101 problems in maths

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National Library of Australia Card Number and ISSNAustralian Mathematics Trust Enrichment Series 1SSN 1 326-0170

101 Problems in Algebra iSBN 1 876420 12 X

THE AUSTRALIAN MATHEMATICS TRUST

ENRiCHMENT SERIES

MiTT E EJ

Chairman GRAHAM H POLLARD, Canberra AUSI'RALiA Editor PETER J TAYLOR, Canberra AUsTRALiA

WARREN J ATKiNs, Canberra AUSTRALIA

Eo J BARBEAU, Toronto CANADA

GEORGE BERZSENYI, Terra Haute USA

RON DUNKLEY, Waterloo CANADA

WALTER E MIENTKA, lincoln USA

N1KOLAY K0NsrANT1N0v, Moscow RussiA

ANDY Liii, Edmonton CANADA

JORDAN B TABOV, Sofia BULGARiA

JOHN WEBB, Cape Town SouFH AFRiCA

The books in this series are selected for their motivating, interestingand stimulating sets of quality problems, with a lucid expository stylein their solutions. Typically, the problems have occurred in eithernational or international contests at the secondary school level.

They are intended to be sufficiently detailed at an elementary levelfor the mathematically inclined or interested to understand but, atthe same time, be interesting and sometimes challenging to theundergraduate and the more advanced mathematician. lt is believedthat these mathematics competition problems are a positiveinfluence on the learning and enrichment of mathematics.

PREFACE

This book contains one hundred highly rated problems used in the train-ing and testing of the USA international Mathematical Olympiad (IMO)team. It is not a collection of one hundred very difficult, impenetrablequestions. Instead, the book gradually builds students' algebraic skillsand techniques. This work aims to broaden students' view of mathemat-ics and better prepare them for possible participation in various mathe-matical competitions. It provides in-depth enrichment in important areasof algebra by reorganizing and enhancing students' problem-solving tac-tics and strategies. The book further stimulates students' interest forfuture study of mathematics.

INTRODUCTION

In the United States of America, the selection process Leading to par-ticipation in the International Mathematical Olympiad (IMO) consistsof a series of national contests called the American Mathematics Con-test 10 (AMC 10), the American Mathematics Contest 12 (AMC 12),the American Invitational Mathematics Examination(AIME), and theUnited States of America Mathematical Olympiad (USAMO). Partici-pation in the AIME and the USAMO is by invitation only, based onperformance in the preceding exams of the sequence. The Mathemati-cal Olympiad Summer Program (MOSP) is a four-week. intense train-ing of 24-30 very promising students who have risen to the top of theAmerican Mathematics Competitions. The six students representing theUnited States of America in the IMO are selected on the basis of theirUSAMO scores and further IMO-type testing that takes place duringMOSP. Throughout MOSP, full days of classes and extensive problemsets give students thorough preparation in several important areas ofmathematics. These topics include combinatorial arguments and identi-ties, generating functions, graph theory. recursive relations, telescopingsums and products, probability, number theory, polynomials, theory ofequations. complex numbers in geometry, algorithmic proofs, combinato-rial and advanced geometry, functional equations and classical inequali-ties.

Olympiad-style exams consist of several challenging essay problems. Cor-rect solutions often require deep analysis and careful argument. Olym-piad questions can seem impenetrable to the novice, yet most can hesolved with elementary high school mathematics techniques, cleverly ap-plied.

Here is some advice for students who attempt the problems that follow.

Take your time! Very few contestants can solve all the given prob-lems.

Try to make connections between problems. A very importanttheme of this work is; all important techniques and ideas featuredin the book appear more than once!

Olympiad problems don't "crack" immediately. Be patient. Trydifferent approaches. Experiment with simple cases. In some cases,working backward from the desired result is helpful.

