/1, Jh,... ,!IT()' hy (-IR.1) there must be t.wo distinct vectors
Hence, we haV0
d t ) _:c 2 )
1 (11.(3: e
IX
U (YJ,Yz,...,Yn),
:l (Zl , Zz, . . . , zn) ,
and so
say, giving rise to t.he sa.Tllt' vector (L], 1. 2 ,.. ., Lm). Set
('J9,l)
!/(:J:) 2J:h(:r) I. .
X,i .:7 Y.i
._"
1<:.1<n.
!\b h(.!:) iti ,I, ra.tional ('tlllrt-ioll WP ma,:,' writ!'
;\, III(' two vector" 11 and l' <1)'1' dioliuct, n(J1 all tilt' ,1'., an' ZPI'O. IV1on'oV('I', as
[1".:')
11(.1'.)
1'( ./:)
t/( .1: I
Li(Y1.!J2,'" ,J/n) Li(zl,z2,... ,zn), 1 <: i <: Tn,
whe)'e p(:r.) and q(.'!:) are »olyoomiaJ:.; with q(;c) not idt'Tltica.lIy wrO, and
GCD(1J(:r),q(:r)) 1. Thp.1I
(J:J, :r2, . . . , :r n ) is a sol utiofL of (48.0). Finally, Ix j I <: ]V, 1 <: j <: n, follows
from the fact that 0 <: l/;i,Zj <: N, 1 <: j <: n.
49.
Liouville proved that if
pi ( :r. )q ( ;t: ) rJ(r)ql ( ;t: )
q(..:)2
and using (49.2) and (.19.3) in (-W.l), we ohtain
hl(.,!;)
( 4H.3)
J f(J')eg(x)rlJ:
(4HA)
ll( l: )q(;r) p( X )ql(;r) 2J:p(:1' )q(;!') q( x)2 .
is a.n elementary function. where f(;c) and .Q(x) are rational functions wit.h
deg]'ee of g( x) > 0, then
If q(x) is a constant polYllornial, sa.\' q(J:) 1.:, t.hen (41A.) bewmes
pl(:r.) - 2;rp(:) I.:,
J f(x)(,y(x)dx h(:!.' )t:,?!r) ,
J . _x 2 1
f: I X
which is d('-arly impost;ible as the dt'grcc of thp- polynomial on the left. side
is at. least one. Thus, q( J:) is a. non-cc.I!Ista.nt polyuomial. Let c denot.e one
of its (compJex) roots, and let. '117, (2:: 1) denote the mlJlt.iplidty of c so that
(:c - c)1r. 'I q(x). Then, we haw (3: c)m-I 11 (/(:1:), and tI'om (;19.4) written in
the form
p(;r)q'(x) (pl(:C) - 2:rp(x) q(3:»)q(X) ,
we see that (x c) I pel;), which contradicts GCD(p(:I:),q(:I:» = 1, atJ. d com-
pletes the proof.
where h(:r) is a rational function. Use Liouville's result to prove that
is not. an element.ary funftion.
Solution: Suppose that I e-:? doc is an dementary function. Then, by
LiouvilJe's rCt;ult, there exi$h; a. rational function h(r) such that
50.
The sequence xo, X],... is defined by the conditions
!
dx h(:l')
(50.0)
Xo = 0, Xl = 1, :1: n +]
n> 1.

116
SOLUTIONS
Determine
L::::; lim X n .
n-ov
Solution: The reCUrrence relation can be written as
n
Xn+l - .'I'n
n+1
Xn-d, n  1,
so tha.t
(50,1) XnH-X n (-l)n(XI a:o)::::; (_1)n
n+l n+l' n1.
The equation in (50.1) trivially holds for n = O. Hence, for N  1, we have
N-I
XN == L(x n +1 - .J;n)
n=O
N-I (_I)n
L n + l '
n=O
and 1i0
L = lim ''CN
lVoo
N-I
lim "
N_ oo 7 n + 1
 (-l)n
 n+l '
that is L = In 2.
5l.
(51.0)
Prove that the only inte g ers N > 3 W I ' th the fioll .
. L _ . . OWIng property:
if 1 < k.s; Nand GCD(k,N) = 1 then k is prime,
a,re
N == 3,4,6,8,12,18,24,30.
SOLUTIONS
117
Solution: It is easy to check that :-1,4,6,8,12,18,24,30 are the only integers
.s; 121 with the given property. Suppose that "v > 121 is an
integer with the property (51.0). Define the positive integer It  5 by
1 5 1.1)
I 
p,. .s; v '\ < P,.+1 ,
where Pic (knott's Ihe I...t.l. primp.. From (51.1) we see tha.t p; ::; :V ,j
1,L,. .., 'Ii, aud w by proppj.ty (;)1.0) we l1lUt have 1'.i IN, lor j = 1,2,.. . , F..
As PI, . . . ,Pn are distinct primes. we must have
( 51.2)
PIP2 .. . pn IN,
and so, by (51.1) and (51.2), we have
(51.3)
]JI]J2 ... pn .s; }v < P+l .
By Bertrand's postulate, we have
pnH .s; 2pn, Pn.s; 2Pn-l ,
and so
(51.4)
2
> Pn >
Pn-IPn - 2 - R
Using the inequality (51A) in (51.3), we obtain
I1-tP2 . . . Pn-2P+d8 < PH ,
that is PIP2'" Pn-2 < 8. Since PIP2 == 6. a.nd PI112P3 = :-10, we must have
n - 2 .s; 2, and It .s; 4, which is impossible, proving that there a.re no integers
N > 121 with property (51.0).
52. Find the sum of the infinite series
1 1 1 1
8=1- 4 + 6 - 9 + 11
1
14 + ."
jlll1l

118
SOLUTIONS
Solution: WE' bf'gitl by observing tha.t
1 1 1 1 1 I
S = 1 - 4 + 6 9 + 11 - 14 + ...
( (1 - :i:: 1 -I :/.:".. :1;" + .,0 If) -. .) d:/.o
J(J
fal (1 ,/1)( I , 1''' + ,/, W 'r ... 1 d.1'
II
x: s
dx
r 1 .1;2 + x -1 1
= 10 x 4 + a:- 1 -+ :1:"1 -+ x + f dx .
Now, tl('composing into partial fractions, we have
:1;2 + x + 1
;4 + :r: i + :/,;2 + x + 1
b
where
+1'
a H
10'
I-VS
C - 2 ,
b -
- 10 ,
d 
2
Tlms, w(' havE'
S aIc + bl,l ,
where
1,= [
dx
+c.r-l- 1
I _- 1 1
d-
o
dx
-I- + 1
Now
J dx 1 ( 'x+t' )
2 F arctan 'r;--- , t < 1,
:;: + 2t:c + 1 v 1 - t 2 . V 1 -, t2
and by th(> fundamental I heot'em of ralculll", we have
[
1
(arctan ( )1  tt 2 ) arct.an (  ) )
a,rctan ( I   : ) -
1
SOLUTIONS
llU
Hence, taking t (1- ,;g1/4 a.nd t (1 1- ../5)/'1, we obtain
II! ) - 2Ji'; , .. ( V i) -I- 2J5 )
/.. - \/ arctan - F
( V fi Vi)
and.
SO\
so thai
r---' '-, ( ' f l. .-----:- Ii: ' )
JI()+'1v' V" 2V,1
t,; = V:) . -- <I.rd,an ''''''-,/,-'--,
fUS( 1!'/1O) (/10 1 -'1>/3 )!--1, sine 1!'/1O) (J- 1)1'1,
( ' 0 J F )/ I ill ( ;{1!' / l()1 "'- (Js+ 1 I/-I ,
cos(3Tt/Hl)" VII v:j"
. 5
t.an(Tt/lOl= r,;:'
Vi)
f", + 2V:
tan(,,/10) ==" h -
VG
Hence, we find t1lat
and so
as required.
'--'--
31!' / 10-2-15
I" = LO V 5' ,
--_ , 1!... . /10-1- 2-./5
la - HI V ,')
s -..
( [ 111 - 20 r.: /10 \- 2vG )
..::. 3(;> f- A)\! " + (5 VG} V 'I'
lOll' V .J '.
.. ( 3(-./5 + 1) / 10 - 2-15 + (-./5 - 1)/ 10 1.2-15 )
100
 ( 6 l 1O + 2-15 t 2[W  2Yi ) '
100
,--- - )
; ( 3 V lO + 2-./5 + y10 - 2-15 ,
dO

120
SOLUTIONS
SOLUTIONS
53. S<:micircle:;; are drawn externally to the sides of a p;iven triangle.
The lengths of the common ta.ngents to tht:'.se semicircles are l, m, and n.
Rl'late the quantity
lm. mn nl
- +.. +-
n 1 m
to tlw lell)!;thH of the ides of the tri<lUgl....
121
54.
f . if R 4 R ha.vin g the properties
Determine all the unctIOns :
(i) H(l,O,O,l) = 1,
(ii) ll(Aa,b, Ac,d) = AH(a,b,c,d),
(iii) H(a,b,c,d) == -H(b,a,d,c),
(iv) H(a + c,b,c + J,d) "'" H(a,b,e. d) + H(e, b, J, d),
WIHt't' (J"b,c,d,t:,f,A a,rc \"teal numhers.
Solution: Lpj 111<' \('rji(',' of I.hl' giv('n tTianle b,c /1, B.C. Lel ,I'.lJ'. C'
h,' 1IIP ('('II!,n", or 111" Pllli('irdp (1,,''1.-; 11".1.\,'0 Oil B('J' 1,111
rpspp('tivply. (,pj Of.',FG.HJ be tlte common tangents to /J and 'J', 'I and (x,
n ;I.nd l' rcspedively. Join JJ'D,C'E a.nd draw C' [( from C ' perpeTl(u('ular to
B'D. IIt:>lIc<" as 1((" RD is a rectangle, we have KC' = DE l.l,et
Solution: Hy (Hi) we have
H(l,l.O.O) -Tl(l,l,O,O), II(O,O,l,l)
IAHI = c, IBL'I = 20" IC Iii = '2b ,
so that
Then, we have
(54.1)
IR'(."I a, !lJlKj 111-cl,
II(l,l,O,O) 1l(0.0,1,1) 0,
'f." ( " 1 - J 2 - (I _ ) 2
Ii.' -. a . c. ,
(54.2)
and from (i) and (iii) we have
1I(0,1,1,0) -JI(l,O,O,l) = -1.
a.nd o
tlwt is
RNlce, we obtain
l=j(a -b+c)(a+b-..).
Similarly. W{ have
{ :2
= IPGI = J (a + b- c)( -a + b + (:) ,
= IHJI = J(=a+b+c)(a b+c),
a.nd EO
mn
T = -a+b+c,
nl
a b+t:,
n
lm
a ,I, b - c,
m
gi \'ing
(5;1.1)
m.n
nl lm
+ +
m n
a+b+c,
that is
so tha.l, t.he left. 8ide of (5;3.1) is the sem.ip(rimeter of the triangle.
H(a,b,r,d)
H(a,b,O,d) + II(O, b,c, d)
all(l,b,O,d) + clI(O,b, I,d)
-" -aH(b, l.d.O) - cIl(b,O,d, 1)
= -a (lI(b, 1,0,0) + H(O, I,d, 0))
-c(lI(b,O,O, 1) + H(O,O,d, 1»)
-abH(1, 1, 0, 0) - adlI(O, 1, 1,0)
-bcJl(1,O,O,1) c.dlI(O,O,l,l)
=-ab(O) ad(-l) be(l) cd(O)
ad bc,
H(a,b,c,d) = I   I.
H(O,O, 1,1),
(hy (i'))
(by (ii»
(by (i'ii»)
(by (i,)
(by (ii))