Even if you can solve a problem, do read the solutions. They maycontain some ideas that did not occur in your solutions, and they

VIII

may discuss strategic and tactical approaches that can be used else-where. The formal solutions are also models of elegant presenta-tion that you should emulate, but they often obscure the torturousprocess of investigation, false starts, inspiration and attention todetail that led to them. When you read the solutions, try to re-construct the thinking that went into them. Ask yourself. "Whatwere the key ideas?" 4How can I apply these ideas further?"

Go back to the original problem later, and see if you can solve itin a different way. Many of the problems have multiple solutions,but not all are outlined here.

All terms in boldface are defined in the Glossary. Use the glossaryand the reading list to further your mathematical education.

Meaningful problem solving takes practice. Don't get discouragedif you have trouble at first. For additional practice, use the bookson the reading list.

ACKNOWLEDGEM ENTS

Thanks to Tiankai Liu who helped in proof reading and preparing solu-

Many problems are either inspired by or fixed from mathematical contestsin different countries and from the following journals:

High-School Mathematics, ChinaRevista Maternatic RomaniaKvant, Russia

We did our best cite all the original sources of the problems in the solu-tion part. We express our deepest appreciation to the original proposersof the problems.

ABBREVIATIONS AND NOTATIONS

AbbreviationsAHSME American High School Mathematics

ExaminationAIME American Invitational Mathematics

ExaminationAMC1O American Mathematics Contest 10AMC12 American Mathematics Contest 12,

which replaces AIISMEARML American Regional Mathematics LeagueIMO International J\Iathematical OlympiadUSAMO United States of America Mathematical OlympiadMOSP Mathematical Olympiad Summer ProgramPutnam The William Lowell Putnam Mathematical

CompetitionSt. Petersburg St. Petersburg (Leningrad) Mathematical

Olympiad

Notations for Numerical Sets and FieldsZ the set of integers

the set of integers modulo nN the set of positive integersN0 the set of nonnegative integersQ the set of rational numbersQ+ the set of positive rational numbersQ the set of nonnegative rational numbers

the set of n-tuples of rational numbersR the set of real numbersIR+ the set of positive real numbersR the set of nonnegative real numbersR'2 the set of n-tuples of real numbersC the set of complex numbers

1C

PREFACE vii

INTRODUCTION ix

ACKNOWLEDGEMENTS xi

ABBREViATiONS AND NOTATiONS xiii

1. INTRODUCTORY PROBLEMS 1

2. ADVANCED PROBLEMS 13

3. SOLUTIONS TO INTRODUCTORY PROBLEMS 27

4. SOLUTIONS TO ADVANCED PROBLEMS 65

GLOSSARY 131

FURTHER READING 137

iNTRODUCTORY PROBLEMS

1 INTRODUCTORY PROBLEMS

problem 1Let a. b. and c be real and positive parameters. Solve the equation

CX-r ax = ax + bi + - cx.

Problem 2Find the genera! terni of the sequence by = 3. x1 = and

=

for alln E N.

Problem 3Let x1. x2 x,, be a sequence of integers such that

(i) n:

(ii) x1 + x2 + + = 19:

(iii) =99.

Determine the minimum aiid niaxiniuni possible values of

Problem 4The function f. defined by

ax + h1(x) = c.r+ d

where a. b. c. and d are nonzero real nunibers. has the j)rOperties

f(19) = 19. f(97) = 97. and f(f(x)) =

for all values of x. eXCCl)t --Find range of f,

2 1. Introductory Problems

Problem 5Prove that

(ab)2 < a+b (ab)28a 2 8b

for all a b> 0.

Problem 6Several (at least two) nonzero numbers are written on a board. One may

erase any two numbers. say a and b, and then write the numbers a +

and b instead.

Prove that the set of numbers on the board, after any number of thepreceding operations, cannot coincide with the initial set.

Problem 7The polynomial

may be written in the form

ao + a1y + a2y2 + a15y16 + a17y17,

where y = x + 1 and are constants.Find a2.

Problem 8Let a, b, and c be distinct nonzero real numbers such that

1 1 1a+-=b+-b c a

Prove that = 1.

Problem 9Find polynomials 1(x), g(x), and h(x), if they exist, such that for all x,

( 1 ifx0.

i. Introductory Problems

Problem 10

Find all real numbers x for which

8x+27x 712x + 18x

Problem 11Find the least positive integer m such that

(2n'\I I

for all positive integers n.

Problem 12Let a, b, c, d, and e be positive integers such that

Find the maximum possible value of max{a, b, c, d, e}.

Problem 13Evaluate

3 4 20011!+2!+3! 1999!+2000!+2001!

Problem 14Letx= 1, a ER.Find all possible values of x.

Problem 15Find all real numbers x for which

+ lix + 12T = 13X + 14x.

4 1. Introductory Problems

Problem 16Let f N x N N be a function such that f(1, 1) = 2,

f(rn + 1. n) = f(rn. n) + rn and f(rn, n + 1) = f(n2, n) n

for all rn, n E N.

Find all pairs (pq) such that f(p,q) = 20th.

Problem 17Let f be a function defined on [0, 1] such that

f(0) = 1(1) 1 and (f(a) f(b)I < a bI.

for all a b in the interval [0, 1].Prove that

11(a) f(b)( f(n) and

f(f(n)) = 3n

for all n.

Evaluate 1(2001).

Problem 58Let F be the set of all polynomials f(x) with integers coefficients suehthat f(x) = 1 has at least one integer root.For each integer k > 1. find the least integer greater than 1 forwhich there exists f E P such that the equation f(x) = rnk has exactlyk distinct integer roots.

Problem 59Let x1 = 2 and

= 1,

for ii i.Prove that

1 1 1 1 11

22 .r1 x2 22

Problem 60Suppose that f : is a decreasing function such that for allx,y E

f(x + y) + f(f(x) + 1(y)) = 1(1(1 + f(y)) + f(y + 1(x))).

Prove that f(f(x)) = x.

15

Problem 61

Find all functions f : Q Q such that

f(x + y) + f(x y) = 2f(x) + 2f(y)

for all x,y EQ.

Problem 62

Let

16 2. Advanced

Problem 67Find the minimum of

log11 (X2-_ + log12

+

where x1,x2,. .. are real numbers in the interval 1).

Problem 68Determine x2 + + z2 + w2 if

22_12+ 22_32+22_52 =1,

42_12 + 42_32 +4252 =1,

62_12 +6232+6252+6272 =L

+ -L + =182_12 82_32 82_52 82_72

Problem 69Find all functions f : R R such that

f(xf(x) + f(y)) = (f(x))2 +

for all x,y E

Problem 70The numbers 1000, iOOl, ,2999 have been written on a board.Each time, one is allowed to erase two numbers, say, a and b, and replace

them by the number min(a. b).

After 1999 such operations, one obtains exactly one number c on theboard. Prove that c < 1.

Problem 71Let a1,a2,. . . be real numbers, not all zero.Prove that the equation

has at most one nonzero real root.

2.Advanced Problems

Problem '72

Let {an) be the sequence of real numbers defined by a1 = t and

=

for n 1.For bow many distinct values oft do we have a1998 = 0?

Problem '73

(a) Do there exist functions f : R and g R R such that

f(g(x)) = x2 and g(f(x)) =

for all x E R?

(b) Do there exist functions f: R and 9: R such that

f(g(x)) = x2 and g(f(x)) =

for all x E R?

Problem 74LetO

18 2. Advanced Problems

Problem 77Find all functions f: R * R such that the equality

f(f(x) + y) = f(x2 y) + 4f(x)y

holds for all pairs of real numbers (x. y).

Problem 78Solve the system of equations:

3x y =3x2 + y2

x+3y =011

Problem 79Mr. Fat and Mr. Taf play a game with a polynomial of degree at least 4:

+... +_x + 1.