122
55.
SOLUTIONS
Let "1,.." "n be the complex roots of the quation
zn + (11Z"-1 +... + {ir. 0,
when' (ll,. .., an ar(' I/. (2: I) ('omplpx Humlwl's. Set
A. wax Iftkl .
lk:Sll, .
l'rov{' thm
!,..I:S'] + ,I. .7
Solution: Set
I .)
P'.H __ fl.
f(::) z"+a.!z"-ll',,,+a.,.
and suppos(' tha1: one of the Zj, j :S' j :S' n, it' such that Iz' l > 1 + A, Then ,
we ll<t.ve .1
() == If(zj) i
:n (1 a1 a". ), '
! "jl + .. . +-
'?j .""5 7 }
I  I " 1 1 + fl.l + an , .
-.1 .' .... 1-.
I Zj zj';
2: I::;jr' ( 1- II
1".11
_ !zjl" (1- II - H.
IZjln ( 1 _ __1.  '"
- iZjt
IZjln ( 1  ) '
1,0.11 1
Iz;I" (Izj; - (.4 + 1»
. IZjl I
> 0,
... _ l (l"l )
IZj!n
/l )
'V :n
!'-J i
Iz" -..-)
SOLUTIONS
123
which is impossible. Thus all the roots z}' 1 :S' .i :S' rt , of fez) must satisfy
IZjl :S' 1 + A.
56.
If m and n are positive integers with 1It odd, det.ermine
(I G(.' D(2 m 1,2" + I) .
Solution: Define integers k and I by
2'" - I = kd, 2", 1 = ld ,
and then. we obtain
2'" = kd + 1, 2T' = ld - 1 ,
and su for integers sandt we have
{ 2mn = (kd 11)" = ,qd + 1
2 m 'lL = (ld - 1)"" = td _.. 1, a.s m is odd,
Hence, we have (,5" t)d -2, and so rl divides 2. nut dearly d is odd, so
that d = 1.
57. If f( x) is a polynomial of degrp.€ 2m f- 1 with integral coefficients
for which there a.re 2m + 1 intep;ers k 1 ,. . . , kZm+I such that
(ii7.0)
f(k 1 ) '.' = f(J'2m+IJ 1,
prove that f(x) is not the product of two non-constant polynom.iab with
illtep;ra.l coefficients.
Solution: Suppose that f( x) is the product. of two non-consta.nt polynomi-
als with int!'gral coefficients, sa.y
f(:c) g(x)h(x) ,

124
SOLUTIONS
where'/'= deg(g(x») and s = deg(h(x» satisfy
r + s = 2m + 1,
1 .::; r .::; 8 .::; 2m .
Clearly, we have r'::; m. Now, for i 1,2,..., 2m + 1, we howe, from (57.0),
1"" I(k,) g(ki}h(k,) .
A;; 9(1,;;) is an integer. we must. have
g(k i ) =
1, 2, .. . , 2r/1. + 1 .
Clea: ly , either + 1 or -1 OCf;urs at least rn + 1 times among the values of 9( 1.,),
1 .::;,  .::; 2m + 1, and we let E denote this value. Then 9(x) - E is a polynollli al
of degree at most m which vanishes for at least m + ] values of 3:. Hence
the polynomial g(x) E must vanish identically, that is, g(x) is a constant
polynomial, which is a contradiction. Thus there is no factorization of I( x)
of the type supposed.
58.
that
Prove that there do not exist integers a, b, r, d (not all zero) such
(58.0)
0,2 + 5b 2 2c 2 - 2rd 3d 2 0 .
Solution: Suppose that (58.0) has a solution in integers a, b, e, d which are
not all zero. Set
{ Tn = GCD(a,b,c,d) ,
al alrn, b l blm, CI = elm, d l dIm.
Then dearly (aI, bI, CI, d l ) is a solution in integers, not all zero, of (58.0) with
GCD(al, b I , CI, dd 1.
SOLUTIONS
125
Hence we may suppose, without loss of generality. that (a, b, c, d) is a solution
of (58.0) with GCD(a,b,c,d) 1. Then, from (58.0), we obtain
(58.1) 2(a 2 + 5b 2 ) (2c + df). + 5d 2 ,
so that 2a 2 (<!c + d)2 (mod:J). Since 2 is a q1Jadmiic Jloll1'esidue (mod j)
we must have
(!).2)
a 1(' + d 0 (lIIod 5) ,
Sd,
a. = )X,
2c+rl J))'
wherc X and}' are integers, so that (58.1) becomes
2(5X 2 + b 2 ) 5y2 + d 2 .
Thus we have 2b 2 d 2 (mod 5). Again, as 2 is (!, quadratic nonresidue (mod
5), we deduce that
(58.3)
b == d 0 (mod 5) .
Appealing to (58.2) and (58.3), we see that a b == c == d == 0 (mod 5),
contradicting GCD( a, b, c, d) = 1. Hence the only solution of (58.0J in integers
is a = b = r d = O.
59. Prove that there exist infinitely many positive integers which are
not representable as sums of fewer than ten squares of odd natural numbers.
Solution: Vile show that the positive integers 72k+42, k = 0,1, . . . , cannot
be expressed as sums of fewer tha.n ten squares of odd natural
numbers. For suppose that
(59.1 )
72k + 42
2 2 2
Xl + X2 + .,. + x 8 ,
for wme k > 0, where XI,''', x. are odd integers and 1 '$ s < 10. Now,
x; == 1 (mod8) for i = 1,2,...,8, and so considering (59.1) as a congruence
modulo 8, we have
8 == 2 (mod 8) .
ii;

fOT n 1,2,. .. , we have
1 ( 211 )
11+1. ,11,
126
SOL{;TIONS
SOLUTIONS
Since 1 ::S . < 10 WI' must, have s '2 a.nd so
(.59.2)
72k + 42
2
+x 2 .
Tr....<!1.ing [!I!),2) a>; it congru01lCe rnollulo :J, we obtain
xi + :t 0 (mod 3) .
SillCf> thl' '''lilaH' of <I,ll illt(('r is <;ong;rIJ1'nt to () or ] (mud ;), we lUust have
;!:I :;:: 0 (mod :!). Finally, T P dlldng (59.2) Illodulo 9, we obta.in the
rnntra.dirtion fi (I (mod !J),
as rCljllired.
61. Prove bhal
60. Eva.lna,t(. the' integra.l
( 60.0)
1 '0'" sin lex rosk x l
(X ,
o x
is a.n illteer for /I . 1, '2, :,. . . .
I(k)
where k is a. positive int0g('r.
Solution:
Solution: By the binomial theof0m, we ha.ve
(60.1)
(t?2i'" + l)k t ( : ) (;:l1'i' .
r=O
As
(e 2ia , + II ekio: (eio: -I- e-i,ry, (cos kx -I- i sin k:r. )2 k cos" x ,
the im1lginary part of (e 2ix + 1)" is 2 k sin kx ros k :1'. Equating imaginary parts
in (60.1), we obtain
2 k sin k:r. cos k x
k ( k )
L r sin 2rx
,'=0
k ( k )
L r sin2r.r..
,'=1
As ( 2n ) and ( 2"+1) are both intep;ers, this shows th,t
n r...
was required to be' proved.
() is an integer, as
Thus usin g f<<> .in x dx =
, Jo 3:
we ha.ve
127
2;1 t ( : )
r=l
...7r ( i< 1
Zk+1 )
7r ( 1 '
2 1 "-'i I )'
1 ( '2n.' )
;:-+ 1 n
21/.! 1
(1/.!)211+1
2n! +
(n!F
'J , (
.....11.. '.
(n! )2 
2n!
2.c-
(n!? -I-
:2 ( 2n ) ( 2n -I- I )
n n
/(k)
1  ( k ) 10 "" sin 21'x
k L rl.r.
2 . 1'=1 r . 0 x
Second solution: (due to S. Elnitsky) For n 1,2, . .' we have
1 ( 2n ) 2n! 1
n+l n (71T)i--1