They fill in real numbers to empty spaces in turn. If the resulting poly-nomial has no real root, Mr. Fat wins; otherwise, Mr. Taf wins.If Mr. Fat goes first, who has a winning strategy?

Problem 80Find all positive integers k for which the following statement is true: ifF(x) is a polynomial with integer coefficients satisfying the condition

for c=0.1 k+1,

then F(0)= F(1)= =F(k+ 1).

Problem 81The Fibonacci sequence is given by

(nEN).

Prove that2n+2 ' 2n2

bf2n

for all n 2.

ced ProblemS 19

Problem 82

Find all functions u : Rfor which there exists a strictly monotonic

function f : R p R such that

f(x + y) = f(x)u(y) + f(y)

for all x, y c R.

Problem 83Let Zn be complex numbers such that

IzlI+Iz2H+IzflI=1.

Prove that there exists a subset S of {z1, Z2,.. . . such that

Problem 84A polynomial P(x) of degree n 5 with integer coefficients and n distinctinteger roots is given.

Find all integer roots of P(P(x)) given that 0 is a root of P(x).

Problem 85Two real sequences x1, x2, . and Yi Y2. . are defined in the followingway:

andyn

Yn+1= 1 + yi +

for all n 1. Prove that 2 < 1.

Problem 86For a polynomial P(x), define the difference of P(x) on the interval [a. b]([a,b), (a,b). (a.b]) as P(b) - P(a).Prove that it is possible to dissect the interval [0. 1[ into a finite numberof intervals and color them red and blue alternately such that. for every

polynomial P(x). the total difference of P(.r) on red intervalsis equal to that of P(x) on blue intervals.What about cubic polynomials?

20 2. Advanced Problems

Problem 87Given a cubic equation

x3 + _x2 + _x + = 0,

Mr. Fat and Mr. Taf are playing the following game. In one move, Mr.Fat chooses a real number and Mr. Taf puts it in one of the empty spaces.After three moves the game is over. Mr. Fat wins the game if the finalequation has three distinct integer roots.Who has a winning strategy?

Problem 88Let n > 2 be an integer and let f R2 R be a function such that forany regular n-gon A1A2 . . .

f(A1) + f(A2) + . +

f is the zero function.

Problem 89Let p be a prime number and let f(x) be a polynomial of degree d withinteger coefficients such that:

(i) f(0) = 0, f(l) = 1;

(ii) for every positive integer n, the remainder upon division ofby p is either 0 or 1.

Prove that d p 1.

Problem 90Let n be a given positive integer.

Consider the sequence a0,a1, with a0 = and

ak = 4TI

fork=l,2..n,Prove that

1 _! < 1.TI

2.Advanced problems 21

Problem 91

Let ai,a2,." be nonnegative real numbers. not all zero.

(a) Prove that = 0 has precisely onepositive real root R.

(b) Let A = a2 and B = ja3.

Prove that AA

22 2. Advanced

Problem 95Let ii 3 be an integer, and let

Xcs={1.2 n3}be a set of 3n2 elements.Prove that one can find nine distinct numbers c7, (1 = 1. 2, 3) in Xsuch that the system

= 0a2x+h2y+c2z = 0a3x+b3y+c3z = 0

has a solution (XO, yo, zO) in nonzero integers.

Problem 96Let n 3 be an integer and let Xi. i2. he positive real numbers.

Suppose that1

=

Prove that

Problem 97Let x1,x2 be distinct real numbers. Define the polynomials

P(x) = (x x1)(.r ..

and (1 1Q(x)=P(x)( + -+. +\.iX1 XX2

Let y1,y2.. . . 'Ynl be the roots of Q. Show that

2.Advanced Problems

Problem 98

Show that for any positive integer fl. the polynomial

f(x) = (x2 + + 1

cannot be written as the product of two non-constant polynomials withinteger coefficients

Problem 99Let 11,12.13 R R be functions such that

a1f1 + a2]'2 + a3f3

ismonotonic for all a1,a2,a3 R.

Prove that there exist c1, c2. C3 R. not all zero. such that

cifi(x) +c2f2(x) +c3f3(x) = 0

for all x Ift.

Problem 100Let x1.x2, be variables, and let . . be the sums ofnonempty subsets of

Let pk(xl be the elementary symmetric polynomial inthe yj (the sum of every product of A- distinct yjs).

For which k and n is every coefficient of Pk (as a polynomial in x1,. .. ,even?

For example, if n = 2. then yi' Y2. y3 are Xl. + s2 and

Pi = + Y2 + = + 2x2,P2 = Y1Y2 + 112Y3 + Y3Y1 = + +P3 = Y1Y2Y3 = X1X2 +

Problem ioi.Prove that there exist 10 distinct real numbers al,a2.. . aio such thatthe equation

(xai)(x_a2)...(xaio) = (X+al)(X+a2) (x+aio)has exactly 5 different real roots.

SOLUTiONS TOiNTRODUCTORY PROBLEMS

3. SOLUTIONS TO

PROBLEMS

problem 1 [Romania 1974]

Let a, b, and c be real and positive parameters.

Solve the equation

\/c+ax= v'bax -f-s/c

Solution 1It is easy to see that x = 0 is a solution. Since the right hand side is adecreasing function of x and the left hand side is an increasing functionof x, there is at most one solution.

Thus x = 0 is the only solution to the equation.

Problem 2Find the general term of the sequence defined by x0 = 3, = 4 and

_2

for all n C N.

Solution 2We shall prove by induction that = n + 3. The claim is evident forn=O,1.Fork 1, if Xk_1 =k+2andxk=k+3, then

kxk =(k+2)2 k(k+3)=k+4,

as desired.

This completes the induction.

28 3. Solutions to Introductory Problems

Problem 3 [AHSME 1999]Let x1. x2, . . , x7, be a sequence of integers such that

(i) n;

(ii) X1 19;

(iii)

Determine the minimum and maximum possible values of

Solution 3Let a, b, and c denote the number of is, is, and 2s in the sequence,respectively. We need not consider the zeros. Then a, b, c are nonnegativeintegers satisfying

a+b+2c= l9anda+b+4c=99.

It follows that a = 40c and b 593c, where 0 c 19 (since b 0),so

19+6c.

When c = 0 (a = 40,b = 59), the lower bound (19) is achieved.When c = 19 (a = 21, b = 2), the upper bound (133) is achieved.

Problem 4 [AIME 1997]The function f, defined by

ax + bf(x) cx+d'

where a, b, c, and d are nonzero real numbers, has the properties

1(19) = 19, f(97) = 97, and f(f(x)) =

for all values of x, except

Find the range of f.

Solution 4, Alternative 1For all x, f(f(x)) = .r, i.e.,

(ax + b'\a(\cx+d)x(ax+b\

cx + dj

3to Introductory Problems 29

i.e. (a2 + bc)x + b(a + d) c(a+d)x+bc+d2

X.

i.e.c(a + d)x2 + (d2 a2)x b(a + d) = 0,

which implies that c(a + d) = 0. Since c 0, we must have a = d.

The conditions f(19) = 19 and f(97) = 97 lead to the equations

and 972c=297a+b.

Hence(972

192)c = 2(97 19)a.

It follows that a = 58c, which in turn leads to b = 1843c. Therefore

58x 1843 1521f(x)=.x58

which never has the value 58.

Thus the range of f is R {58}.

Solution 4, Alternative 2The statement implies that f is its own inverse. The inverse may befound by solving the equation

ay + bcy + d

fory. Thisyieldsdxb

cx + aThe nonzero numbers a. b, c, and d must therefore be proportional to d,

c, and a, respectively; it follows that a = d, and the rest is thesame as in the first solution.