128
SOL UTIONS
==
2n!
n!(n + I)!
"n'
n!( 1)! (11.+1)-71.)
2n! 2n!
(n!)2 (:;-:- l)!(n+ i)!
(2n ) ( 2" )
\ n 71. l
:b {2 , :' } ' aud ( ') 1 tl '
. ,,,:v iU'!, ,>0 I lItlegl't;" this ;.lIOW,; that I ( "" ) '
"IT .;" j an inl!'5!;pr.
62. Find the sllm of the inlinite series
cv':' 2 n
8 == L  2n -/ _ 1 '
n=O
wlwre a. > I.
Solution: We hav for 0. >
2 n
n'ln + 1
:,In(0.2n 1)
0.2n+1 1
2n+1
2 n
a'ln+1 _ 1 '
,,0 that
==  ( 2" 2'f1.+I )
S L.- - '
n=O a 2n 1 a 2n + 1 1
1
a-I'
63.
Let k be an inte g er. Pr ov p . 1 t th f, 1
' .. L la c onna power series
v i + kx 1 + alX + 0,2:(;2 +...
SOLUTIONS
129
has integra! co<?fficients if and only if k 0 (mod 4).
Solution: If I.' == 1 (1110<1 2) thpn at '"' 1.:/2 is n01 art iute/!;er and if' k
:,I (mo{l4) th,'n tzz _1.:2/iI, is not an intl\ger. Wlwl} I: -
o (mod 4), we have for n == 1,2,...
( '1/2 ) I. , "
o,'u 
n
I )( :.' ,J " , _ I .. _ (II 1\ i
.  i __'.. 'L'___ ln
n!
1 3 ,, ) ' --. ( " 11 - " )
(_.1)"-1 ""'-' ' )-/"
. 2" n!
-1 n--1 (211 - 2)! k"
( ) 22n- I n!(n-1)! '
2 ( 1)n-11 ( 2ft - 2 ) (  ) r,
n n 1 41
\vhich is a.ll integer since k/4 is an int(ger and e:'-='12) is an integer by
Problem 61.
64. Let rn be a, positive integer. Eval.uatt:' the determinant of the
m X Tn matrix ]"lm whose (i,j)-th entry is GCD(i,j).
Solution:
Let C1,..' ,Cm denote the columns ofthe matrix kIm.' \Vt' <tefitH-!
N m t.o the matrix whoe columns D1"'" Dm a,re given by
{ D,
D:n
Ci, i= 1,2,...,m 1,
Ld(-l)"'( dj C m /d,
where the sum is taken over those squarefree integer d whil'h divide m,.
Clearly, as Dm :::: C m + J, where J is a lineal" combina.tion of the C:i, 1  i 
in - 1, we have
det .i.'v1 m :::: det N m .
.

130
SOLUTIONS
SOLUTIOKS
For 1 5- i :S 'm, I.he entry in the i- tIt row of D", is (writing (i,.i) for GC D(i,.i»
L (-1y(.1){i,mfd)
II
p"lIm
L ('-1)T(d)(i,p') fd)
.11f.."
131
,( f>qnJrefr
d s<{uar(:fn'f:
m. :::: fil + 8(21.lv .!- 1,2) ,
h- t ( 6" 0 \ 1 )1ds. Thml
Suppose there exist integers u a.11d l' such t a, .). i It .,
we have
Solution:
dfm
II (I,pO) -, (i,JlL>-l))
sn that m "" f)/ (mod R). Hence, we must have
.. '" 2 ,.. 2
:1'2 !- :'C:IJ t 17:/}' "" 'J:C + :);J
(mod x) ,
P'\iFlt
J 0""11"" (p" - ,/'-') , if i 111,
1 (I . Ii III
{ d>( 111) ,f i  Tn.:;
o , If 1 ":, Z :::0 m .- 1 .
1 [lat j"
4.<;2 + 'l:lJ 2 =0' 0 (mod x) ,
and so
X" ..!- y" == 0 (mod 2) ,
which contnHlicts the condition that I is o,Jd.
lIelIG', expanding the dderminant of I"i m hy ib; T/I.'( h columIl, wp obtain
det IV m. = ,i>( '/TI) det !If'n-l
a nd so
66.
Let
del. lU n ...
(_1)"-1
!-.'--ln2.
. n
<i>(m)detM m _ 1 .
an = 1
1 1
- !- ,-
2 . 3
Tlm'3, as <let ilh = 1 ( 1), Wt!, find tha.t
dctM". =: 'Pfm);,'>(m, - 1).. .?(2)q.!l).
P tl . t \',,:- a converges and determine its sum.
TOV{ la L-rTf:::::l n .
65. Let land 171. bE' posit.ive integers with 1 odd a.nd for which thel'C
11,1'(; integers x and 11 with
Solution: \Ve have
1 1 .. ' 1 ' ) n-l n-l) dx
(1 x !-:c -...t(-. :r. i
(1
1 ( 1 + (_It-IX: ) dx _ t d:
1 1 + x 10 1" X
1. 1 x"
dx.
o tx
Hence, for any integer IV  1, we have
N ,v, fl (_1)n-lxn dx
L: (I,,, := L.. /0 1 j- X
n=1 ,,=1 .
an
{m
"-,,x2+y2,
=: x 2 + 8.1:Y + 17y2.
Provl} that there do not exist integers 'U <tnt! 'V \vith
(6,').0)
{ I '- u 2 + 'v 2 ,
m= ,')v," + 16'11:11 ,. 13v".
t _ (1:.£
10 1 +;£

132
aHo:.! o
SOLllTIONS
[1 1 N
10 1 + L(_J)"-lxn dx
o X n=1
[1 ( X + (_I)N+1x N +1)
10 (1 + :/")2 dx
1 1 iC N 1 1 x,'\'H
( - 1 ..- ) 2 dx + (-1) H .  dx .
o . +.t 0 (1 + .1' j2 .
[II,' 1 1 X I
L an - d.r
In=1 0 (1 + x)2 I
l (t:: 1 )2 dx
s 1 1 x NH d:1;
J
+ 2'
Letting N -> oc. we see that
an converp;es, and has sum
1 1 X
dx
o (1+;r.F
1 1 C  ;1 : - (1 :.)2 ) dx
In2 1/2,
67.
Let A = {a; I 0 S i S 6} be a sequence of seven integers satisfying
o = aD .$ 0,1 .$ ... S 0,6 .$ 6.
For i -= 0, 1, . . . ,6 let
N; = number of aj (0 S j S 6) such that aj == i.
Determine all sequences A such that
(67.0)
Ni !16-i, i == 0,1, . . . ,6 .
SOLUTIONS
133
Solution: Let A be a sequence of the required type satisfying (67.0) and
let k denote the number of zerOs in A. As Go = 0 we have
k  1, and a k = No = a6 we have k .$ 6. If k = 6 then it follows
that..t {O, 0, 0, 0, 0, 0, 6}, contradicting Nn 0,0 = O. Hence, we have
J S k an .$ 5, a.nd so
(67.1 )
Nk  1, Nk+1 ... == N e O.
TbuH, oy (67.0) and (67.1), we obtain
(b7.2)
lIO =.. (II ' .. ,= O';-(!:+.IJ II, tl";'k'C'
and so
k No 6-(k+l)+I,
that is k = 3. This proves that A is of the form
(67.3)
A == {0,0,O,a3,a4,as,3} ,
where
(67.4 )
1 S 0,3 S 0,4 S as .$ 3 .
Clearly, we have 0 < N 1 S 3. If Nt 0 then, by (67.0), we have the
contradiction as N 1 O. If N 1 1 then, by (67.0), we have 0,5 1, and
80 (67.'1) implies tl1at a3 0,4 == as 1, giving the contradiction N I = 3. If
N I = 3 then a3 a4 = as = 1 and so, by (67.0), we obtain the contradiction
as Nl = 3. Hence, we see that NI = 250 that 0,3 = a4 = 1 and as = N J = 2.
The resulting sequence
A= {0,0,0,1,1,2,3}
satisfies (67.0), and the proof shows that it is the only such sequence to do
so.
68. Let G be a finite group with identity e. If G contains elements 9
and h suc.h that
(68.0)
g5 e, ghg- I = h 2 ,
II'
II:

134
determine the order of II.
Solution:
SOLUTIONS
If h e theu tIw order of hi;; 1. Thus We' may suppose that
h f c We have
g211g-2
g3/ur 3
!/lhy.4
!/5I1g-5
gighg-1 )g-1
= !1(q2/lfI-2)tj-1
g(g:1hg-")g-1
.Qi.q'l h g-1 ),q- j
c-: g/1 2 [1-1
,= tjll4tF 1
!1//"g-l
= ,q!JH;y-1
= (gh.q-l )2
(y/iy-l)1
:- (g/Ig-I)
= (!lhtF I)'"
11 4 ,
-- hi! ,
., h 1 (; ,
-'-" h'{2 .
and dO, as g5 = t, we obta.in h = h 3 "2 , that. i s /1 31 ' 1. ' 1
e. " IUS th0 ord",r of h is
31 a h f f: and 31 is prime.
69.
Let a and h be positive integers uch tbat.
DCD(a,b) = 1, t1 =t b (mod 2) .
If the set S II<IS the ioUowing two properties:
(i) a,bES,
(ii) ;I;,y,z E S implips x +:1+ z E S,
prove t.hat evcry integer> 2llb belongs to S.
Solution:
IAt N he an integer> 2(lb. As GC D(a, b) -= 1 there exist integers
k and 1 I'uch that
aA + bl=. N
Furthermore, as
 _ ( l ) 
there exiRts an integer t such that
ak + bl
ab
-k
b U <;t < t, + 1 <; a
IV
I > 2,
l,tJ
J ,Ij
SOLUTIONS
U5
Define integers 11 and -v by
'II, = /... + bt,
" = I - at ,
;J.nd intep;prs : and y hy
{ : c= :: 'I' " ,
y = v,
Y I' f/
if'll + t' = 1 (mod 2) ,
if /I + n '" 0 (mod 2) .
It. is <'asy to ch l'ck tlHt 1.
iY=:ra+yb, :1:>0, y2:0. 3;+y=l(mod2).
\V(! show bdow that S cOllIains a.1l integers of the fortI!
:W + yb , x? 0, y 2: 0, ;c + y =' I (mod ) ,
completing the proof that IV E S.
For Tn an odd positive integer, let, Pm be tbe a.%ertion that XQ. + yb E S
for all integers x and !I satisfying
;r: 2: O. y? 0, :r: + y 1 (mod 2), x + Y Tn .
Clearly PI is true a<l a,lJ E S by (i). Assume that P,n is true and considcr an
integer of the form X a + Vb, where X and Yare integers witb
X ? o. Y? 0, X + Y =' 1 (mod 2) , X + Y m + 2.
As m -I- 2 2: 3 at leal't one of X and Y is ? 2. Then, writing X Q + Yb iu the
lorlll
{ ((X - 2)a + Vb) + a + a ,if X > 2 ,
(Xa+(Y-2)b)+b+b ,ifY?'2,
we see that Xa + Yb E S, by the inductive hypothesis, and so P m +2 is true.
Hence, by the }Irinciple of mathematical induction, Pm is true for all odd
positive integers rn.
70. Prove that every integer can be expressed in the form XZ+y2. 5z 2 ,
wher!' x,y,z are integers.