Problem 5Prove that

(ab)2 < a+b (ab)28a 2 8b

forallab> 0.Alternative 1

that

) -


i.e.

4a 4b

i.e.(ab)2 < < (ab)2

from which the result follows.

Solution 5, Alternative 2Note that

(a-t-b\2 -ab (a-b)22

+ 2(a + b) +

Thus the desired inequality is equivalent to

4a a + b + 4b.which is evident as a b> 0 (which implies a b).

Problem 6 [St. Petersburg 1989]Several (at least two) nonzero numbers are written on a board. One may

erase any two numbers, say a and b, and then write the numbers a +

and b instead.

Prove that the set of numbers on the board. after any number of thepreceding operations, cannot coincide with the initial set.

Solution 6Let S be the sum of the squares of the numbers on the board. Note thatS increases in the first operation and does not decrease in any successiveoperation, as

(a+ a)2 =

with equality only if a = b = 0.This completes the proof.

to Introductory Problems 31

problem 7 [AIME 1986]

The

may be written in the form

ao + aiy + a2y2 + . - + ai6y'6 + a17y17,

where y = x + 1 and are constants. Find a2.

SolutiOn 7, Alternative 1

Let f(x) denote the given expression. Then

xf(x)=xx2+x3 .x18

and(1 + x)f(x) = 1

Hence1 (y 1)18 1 (y 1)18f(x)=f(y1)= l+(yl) y

Therefore a2 is equal to the coefficient of y3 in the expansion of

1 (y 1)18,

i.e.,

a2=

= 816.

Solution 7, Alternative 2Let 1(x) denote the given expression. Then

f(x) =f(y1) l(y l)-f(y

Thus

/2\ /3\. (17\ (18"

Here we used the

I 7i '\ /n+1"

and the fact that


Problem 8Let a. b, and c be distinct nonzero real numbers such that

1 1 1- = b+ - = c+ -.b c a

Prove that = 1,

Solution 8From the given conditions it follows that

bc ca abab=.bc,andca=.bc ca ab

Multiplying the above equations gives (abc)2 = 1, from which the desiredresult follows.

Problem 9 [Putnam 1999]Find polynomials f(s), g(x), and h(s). if they exist. such that for all x

( 1 ifr

colutions to Introductory Problems 33

Problem 10

Find all real numbers x for which

8X 7

12x + 18x 6

Solution 10

By setting = a and = b. the equation becomes

a3+b3 _7a2b+ b2a 6

i.e. a2ab+b2 _7ab 6'

i.e.Ga2 l3ab+6b2 = 0.

i.e.(2a 3b)(3a 2b) = 0.

Therefore 2x+i = 3x+i or 2x1 = 3x1, which implies that x = 1 andx = 1.

It is easy to check that both .r = 1 and x = 1 satisfy the given equation.

Problem 11 [Romania 1990]Find the least positive integer in such that

(2n\

Thus in


Problem 12Let a. b, c, d, and e be positive integers such that

abcde=a+b+c+d+e.

Find the maximum possible value of max{a, b, c. d, e}.

Solution 12, Alternative 1Suppose that a b c d e. We need to find the maximum value ofe. Since

c


13

Note that

k+2 k+2= k![1+k+1+(k+1)(k+2)]

1

k!(lc+2)

k

(k+2)!

(k+2)1 (k-i--2)!

1

(k+1)! (k+2)!

By telescoping sum, the desired value is equal to

2 2001!

Problem 14

Find all possible values of x.

solution 14, Alternative 1Since

\/a2 + al + 1> aland

2a

Va2 +a+ 1+ Va2 -a+ 1'we have

lxi < 12a/ai = 2,

Squaring both sides of

Va2+a+1

Yields

a + 1 = 2a x2.