136
SOLUTIONS
Solution:
(due to L Smith) If rn is even, say m
m (n--2}'+(211 ])2 5(n
whereas if m i;, odd, "".V lit == :In + 1, then
2'11., then
1)2,
m. (n+1?+(2nf--Sn2.
71.
EVi:l,luatt! the sum of the infinite series
In2 In3 In-i In II
 a +T 5+'"
Solution: .For x > I we have
In x r dt < r dt
J 1 t J 1 .Ji 2.J;/: 2 < 2.J;/:
and
-1/2:5 x [x] -. 1/2 < 1/2,
so that for any a 2: 1 we have
1 " I (In.'f. I) .
" (:1: - [x J
] x-
1/2)/ dx
< 1 " ( 2VX + 1) 1 .
2 _ 2 ' d,c
I :c
<  {" dx
2 J 1 x 3 / 2
== 3 ( 2  )
2 vf(i
< 3.
Thus, th integr<il
1 == r' (In x - 1) X
Jl ,2 (.
[x]
1/2) lix
SOLUTIONS
137
is absolutely conv(rgent.
Kow, one form of the E:ulel'-MacLaurin summa,tion formula. aggerts that .if
f(;c) lias a continuous derivative on [1, n], where n (> ]) is a positiv( integer,
then
"1 f. " f. "
:L I(k) == -(fen) + 1(1)) + f(.,,) d;c + J'(x)(:J: - :I:] ,- 1/'2) d;".
I.,I 2 .'1 . I
Taking J(x) In :,./:,:, Wf' oh1.aill
th
k=l Ii:
lll + n +!. "Li. -Ia.,,).
-.- (:1:
211 2 .1 3,2
1/2) tLL: .
(a;]
Setting
In 2 n
n In/.'
E(n)==:L k
k=l
"
L
and letting n --;. 00, we see tha.I lim n -+ oc , E( n) exists and has the value -I.
rhus
lim (E(2n) F,(n))
'n--O!:,
exists and has the vahle O. N",xt, we have th(' following
2"
". l) T.1n r
LJ--
r=2 r
In 2 In 3 In 4 In 2n
2 - :C +4 .. . + 2n
( In , 21n4 In2n ) ( In2 In:) In2n )
]+ 2+"'+ 2+ 3 +...+
(In2+1n1) (In2+ln2) 2+1n1l) n Ink
+ +...+ n --:L.
1. 2 k=]
( 1 J ) n.lnk 2-n In/.;
In2 1+. +...+-:- +:L--:L-
2 11 kd k k=l k
In 2 (] +  + ... + ) + ( E(n) + In;  )
_ (F,(2n) + Ir / 2 2n )

138
.--- ---
SOLUTIO;'IJS
SOLUTIONS
139
( 11
:::: In 2 1 + ? + '. +
 n
' ) In:.! 2
Inn - 2 + (1';(1'1.) - E(2n)) .
n-I . .'
I: \j vf 16A:3 + 24A2 + 91.' + 1- \mk 3 + 24k2 + 9k
k=O
Lettill n =, a.nd rem(m bering tbat
limo ( 'f.! - In It ) = l'
11.-+<"., .A''''I k .
n-1
)(Vk+1 - Vk) = ..;n.
"--<I
k=O
lIence we may take II, H1, () 24, a.nd c 9.
wlWfe l' ::::: O.577:n is Euler's conjnt, WI' ohtail'
73.
Ld,lI 1)(' a poil.ive Int"I';"1' a,nd a,b in1."1!,r-J'B such that
, T /Ill
2:)-1) -;""l'In2
r=2
1 "
In  ')
2 - ,
(,'(.'/)(fI,l>,lI) = 1
Prove thaI there fxist illt,ef:',ers aI,b l with
a] a (mod n), b1==b(modn), GCD(Il].bti I,
72. Determine constants a,b and c such that
vii  (J;;k 3+-b k +] - -laP + "J.2 + C" .
k=O .
Solution: \Ve ('hoose (/.1 to be any nonz(!ro integer such that
for n :::: 1,2,. .. .
U:U)
0,1 == tl (mod n) .
Then we set.
Solution: J;or k 0,1,. ,. , 1.ve have
( Vk + ( vi;r
(k + ) )Vk+l- :!(k + I)Vk + ; hlk -t 1 - k.Jf
(4" + 1 )vk+l- (4A + 3)v'1.:
f(4 k + -i Rk+l) ';( 4k + 3)2k
Vi6k3-+24'k2 + 9k + 1 - -I 16k 3 + 24k 2 + 9k
b] b + 1'1'1. ,
when' r is the product of those primes which divide a] but which do not divide
either born. If there a.re no such primps then l' I. Clea,rly we have
b] = Ii (mod n)
\Ve now show that
GCJ)(a1,b]) = ] .
Suppose tha,t GCD( 0,] , b]) > I, Then there exisls a prime q which divides
both III and b l . We consider three cases according a.s
so tlJ a.t
{j V I 6P -I 24i: +9k +  =- + 4 Vk + ) ._ v'k ,
and thu:>
(i) q divides b,
(ii) q docs not divide b but divides n,
(Iii) q divide;; neither b nor n.

140
SOLUTIONS
Case (i): As q I b, q I b 1 and b 1 - b = m, we have q I Tn. Now, by (73.1),
GCD(aI,b,n) GCD(a,b.n) = 1 .
Sine" q I 0,1 and q I b w£' "ee that q does not divide n, Thus we have q I r,
contradictinl!: the definition of T.
Case (ii): This case clearly cannot occur as b 1 = b + rn, yet q divides both
b l and n, hut does not diYit!e b.
Case (iii): As q I 0,1 but does not divide born. we haW' q i T. Sinn.. q I b 1 ,
q I rand b l = b + 1''1) , we musl. haw' t] I b, which is impossi hl£'.
This completes the solution.
74. For 11. = 1,2,... let s( rt) denote the sum of the digits of 2 n , Thw;,
for example, as 2 8 = 256 we have 8(8) 2+5 +6 13. Determine an positive
integers 11. such that
(74.0)
s( n) = s( n + 1) .
Solution: 'Write
2 n = 0,11110 111 + a",_I 1Om - 1 +... + 0,110 + ao ,
where 0,0, 0,1, . . . , am are integers such that
1 ::; am ::; 9; 0 S ak ::; 9, 0::; k S mI.
then
2 n a m +a m -l+..,+al+aO s(n) (mod 3) ,
and so
sIn + 1) == 2n+l == 2.2 n 2 s(n) (mod 3) ,
Hence, if 8( n + 1) s( n), we must have
8(n) 0 (mod 3), 2 n == 0 (mod 3) ,
SOLUTIONS
141
which is impossible. Thus there are no positive integers satisfying (74.0),
75. Evaluate the snm of the infinite serit,s
Solution: \Ve ha,ve
"" 1
L m(m+n )
7n.n=1
On the other hand, we have
00 1
L m11.(m + 11)
m,n=l
:1U1
00
1
mn(m+11.) .
s
L
"",1'1:;::,]
d(:L1(m,r101::;;;1
t !..-11 xm+n-l dx
rn,n=l rnn 0
1 ( 00 x m ) ( ':><) x n ) dx
1 L  L-;- x
o 111=1 n;::;1
rr In 2 (1 x) dx
J o x
1 00 u2e-"
du
o (1 e-")
{"'" u 2 f e- nu du
Jo n=1
= f 1= u 2 e- n " d'u
71.=1
co 2
L n3
n=1
00 1
2 L n 3 '
n=1
(x = 1 - e- U )
00
L
d=1
00
L
1
rnn(m, + 11.)
1H,n=1
GClD(m,n)=d

142
SOLUTIONS
'x'
00
L
1
ri3qr(q+ rj
L
d=l
q,r=]
GCIJ(g,T)=1
(E :3)
'oX
L
I
qr( q + r)
'1,"=1
Or' 0("",)=,
( 'X I )
S L;j:,
<1=1
so that 8 2.
76. A cross-country racer ruIIS a 10-mile race in 50 minutes. Prove
that somewhere alon/!; the COl1rse the racer ran 2 miles in exactly 10 rnimltes.
Solution: For 0 « x :::; Slet T(:) denote the \,ime (in rnimltes) taken by the
, , ". raer o 1'Iln )etween points x and J' + 2 mileii along the course.
Ihe functIun 1 (x) 1M COIltmuol1s on [0,8] and has the property
(7(>.1)
T(O) + T(2) + T(4) + T(6) + T(S) = 50 .
1'h<, equation (76.1) shows that not all ofth<, values T(0),T(2), T(4), T(6) and
!(8) ar('> greater than 10 nor a.re aU of them less than W. Hence, there exist
mteg<'rs rand s with 0:::; r,.s :::; 8 such that
T(r) s: 10:::: T(..,).
Then, by he int.erml.'diate value t.heorem, there exists a value y, r s: ?! s: oS ,
snell t.hat T(y) = 10, and this proves the assertion.
77. Let AB be a line segment with midpoint O. Let R be a point OIl
AB between Ii and O. Three semicircles are constructed 011 the same :;ide of
AB as Iollows: 8 1 is the semicircle with centre 0 and ra.diuslOA I = l OB'. 8
. tl . . I . h " 2
IS ,Ie Sl.'tIJJclrc e WIt centre R and radius I AR I meetin g R B at e. S ' . tl
, , . < -',' 3 IS . lC
SOLUTIONS
143
semicircle witll centre S (tlw midpoint of CH) and radius IC81 = 18BI. The
common tangent to ''''2 and 8 3 touches 8') at P and SJ at Q. The PNpendicular
to AB through C meets HI at D, Prove that PCQD is a rectangle.
Solution:
\V" give a <;olution using coordinate geometry. The coordinat(-
svstPTTl is dl<,'';''fI so 1.ha,t
,Ie ( 1,0), 0 -c: (0,0), n (1.0)
Th<,n w(' h.ave It := ( -Il, U), w IWf(' 0 <. It <. I, ami hClln-
C' = (1- 2a,0), S -= (I. - tt,O)
Tile <,quatiolls of the t.hreE! >'t)mjdrc)(.;; arE' givE.'T! as IoJlows:
!h ;r2 + !J2 = 1
5'2 (3: + a)'l + y2 = (1- a?
8:\ (:r+a__l)2+.y2=a'l
The perp('uJicula.r to .41J through C meets 5'1 at
D (1 2a,2 Va - a 2 ).
The equation of the common tangent to 8z and 8: 3 is
;dl- 2 a)+2 y Ja-a 2 = 1-2(1.+2(12,
and this line touches !h at the point
P (2a? - 4a + 1,2(1 a) a 2 )
and 5'3 at tlw point
Q (1- 2a?,2a va - tP ).
The slope of P D is
/--.., 
2a\i Q. - a.-:= - l -- _ r ( ' I - .
- 20. - 2a 2
,Ii