Squaring both sides of the above equation gives

2 2 2 2' 2 x2(x24)4(x 1)a =x (x 4)ora = 4(21)'Since a2 0, we must have

x2(x2 4)(x2 1) 0,

Since xI < 2, x2 4 < 0 which forces x2 1 < 0. Therefore, 1 < x < 1.Conversely, for every x C (1, 1) there exists a real number a such that

x Va2 +a+1 a+1.

Solution 14, Alternative 2Let A = B = and P = (a.O). Then Pis a point on the x-axis and we are looking for all possible values ofd=PA-PB.By the Triangle Inequality, IPA PBJ < = 1. And it is clearthat all the values 1

Solutions to Introductory Problems 37

Comment: More generally.

a2+(a+1)2+.+(a+k)2=(a+k+1)2+(a+k+2)2+"+(a+2k)2

forak(2k+'). kcN.

problem 16 [Korean Mathematics Competition 2001]

Let f: N x N N be a function such that f(1, 1) = 2.

f(m + 1. n) = f(ni, n) + ni and f(772, 72 + 1) f(n2. n)

for all rn, n c N.

Find all pairs (p, q) such that f(p. q) = 2001.

Solution 16

We have

f(pq) = f(pl,q)+p1=

=

=2

q(q1) p(pl)= f(1,l)2 2

= 2001.

Therefore

p(p 1) q(q 1) 1999,

2 2

i.e.

(pq)(p+q 1) =2. 1999.

ROte that 1999 is a prime number and thatpq < p+q 1 for p.q c NWe have the following two cases:

1 = 1999. Hence p= 1001 and q= 999.


Therefore (p. q) = (2000, 1999) or (1001,999).

Problem 17 [China 1983]Let f be a function defined on [0, 1] such that

f(0) = f(1) = land f(a)f(b)I < labi,

for all a bin the interval [0, 1].Prove that

f(a) f(b)j 1/2. By symmetry, we may assume that a> b. Then

f(a) f(b)I = If(a) 1(1) + f(0) f(b)I< f(a) f(')I + 1(0) f(b)I< IalI+I0bI= 1a+b0= 1(a-b)

as desired.

Problem 18Find all pairs of integers (x, y) such that

+ = (x + y)2.

Solution 18Since x3+y3 = (x+y)(x2xy+y2), all pairs of integers (nri), n C 7Z.are solutions.

Suppose that x + y 0. Then the equation becomes

S2 Xy + y2 = x + y,

i.e.

3solutions to Introductory Problems 39

a quadratic equation in x, we calculate the discriminant

A = y2 + 2y + 1 4y2 + 4y = 3y2 + 6y + 1.

solving for A 0 yields

3 3

Thus the possible values for y are 0, 1, and 2. which lead to the solutions(1,0), (0,1), (1,2), (2.1), and (2,2).Therefore, the integer solutions of the equation are (x. y) = (1, 0), (0, 1),(1,2), (2,1), (2,2), and (n, n). for all nEZ.

Problem 19 [Korean Mathematics Competition 2001]Let

2f(x)= 4X+2for real numbers x. Evaluate

/ 1 \ / 2 '\ /2000

f has a half-turn symmetry about point (1/2, 1/2). Indeed,

2 2.4x 4X

from which it follows that 1(x) -t- f(1 x) 1.Thus the desired sum is equal to 1000.

Problem 20Prove that for n 6 the equation

1 1 1

x1

has integer solutions.

Solution 20Note that

1 1 1 1 1


from which it follows that if (x1, x2, , = (a1 .a2, is an inte.ger solution to

1 1 1

x1 x2

then

(x1,x2,. .

is an integer solution to

1 1 1

xl x2

Therefore we can construct the solutions inductively if there are solutionsfor n = 6, 7, and 8.

Since xi = 1 is a solution for n = 1, (2,2,2,2) is a solution for n = 4,and (2, 2, 2,4,4.4,4) is a solution for n = 7.