144
SOLUTIONS
and the slope of PC if;
2(1 -, (l)
2a 2 20,
_ (I a .
V - a
The product oflhese slopes is -1, slJOwin that PC ;}.nd l'D iU!:' perpeudic111a.r,
that is IC}'!) = !.I(I". Similarl.\',
L P IJq = L nq( , == I <J( . I J = !)()" ,
:'0 thaI. P LJ(J{' i: a H'<'tan,d".
78.
Determine the inverst" of the 11 X 11 matrix
(71<.0)
S[;: 1]
where n;?: 2.
Solution: Sel.,
I[r:: f] u[j:: I]
so that
S = {j - I, U 2 = nU .
For any real number e, we have
(U - I)(cU l)
('.[;2 (c + 1)F + I
(en. (c + 1»[' + 1.
SOLUTIONS
145
Thus, if we choose en - (('. + 1) == 0, that is c 1/( n - I), we have
S-l = (TJ
I)-I =F
'f! - 1
I
r Z-n
X__l
n---l
---1-.
_ /1-1
1
1 L-l
2-n
n::. 1
",] 1
n.l
fi-.
n-
79.
l':vaJHa,te the sum
(79.0)
n-]
S " "( l) k. n f I.. / )
,(11.) == L.,\- 10S .1'11" n ,
k=O
wh'l'e fI. is a positive integer.
Solution: Set w exp(1I"i/n) so that
S(n) I:(-nk ( w k .)w-k ) T1
k=U ..
Hcncf', by the binomial theorem, Wt" obtain
8(1/.)
I " I: ,-l kn I: " ( n ) . k(n-ZI)
-- W \,01..'
2 n 1
k=O 1=0
" ( ) n-]
 I: 1/. I: w k (2Ti--Z!)
2 1=0 1 k=O
2 ((  ) 11 + (:) n) ,
that is S(n) = 11/2'1'1-1.

146
SOLUTIONS
SOLUTIONS
147
(80.0)
80.
Determine 2 X 2 matric<:s Band C with integral elltri0S such that
Solution: Let tbe two triangles be IIBC and DEF. We suppose that
[ -1 ] ]
o -2
B 3 +C 3 .
IABI
IDEI
a, 1,1C 1 b, I/fCj=c,
h, IDFI = c, [J>'FI = tl,
awl that
(HLl)
(l < b .
Solution: Let
so that
and thus
giving
lIenee, we have
and so
.1
[
1 .,
:: J
1
()
,\ j\",t/j(' a1l(] 1\/)1"1" ar" Rimila.r, w.' havt'
a b t.'
A 2
[ 1 ".3 ]
o 4 '
b
d '
c
RO that
/1.2 + :lA + 21 = 0 ,
(iiI.2)
c = l? I a, Ii = b:11 a 2 .
F'rom (B1.1)we ha\'e
A 3 + 3A 2 +2A O.
(81.:1)
1 < bin,
(A + I)J A 3 + 3A + 311 + I = .4 + I ,
a.nd from (81.2) and the inequality c < a + b we have
b 2
<a+b
a
A (A + 1)3 - 1 ,
so that
and we may t.ake
B /1+1 [ _], (:=-1=[ - -n.
 <' 1 +  R; 1.618 .
(g]..t) .,' 2
To satisfy (81.;1) and (81.4) we choose bla :3/2. say n = 2t and b = 3t.
Then, by (81.2), we have
9t
c = "2'
d == 27t .
4
81. find two llon-conf,Tuent similar triangles with sides of integral
length baving the lengtbs of two sides of olle triangle equal 10 the lengths of
two sides of the other.
To ensure that c a.nd d a,re integers we choose t = .J so that
(I 8, b = ]2, c ]8, d 27.

148
SOLCTIONS
The triangles with sides 8,12, 18 and 12, HI, 27 respectively, meet the require-
ments of the problem.
82. Let a, b, c be three real numbers with a < b < c. Th fmlctiOll
f(x) is mntinuous on [a,c and differentiable on (a, c). The derivative f'(x)
is strictly inereasing on ( a., c). Prove that
(82.0)
(c.. b)f(a) + (b - o)f(c) > (e  a)f(IJ) ,
Solution: By the mean-value th",orem there exist.s a, real number 'u such
that
f(b) - f(a) = f ' ( 'I1, ), b
b-cL ,a<'I1,<,
and a real number v such that
f(c)-f(b) ' (
b = f v),
c
b<v<c.
As a < 'Ii < V < (' and f' is increasing on (a, c), we have
f'(-lL) < f'(v),
and so
f(b) f(a) < fee) - f(b)
b-(! c b
Rearra.nging this inequality gives (82,0).
83. The sequen.ce {am i m = 1,2,...} is such that am > a m +1 >
O,m = 1,2,..., and am converges. Prove that
<X>
L m(a". a m +1)
m=l
SOLUTIONS
149
converes and determine its SI11ll.
L t l) . \ . \"" a . is a conver g ent serit's of positive terms,
C f > ..1 S /nt:=::;:;l Tn '_
t.here ,-,xists it positiv., integer N(c) snch that
( , ] 0< tl m + l + Um+Z + .., < (./3 ,
g:,. )
for a.JJ 111 2: IV( d. Let 11 2: :L!\ (d+ L If'11 is even, say 11 = 21.', whpre f.: > X( (),
from P":!. II \\'t' h"vp
Solution:
ka2k < akH + ak+2 + . . . + a2k < C.n ,
so that
'WI" = 2ku2k < 2E/;J < ( .
If n is odd, say 1) = 2k + 1, where k 2: N(i), from (8;J.1) we have
kaZk+1 < a/,'+2 + (l,k+3 + . . . + aZk+1 < {/3 .
so that.
nUn .= 2kaZk+J + a2kH < 2E/3 + €/3 = ( .
VV", have shown thai
0< na n < €, for all n 2: 2N(E) + 1,
and thus
lim na n = ° .
)}.--4OC'
Next., set
n
Sn = L k( Gok - ak+1), n 1,2, , .. .
k=J
vVe hay.;
S"
n n
L kak L kak+l
k=1 k=l
n 71.+1
L kUk - 2)k l)ak
k=1 k=l

150
SOLUTIONS
n
(k - (k l))ak na n +!
:=1
n
 (/k na1/. T l'
k=l
Letting n ---, 'JC, we ee that lim,,,,:, Sn exi;,ts, anti hag the valu" Y:kl H;,., aR
)0,<> nanH ,,10,: ((11 + l)ft,,+1 - (l.n+d () - 0 () .
llepce, > I k(ak - Ilk+l) COli verges, aud its sum is ) ::;::;;01 Uk-
84. The continued fraction of ,(D, where D is all odd nOl1squal'e
imeger > 5, has a period of length one. 'What is the length of tile period of
the continued fraction of Ml + ..Ji5}?
Solution: The continued fraction of yf[j is of the form
ViJ' [a; b] ,
where a and b are positive intpgers, so that
- (I
1
b I
+ 1
b+-
b+...
1
b+..Ji5 a,
giving
D + a 2 - (lb 1
2a
As D is not a square, ,jjj is irrational, and we mllst have
b 2(1, D::: (lz + 1 _
SOLUTIONS
151
I-'urthcrmore, a. D is odd and greater than , we have a ::: 2c. c 2: 2 and
D 4c 2 + 1. It is easy to check that
[ (L+ !) , ]
[( '""I .1]
[ (u,!';) - I ]
[ (-'1"'ij (2'-') ]
[ L :t 2 V1J ]
- I + yf[j 
:!,. ,\
[ I + In ]
21"
::: [ 3.C 1 + J iJ ]
2
::: [ 2C - 1 + -Ji5 ]
'lc
1 ,
== c ..
1 ,
I.
2c - 1 .
so tha.t the nmtinueJ fraction of (1 +..[i5) is
as 2(: I > ;, and its period is of length 3.
85.
(85,0)
Let G be a group which has the following two propertie5:
(i) G has no elem<,nt. of order 2,
(ii) (,,'y)2 (yx}2, for all :C,y E G.
Prove that G if' ahdian.
Solution:
, , .
For x,y E G we have
,,:2 11 ::: ((:I:y-l}y)2 y
(y( xy-I »)2y (by (8!J.0)(ii)
(1/:r.,Ij-l)(Y3:y-l)y,

152
SOLUTIONS
SOLUTIONS
153
that is
for all i = 1,2,. . . ,n. Prove tha.t 0  det A  1.
(85.1)
x 2 y = yx 2
:1:.- 1 J,-1. 1 ;
:r(};-1)2y- I x
;q,-l(.r,-I )2x,
Solution: tet A denote 0111' of the eig'?nvahws of ct and It,1; ;!2 (Ie Q) h. an
eigellvector of A r.orr<;spolHling to >., 50 that
K'xt, we have
(hy (Sfi.1))
(86.1) .43;= ),;r .
that is
SiC1- :I, 0= (:II: _.. ,.1:"f "lid choose i, 1 <:: .; < 1/., o I hat
(R!).2 )
,/,' ly-1:r:::: ;1;.tJ- l. r -1
"rtl
IHax 1:/'11 f 0,
l::O;.1:5 n .
Similarly, we have
(85.3)
y- 1 x- 1 y y:r.- 1 y-1.
From the i-th row of (8tU) , we obtain
(p,:r.- 1 y-1 yl :::: xy( x-1y-1x )y;c-1y 1
xy(xy-1 x -1 )yx- 1 y--I (by (85.2))
xyx(y- 1 x- 1 Y)X- I y-l
xyx(y:j;-ly-1 )x- 1 y-1 (by (85.;1))
(xy)2(X-1 y-1)2
(xy ?(y:t' )-2
(yx )2(yx )-2 (by (85.0)(ii»
n
2.::: ai33:j :::: AXi ,
j==1
Then we obtain
so tha.t
n
(>' - l)xi 2.::: aijX 3 ,
3=1
#i
a.nd thus
:::: 1,
'"
I>' 111 x il 12.::: aijxj\
j=l
#i
and thus, as G has no elements of ordcr 2, we have
xy.r:- 1 y-1 = 1 ,
that is ;!:y = yx, proving that G is a.beJian.
n
<:: 2.::: \a;. j II): 31
j=l
#i
86. Let /1 :::: [aij] be an n X n real symmetric matrix whose entries
satisfy
n
S, 1;l:i l 2.::: !aij!
.io=l
j#i
< !.r.;!,
(81:5.0)
n
aii = 1, 2.::: la;jl  2 ,
j==l
showing that
(86.2)
\>. - 11  1 .

154
SOLUTIONS
Since A is a, real symmetric tnatrix, ). is rea] ami from (86.2) we see that
(86.3)
0:SAS;2
Lei. /\1,"" An dt'liote the 'Ii eigenvalues of it. I'a.ch Aj is nonntga.tive by (81).3).
Thu& we have
o <: dt't il
'\1'\2'" An
( I _R
-( AI + ).'2 + ... + An) )
.n /
( trace A ) "
( 1 ) ..
-n
\n
-- I
<
87. Let R be a finite ring containing an dement T which is not a
divisor oJ zero. Prove that R must haV( a multiplicative identity.
Solution: As II is a finite ring there exist integers 111 and 'Ii such that
(R7.1)
1'''' Tn, I S; Tn < n
We wish to show that
(87.2)
l'
1'1:
,
for some integer A 2:: 2. If m 1 we may takt) k
we have
1'(r m - 1 _ rn-l) O.
As l' is not a divisor of zero, We must have
n. If m 2:: 2, from (87.1),
(i.:)
1'",-1 1' n - 1 = 0 .
If m - 2 we may tali:e k 'Ii - 1 (2:: 2). If 'In 2:: 3, f.rom (87.3) \W have
r{r m - 2 _ 1'''-2) = 0 .
SOLUTIONS
155
As l' is not a divisor of zero, we must have
1' m - 2
n.-2
l'
o.
If m ,,_ 3 we may titke k -c /l. . - 2 (:> 2). Continuing in this wa. y .' we S('(' that
(87.2) holds with k It - m+ 1 (2: 2). For any x E R. we have from (R7.2)
:1'1'=,0.1.
and 50
(.1' .I'l'l. 1)1'
0_
Ali r is not a, divisor of zero, we SE'€ that
:1: = :1.'1'/;-1 .
(87.4)
Similarly, we have
(87.5) x r.l"- 1 x.
From (87.4) and (87.5) we selJ tha,t 1'k-l is a multiplicative identity for R.
88. Set J" {I, 2,... , n}. For eadl non-empty subset S of .In define
It'( S) = max S - mill S .
. BES BES
Determine the averi1ge of wl.S) over a.1l non-empty subsets 5' of I n .
Solution: I'or 1 < k S; 1 S; n let S( k, l) d()f\ote the set of' subsets of Jr. with
miTIS
BE,';
k,
maxS
sES
1.
We have, for aU S E 5'(k,l),
w(S)=l-k,
and
{ I, ifk t,
IS(k,l)1 = 1-k-1, if k < l,
'II'
illl

156
Then we hav('
L.: toeS)
','cfs.Jn
SOLUTIONS
L.: L.: w( S)
1 </"<1<,,, 8(;.'I(k,l)
L (1- k)IS(k,l)l
1 :$k<I:5"
(l I.: 1'21-;:- 1
I:$/"<::l>
11-1
11.-1
z=h"l
k=1
"
I L l:i
1=/;+1
n
'\
L..,i
:: 1
k=1
l=k+1
1"t-l
L.:
k=1
('17 - 1 )2 n +1 - (k _1)2 k + 1 )
n-I
L k2- k - 1 (2 n +1 - 2k+1)
k=1
'fl.-I
(n 1)2 11 L.: T k
k=1
n-l
L.:(k - 1)
k=1
('17
n-1 n-1
-2" L.: k-r k + L.: k
k=1 k=1
1)2" ( 1 -  )
, 211.-1
...2" ( 2 _ ( '17 + 1» ) + n - 1
2 n - 1
l):t' 2('17 - 1)- 211.+] + 2('17 + 1) + '17 .... 1
3)2" + (n +{),
= (n
= (n
so that the required average is
('17
n 1,2,....
89. Prove that die number of odd binomial coefficients in each row
SOLUTIONS
157
of Pasca.l'" triangle is a. POWt'l' of 2.
Solution: The ellt.ries in t.he n-th row of Pascal's triangle iln the mdtidcnts
of tlit' IH)Wf'I"S of x in 1.11(' expansion of (1 + 3)". vVe write n in
binary notation
p.m. I )
'1/ 'c 2'" + n2 ..J . _ . + :J"'" .
(9.2)
W]IC1'£ 0.1,' .., f1.k arc intcgers such thai
III :> (l2 :> ". :> (lk ::: [) .
Now
(1+;rf
(I + :1:)'1
(1 + 3f
- I + 2x + :1: 2
_ (:I + 3;2)2
(1 + :'1;4)2
== 1 +x 2
== 1 + :,/;"
== 1+:1: 8
(mod 2),
(mod 2;,
(mod 21,
and so generally for cl.ny nonnegative integer a we have
Thus, we have
(1 + ;/;)'"
(I + .r. fl" == 1 + x 2 " (mod 2) ,
(1 + x)2"1+2"2+...+2'"
(1 + x)2"1 (1 + x)2"2 ." (1 + x)2"k
(1 + X 2 "1 )(1 + x2"2)... (1 + X2"k) (mod 2)
1 + (X2"1 + X2"2 + . . . + :t 2 "k)
( 2"1 +2"2 + 2"k-1 + 2 0 k )
+ x +... x
+...
+:,/:2"1 +2"2+...+2"k (mod 2) ,
and the llumlJer of odd weffidents is
1 + k + () +... + G:) 2 k .

158
SOLUTIONS
90. From tbe n x n array
[ (n 1 2 3 'II
n+l n+2 n+3 2n
2n";'" 1 2'11 +2 271 -L;3 3/1
I)n + 1 (n. I)n +2 (11 1)11 +;1 ')
n-
". number :1:1 is sek.ctcd. The row and column cont.ainhll!, ;1:1 <'In' t.hen delet.ed.
From tb., resulthg array a numtJP1' .1'2 i, s('lecjp,L and it,., 1'0\,' ilild colu1Il1l
deleted a>; before. The :=:election is continued until only one number :e'l remains
available for selection. Determine the sum Xl + :r2 -I- .. . .j... X n '
Solution: Suppose that :r., 1 ::; i ::; n . belongs to the t'i-th row and the
-5i-th column of the array, Then
Xi = (1'; .- 1)n 'S;, 1::; i ::; n ,
and so
n
LXi
;=1
n
nLt'i
;0=1
n
n 2 + L S;
.i=l
;'oJow {1'1"", 1'n} and {Sl,.... .s,,} a.re permutations of {l, 2, . .., n} and so
n n n (, J)
L"'i=Ls;=Li= nn
;=1 io=l i=1 2
Thus
n
LXi
;=1
2 , n(n+ 1) n(n 2 -!..1)
-n _._=
2 2
91. Suppose that p X's and q O's are placed on the circumference of
a circle. The number of OCCIHrences of two adjacent X's is t1. and the number
of occurn>nces of two adjar.ent O'sis b, Determine n b in terms of p and q.
'I
:1
,j
SOLUTIO:;\J'S
159
Solution: Let
N x : e , N xo , N ox , N oo
denote the llnwber of oc.curreTlct's of XX. XO, OX, 00, respectively. 'fht'u
dearly we have
{ N"x = a ,
lv'"" = b,
T . r __.
\x,. + ;\ ,,'>. - P ,
_'V OII -!- N o . l ; -.::: fJ 
so that
0, b
IV:rx - ;"V oo
(N x >; ,L Noo) (N oo + N",.) + (No' .. N;",,)
p - q + (N ox - Nxo) ,
Finally, we show that N ox = N xo , which giv<:'$ the resuU
a.-b=p-q.
To see that N ox = N xo we consider the values of a function S ;<1> :ve .make one
clockwise tour of the circumference of the circle, starting an.d nmshmg at the
, 1. I . t ' al1 m e let C'_ 0 Then as we tour the CIrcle, tbe value of
same potn. 111 1 y,.., , ,J - . ' .,
S is changed as follows as we pass from ear.h X or 0 to tht, next. X or 0:
new value of S old value of 5 1- €,
where
€
{ -
, in going form 0 to X ,
. in going from X to X or 0 to 0 ,
, in going from X to 0 .
Clearly the value of S at. the end of \,he tour is N ox N,.o. owever, S must
be 0 at' the end as we have returned to the starting point. ThIS completes the
proof of N o3 ; = N"", <Uld the solution.

160
SOLUTIONS
92.
In the triangular array
1
1 1 1
1 '2 3 2 1
1 ;) 6 -; 6 a 1
4 10 Hi 19 16 10 4
(92.0)
('VNY (,JltfY (exc('pt tilt' t(IP 1) is tll(' slim of thp ('Id,r.... t1 illlI1H'di".!,('jy all"v!' it,
and the entrieli II and c immediately to the left ,wd right of 0.. Absenee of a.1I
entry indicalcs zero, Prove that every row after the secon<1 row contaiJU; an
entry which is even.
Solution: The first eight rows of the triangulaf array taken modulo 2 are
given in (92.1).
1
1 1 1
1 0 1 0 1
(92.1) 1 ] 0 1 0 1 1
1 0 0 0 1 0 0 0 1
1 1 1 0 1 1 1 0 1 1 1
1 0 1 0 0 0 1 0 0 0 1 0 1
1 1 0 1 1 0 1 1 1 0 1 1 0 1 1
The first four entries in the fourth row of (92.1) a.re 1 1 0 1, which aJ'e exactly
the S<J.llle as the first four Dutries in the eighth row. Thus the pattern
(92.2)
1
1 1
101
1 1 0 1
000
1 0
o
repeats itself dow1l the left-hand edge of the array. As each row of (92.2)
con/.ains at least one zero, every row from the fourth 011 down contains an
161
SOLUTIONS
r
r........
1 ' h ' ]  , " ompl tes the P roof, as the third row contains an even
even l1I1Inbcr. "  "
number.
. I
;
01
I' I nb ' I '" Xl r s<I,lis!ies
A sequcnce 0 n rea. mn .... ' "..,"n '
93.
i\
{ Xl 0:: 0,
I \ + I 2..::i::;n.
i J ; X._I r,
. " t ' '0 "0. 1 numb,'f IktNmilll' a,lowN hound for tlw avr-rage
wlwl'<' e IS o. pOST He 1(',1.1 - .
f " .' .\::; a. runc.tioll nf (' only,
o Xl,.....&Jn 4:.
(93,0)
Solution:
I t be an'i . ' rea.l number such tha.t
. ,e Xn+I
IXn+ll:::: \x n +cl.
Then, we have
:1
n+l
2
L,.; ..f. i
i=l
n+l 71.+1
L Ix;l2 == L !Xi-l + cl 2
i=2 i=2
n+1
L(Xi-l + ci
i=2
,,+1 n+l
L X;_l + 2c L Xi-I + C 2 tl.
i=2 i=2
n
LxI + 2c
i=I
n
Xi +c 2 n,
so that
n
L 2
O <: X 2 ' 1 == 2c Xi + c 7/. ,
.....2 '1l+
i=l
and thus (a.s c> 0)
1 n C
L X i;:::-2'
n i=1

162
SOLUTIONS
94.
Pro\."<) tha.1, the polynomial
(9/LO)
f(:I: I = ,1''' I :/:,1 -t 3: 2 .j :1: I- 5
is iITNludblt, over Z If,r n » .1.
Solution: Suppw,(' f('1:) i., rl'rhldbl0 ov(']' Z T/1"1I (/1pl'l' t'xij llIonic poly-
nomials f,(.r) alld h(.r) witb ;uJ"'r:.ral c()eflicj('trl, ,-,lIdl I ha',
UI'LI)
/, r) !J!'./';!I(:J"), dt').'; r, >' dP1', It
Thus, we havE:'
.') f(O) [1(0)11.(0),
and, a.s 9(0), h(O) a1'(, illteg('rs <.Iud ,1 is primc, we lrav!) witJlOu1.ln,;fi ot generality
9(0) :1:1, h(O) H,.
Lt
,.
g(x) III:r- ii,i)
_';=1
be the facf.orization or g(:/:.) over C. TlI';n, \'IrE; JJ,w('
,.
1 19(0)1 .. II IfJ.i1 ,
j=]
and o at least one of the 1;1 j l is less than oj' equal to 1, say
/;31/  1, 1  1  r ,
Henc('
/.f(;ll Ii
/,6 1 '/3j! + fir I ill I !')i
> !) - IfJLI -- liW rfJ/ 1 3 d"
:::,51]11
1 ,
SOLUTIONS
I
!II
whidl t:ontr;1dicts
,
163
Thb prows tlli.t /(.1') is irreducible over Z.
f«(3z) g«(3/ )h(tl[) 0 hUJz) 0 .
. 1 I I )Nt>rrn;ll(' 1;11<'
9 r:: I ' I (/ hp 11 ( > ,'1 \ dbt.inct r0<l !lI11Il )l'l'fi.
.t). Jt- , at,..., -71 ,.._ I
II 1 .' f 1 ' 11( ' ""RtA {11 of '11 2 lilH'ilr e(jlJatiolls
r;c'lwra so 11 Ion 0 , . ",' ,"
(9,5.0)
{ . :"1 ..I '/'2 1"" -I- ,/'"
a , IX] l rL2J:2+ ...., (t".X n
a 2 . c _ , ,,2' Cl ' I ... 1- a2xn
I'] "2" ".
' I ' 1'1.-3 + n-:I,
a-' ;.{:] 1- (t2 ;C2 + . , . a" wn
= 0,
= 0,
0,
= 0,
ill the 11. unknowns XI,. .. ,In.
Solution: Set
For k
0,1,... ,n
(95.1)
f(x) = (;/;- al)(:r a',!)...(x-u n ).
1 the pa:rt.ia.l fraction expansion of ;r k If( x) is
Xk
f(a:)
n
?= ;r
$=1
(Ii
Multiplying both sides of (95.1) by f(x), and equating coefficients of xn-l, we
obtain
(95.2)
This shows that
n a
L: - f ' ( ' )
i=l '1,
{ 0 , k
1 ,k
0,1,...,11. - 2 ,
n-1
if = (f/(l) "'" f'(n) )

16.1
SOLCTIONS
SOLUTIONS
and
96.
Evaluate the sum
( 0,1 an )
:/; == jl((ld "'" f'(a n )
N 2,3,...
L
H( r'lT) ==
mn
are two solutions of (fl!!.O). Thl'se twn solutions a.re lint:'ady in <leJl!"lIdt'u I ; 1i:11'
otherwise tht'n wouJd pxjt I'''al 11111111)(']'<; ," and i ( not both zero) such thai
ji1t<n'SN
nt+n>N
IX' D(m ,>I )=1
-"11. + tp == (0. ...,0) .
that if<
(95.3)
s + Ill,; == 0, i == 1,2,. . . , n .
Solution: For iV :>. :1 we ha.v£'
If t == 0 theIl from (95.;H we havi' B 0, which i:; a cont.radict.ion. Thus, I 1: (I
and (95.3) giw;;
L 1
H(.'V ) L + rnn
m.1l
1m<n:$N.-1 l1tJ.<n=N
rn+'1>N 1n+n>N
GCV(rn,n)=1 GCV(m,n)=1
1 L 1
L r1/.n mn
111t<nN-l ln.<niV--]
m+n>"'--1 m+n=.;\'
GCD(m,,,)=1 GCD(m,n)=l

i==1,2,...,n,
(1;
-1:'
which c011tradict.s the fad that. th! a; aI'e dislincL TllU tILe solutions l a.nd
]1. are linearly ind'pendent..
:'{(1xt, as the ai art" distinct, the Vandermonde determinant
1 1 1
l!l °2 {/,n-2
a? 11. 2 a_2
2
n-:1 n-3 (L= I
tt 1 11. 2 I
does 1101. vanish, and so the rank of the coefficient. matrix of (95.0) is 1/. - 2.
Thus all solution,; of (!J5.0) are given as linear cornbinatiomi of a.ny two linearly
independent solutions. Hence an solutions of (95.0) are given hy
H( N - ])
L
1:$m<N/2
GCD(m,N)=1
1 L 1
N m,
l:$m<N/
GCD(m,N)=l
] L
+ N m
l:$rn<N
GCD(m,N)=l
S(N - 1)
(Xl,...,Xn)
t).JJ.. + ill
( Q + (3(1'1 Q + !3 an )
-f'(al) ,...,- f'(a)
for real numbers n and (i.
165
+
L
m.N
l'in<X
GCD(rn,N)=l
1 L 1
+ N m
l:$m<N
GCD(:r.,N)=1
]
L
Ne-m
N
1:$rn<N/2
GCD(m,N )=1

5'(A' -- 1)
1 '\' 1 1 L 1
-\
1V L-o m N m
l':S:m<N 1::;'111<.N
'iCD{m,N)=l aCf)(m)\')=l
Ccc S l IV - 1) ,
remembering that GC IJ(.N 11, N) > 1 fOJ' even ;V (2: <1). Thus, we havc
8(1\') S(A -- I: - SU\ J-
8(1)"'1/2.
97.
FV<lJna,te the limit
(97.0)
1 nit.
L = lim '\. ''\' 
"_ex, n  L-o J '2 .L 1.:2 .
J=1k=1
Solution: Pa.rtition the unit squa.re [0, 1J x [0,1; into ,,2 suhsqlJ<l,r"s by tIt('
partit.ion points
{Uln,k/n): 0  ),1,;  n}.
Then a IU"ma,un sum of the functiorL x/(;r.2 .\. y2) for this partition is
\.' jln 1
L..- Uln' ) 2 + (k/n}2 n 2
1t,""
)
"" :-- 2 1'"
nL..- J + 1\'-
15:.lt k 5Jt
a.n d al so
',' jln 1
11m L..- ( ,/'2 ("/F 2
n. '00. J n J' -\ : n 'it
lJ,k:5n
it. x
'd:t'd!J,
, [O,Jlx [0,11 ;<2 .1. y-
sO tha1 (9..0) becomes
!o l !o l :r
L = -.' dJ: d'l}
, 0 ,0 X 2 .1. y2 .
SOLUTIO:;\lS
167
t'!4 1 "ce cos () dl' dO .1. 2 l : ce COB e dr- de
.-- 18=0 '1"=0 Jli_'Jr/4 T-O
/4 1 /2
[ nO -\- cot 0 ,10
in IT/I
To /-1.1 [In sin 0 ]:
r./I -In(l/Ji,
that is r .= 7[/1 + (1112)/2.
98. Prove th<l.t
(98.0)
:.lr. . 2r. / -
tan .L 4 SUI Ji = " 11. .
Solution: :For col1vetlien('c we let 7' = 'ir /11, a.nd set
r; =: co,:;p, . = sin]'.
. pi I ( + is ) l1 - .-1 that is
Th.en, we ha.ve c -1- zs = co an( so (' -,
8 'I' 3 'J O 74 I If ' 2 / j s"i
11 1 ' 1 JO'  5 c.\1 02 - 165e s'  -\- ," e s ," I (.
e + e S -. ;:h - :;, c _. 10 11 . 1
_ 1 " 2 fi (; 3 'J O 4"7 i + 165e3 .8 + 55r;2 .\). 11eo5 05  -.
-«,) (' S " .. ..
Equa.ting imaginary parts, we obtain
" :; 6 5 'i30 4 7 .1 55(;2 S9 _ 05 11 0 .
(fiR-I) 11e 1O s 165e s + 462c s' -. r;. .
.From (mU), as . i 0, we have
. 64 . 0 46 -28 JO 0
( , 11c1O-165e8,2-\-462e's .-33e. +0;)('05 .
f!8.k)
::-\I:xt, as
(fl8.3)
C2 1 .2,
the equation (98.2) becom!'s
11 22005 2 + 123205 4 - 2816s 6 .1- 281605 8 ,- 102405 10 = 0 ,
(%.4)
and thus
;

168
SOLUTIONS
SOLUTIONS
169
(lIs - 41.q3 - 32.'<;) 11r. 2 (1 4.q2)2
121.q2 - 968,q4 ./- 2640i; 2816,g8 + 1024,g1O
-.11(1 $2)(1 8,q2+,g4)
-11 + 220,q2 1232.q.l + 216,g6 2816,<;8 + 102.1s 1O
() ,
Evaluate the sum
00 c
s 1: n(n'/- 1) .
n=1
by (98..1). This proves that
. I. ,pt 1.: he it ! wsitive integ£'L We have
SolutIon:
(mu))
I J" ,I.\.g:' - ;t.4"
= -L/II
('( I.
k
\1::.., ( ':.!>
 '/I
n=-I
k
L Cn _0'
n:] n
A,+l .
L (.n -]
n=2 n
'/1('/1\ 1
n=l
('/I. \
II 1.1)
Xext, we have
tall 3p -I- 4 sin 2p
+ 8sinp cosp
A' ( ) C
"'\ c? :, C. 1l ---1 _ 
Cl+L...-- n k+1
n=2
that is, using (98.3),
k 1. Ck
1/-"---
. L...- n 2 k + J
n::;:::;;:l
( 98.6)
tan 3p + 1 sin 2p
k 1
= L -1:'+1.
n=l
 ( cA-lnk)
L...- n 2 k + 1
n=l
Ink
k+1.
Then, from (98.5) and (98.!i), we obt(1,in
ta;n 3p + 4 sin 2p ::!:J1i.
311" . 21T r;-:;
tan + 4 SID = v 11 ,
11 ] 1
Letting n 00, and using tIw fact that
lim (Ck --In k)
k-+oo .
As tan 3p > 0, sin 2p > 0, we must have
exists, and al,,;o
Ink
lim - =0,
k-+= k + 1
a s required.
99.
For n
1,2, _ _ . let
we lind that
00 1 1T 2
S=L n 2=fi'
n=1
C n
1. 1 1.
1.+ +-+"'+-.
2 3 n

170
100.
SOLUTIONS
For x > 1 determine the sum of the infinite series
x  
+- +
;1' + 1 (:1' + 1)(;1;2 + 1)
Solution: For 11. it positive integer, BPI,
S"Jx)
so that
.1: :C 2 :l2J"/
;:-+1 + ( .1' + 1 :I( .l';; + 11 . . +...j C -" ..f
« , 1)(.:1':' + 1)... (:1'"" + 1)
.1'
1
X x 2
+--+""'+
- 1 :1: 4 - 1
( - . -- -  ) +
x - 1 x 2 - 1
1
+... +
Thus, as x > 1, we have
x - I x 2n +1 - 1 .
giving
I " Sn(X) 1
Ull-
n-.oo x-I x 1
:it!
:r.
-. 1
1
- x 4  1 )
I
:r. x 2
x + 1 +(x + 1)(x 2 + 1) +... = J1:, 8,,(.1:) = 1 .

,
Problem
THE SOURCES
01: Gauss, see Werke, Vol 2, Gottingen (1876), pp.11-45, showed that
( E=.!. ) 2
,,:Tl + 1..)"2 + . . . + W"(p-l)/2 = (-1 + i 2 -jP)f2 ,
( E=.!. ) 2
w''' + W1!2 + ... + w1I-(p-l)/2 = (-1 - i 2 .,jP)f2 .
04:
05:
09:
This result is implicit in the work of Gauss, ('(' Wet'ke, Vol '1.,
Gottingen (1876), p.292.
The more general equation y2 x 3 + « 4b - lYI 4a l ), whcr<. (1, has
no prime factors 3 (mod 4), is treated in L.J. Mordell, Diophantine
Equations, Academic Press (1969), pp.238-239.
This prohlem was suggested by Problem 97 of The Green Rook. It
also appears as Problem E2115 in American Mathematical Monthly
75 (1968), p.897 with a solution by G.V. lcWi1lia,ffis in American
'Mathematical \1lonthly 76 (1969), p.828.
10:
This problem is due to Professor Charles A. Nicol of the University
of South Carolina.
11:
Another soh]tion to this problem is given in Crux \ifathemat.icorum
14 (1988), pp.19-20.
14:
The more general equation dV 2 - 2eVW dW 2 1 is t.reated in
K. Hardy and K.S. Williams, On the solvability of the diophantine
equation dV 2 - 2eVW - dW 2 = 1, Pacific Journal of Mathematics
124 (1986), pp.JL15-158.
17:
This generaJizes the well-known result that the sequence 1,2,...,10
contains a pair of consecut.ive quadratic residues modulo a prime
2:': 11. The required pair can be taken to be one of (1,2),(4,5) or
(9, 10).
19: Based on Theorem A of G.H. Hardy, Notes on .'?ome points in the
integral calculus, Messenger of \1Iathematics ..18 (HJI I), pp.107-1l2.

20:
This identity can be found (eqn. (4.9» on pA7 of H.W. Gould,  ";
Combinatorial Identities, Morgantown. W. Va. (1972). :1
The more general equation aIxI + ... + anx n k is trcated in Hua :j
LooI<eng, Introduction to Ntlmber Theory, Springer-Verlag (1!J82),
see Theorem 2.1, p.276.
21:
22:
Finite sums of this type are discussed extensively in Chapter 1.5 of
W.L. Ferra.r, IIightT Algebra, Oxford University Press (H):;O).
25:
See Problem 2 on p.113 of W. Sierpinski, Eft'menta!"?! Thmrll of
Number.'!, WaIsaw (196,1).
Suggested by Problem A-3 of the Forty Seventh Annual William
Lowell Putnam Mathematical Competition (December 1986).
The discriminant of f(x"), k  2, is given in terms of the dif;.
criminant of f(x) in R.L. Goodstein, The. discriminant of a certain
polynomial, Mathematical Gazette 53 (1969), pp.60-61.
H. Steinhaus, Zadanie 498, Matematyka 10 (1957), No.2, p.58 (Pol-
ish).
26:
29:
30:
34:
This problem was given as Problem 3 in Part H of the Seventh An-
nual Carleton University Mathematics Competition (1979).
37:
Based on a question in the ScholaIship and Entrance Examination
in Mathematics for Colleges of Oxford University (1975).
38:
Based on a question in the Scholarship and Entrance Examination
in Mathematics for Colleges of Oxford University (1972).
39:
Based on a question in the Scholarship and Entrance Examination
in Mathematics for Colleges of Oxford University (1973).
40:
Based on a question in the Scholarship and Entrance F,xamination
in Mathematics for Colleges of Oxford University (1973).
172
41:
45:
47:
48:
49:
52:
53:
56:
59:
62:
...
" 63:
:t
64:
This is a classical result, see for example H.S.M. ?o:,eter and S:L.
Greitzer, Geometry Revisited, Mathematical Association of AmerIca
(1967), pp.57, 60.
" t d b T S Chu An g les with mtional tangents, America.n
imgges e y .. , .
MathematicallVlonthly 57 (1950), pp.407-408.
S t d I y W Cross P Hilton, J. p,'ders"II, ICY. Yap, An alflo-
uggcs ,e I . 1 , . . . .. M h' t' Maga-
rithm fm' 1mdlil}lication in modlLlar arzthmetzc, i at (.ma ICS l .
zinc !i!) (1!)Rf5), pp.Hi7-170.
d S t 3 P 8 of Th Skolem Diophantischc Gleidmngen,
Base on a z on . _ ,
Chelsea Publishing Co., New York (1950).
d E I 1 . n D G Mead Integration, American Mathe-
Base on ,xamp e. I .. _ .
matical Monthly 68 (1961), pp.152-156.
Suggested by 5.4.5 of L.C. Larson, Problem-Solving Thr'olLgh Pr'ob-
lems, Sprinf!;er-Verlag (1983).
See Problem 48 of Lewis Carroll's Pillow Proble.ms.
See Problem 1 on p.13 ofW. Sierpinski, Elementary Theory of Num-
bers, Warsaw (1964).
See Problem 12 on p.368 of W. Sicrpinski, Elementary Theory of
NlLmber's, Warsaw (1964).
This problem was suggested by I)roblem A-4 of the r'irty Eighth
Annual William Lowell Putnam Mathematical CompetItion (Decem-
ber 1977).
Th ' P oblcm was shown to us by I'rofes.<;ors David Richman and
IS r . C r
Michael Filaseta of the University of South aro ma.
Th ' It ' d e to II J S S mith On the vallLp of a certain arith-
IS resu IS u . . . , . al S
metical determinant, Proceedings of the London Mathematic 0-
ciety 7 (1876), pp.208-212.
173

68: This is a well-known problem, see for example 4.4.4 in L.C. Larson,
Problem-Solving Through Problems, Springer-Verlag (1983).
69: This problem was suggested by Problem 3 of Part A ofthe Fifteenth
Annual Carleton University Mathematics Competition (1987).
70:
Forms ax 2 +by2+cz 2 which represent every integer have been charac-
j..rized by L.E. Dickson, The forms a,;r2+by2+t.::;2 which rqlresent all
integers, Bulletin of the American I\.1atht'matic:tJ Society, 35 (1929),
1)p.fi i 'i-59.
86:
Suggested by ideas of 97.5, Estimates of characteristic roots, in L.
Mirsky, An Introduction to Linear Algebra, Oxford Universit.y Press
(1972).
I. .. A c:(()[ ED
IN SCIENCE AND MATHEMATICS 
l'
In
75:
82:
This problem was suggested by Problem 95 of The Green Book.
Suggested by K.A. Hush, On an application of the mean value theo-
rem, American Mathematical Monthly 62 (1955), pp.557-578.
89: This is a well-known problem. A generalization to the multinomial
theorem is given by H.D. Ruderman in Problem 1255, Mathematics
Magazine 61 (1988), pp.52-54.
94: Suggested by an example given in a talk by Professor Michael Fi-
laseta at Carleton University, October 1987.
95: See Problem 2 on p.219 of W.L. Ferra:r, Higher Algebra, Oxford
University Press (1950).
98: See Problem 29 on p.123 of E.W. Hobson, A Treatise on Plane and
Advanced Trigonometry, Dover Publications, Inc. New York (1957).
174