It is easy to check that (2, 2, 2,3, 3, 6) and (2, 2. 2, 3,4,4, 12, 12) are solu-tions for n = 6 and n = 8, respectively. This completes the proof.

Problem 21 [AIME 1988]Find all pairs of integers (a, b) such that the polynomial

ax17 + bx'6 -i- 1

is divisible by x2 x 1.

Solution 21, Alternative 1Let p and q be the roots of x2 x I = 0. By Vieta's theorem,p + q = 1 and pq 1. Note that p and q must also be the roots ofax17 + bx'6 + 1 = 0. Thus

ap17 + bp'6 = 1 and aq17 + bq'6 1.

Multiplying the first of these equations by q'6, the second one by p16,and using the fact that pq = 1, we find

ap+b=q16andaq+b=p'6. (1)

Thus

a = I q'6 = (p8 + q8)(p4 + q4)(p2 + q2)(p q).

solutions to Introductory Problems 41

since

p+q = 1,p2+q2 = (p+q)22pq=1+2=3,p4+q4 = (p2+q2)22p2q2=92=7,p8+q8 = (p4+q4)22p4q4=49247,

follows that a 1 3 7 . 47 = 987.Likewise, eliminating a in (1) gives

b = p'7q'7pq

= p'6+p'5q+p'4q2++q16= (p'6 + q16) + pq(p'4 + q14) + p2q2 (p'2 + q'2)

+ p7q7(p2 + q2 ) + p8q8

= (p16 +q'6) (p'4 +q'4)+ (p2 +q2)+ 1.

For n 1, let k2 = 3 and k4 = 7, and

1. 2n+4 2n+42n+4 P= (p2fl+2 + + q2) + q2Th)

=

for n 3. Then k6 18, k8 = 47, k1,, = 123, k,2 = 322, k,4 843,k16 = 2207.

Hence

b=2207843+322123+47 18+73+1=1597

or

(a, b) = (987, 1597).

solution 21, Alternative 2The other factor is of degree 15 and we write

(C15x15 C14X14 + .. + cix x 1) = ax17 + bx'6 + 1.

Comparing coefficients:

x0: c0=1.c0c,=0,c,=1

x2: coc,+c2=0.c2=2,andfor3

42 3. Solutions to Introductory

It follows that for k 15, = Fk+1 (the Fibonacci number).Thus a = = F16 = 987 and b c14 c15 = F17 = 1597(a, b) = (987, 1597).

Comment: Combining the two methods, we obtain some interestingfacts about sequences and Since

= = =

it follows that and satisfy the same recursive relation. It ,i5easy to check that k2 = F1 + F3 and k4 = F3 + F5.Therefore = + and

I. Lb Lf L12n+1 2m 2n2 -r 2n4 2

Problem 22 [AIME 1994]Given a positive integer n, let p(n) be the product of the non-zero digitsof n. (If n has only one digit, then p(n) is equal to that digit.) Let

S=p(l)+p(2)+ +p(999).

What is the largest prime factor of 8?

Solution 22Consider each positive integer less than 1000 to be a three-digit numberby prefixing Os to numbers with fewer than three digits. The sum of theproducts of the digits of all such positive numbers is

(0O.O+OO.1+...+9.99)O.O.O

However, p(n) is the product of non-zero digits of n. The sum of theseproducts can be found by replacing 0 by 1 in the above expression, sinceignoring 0's is equivalent to thinking of them as l's in the products. (Note

that the final 0 in the above expression becomes a 1 and compensatesfor the contribution of 000 after it is changed to 111.)Hence

S 1 = (46 1)(462+46+ 1) = 33 5 7.103,

and the largest prime factor is 103.


problem 23 [Putnam 1979]

Zn be a sequence of nonzero real numbers such that

Xn_2Xm_1xn '

for n = 3,4establish necessary and sufficient conditions on x1 and x2 for to bean integer for infinitely many values of n.

Solution 23, Alternative 1

We have1 2

Xn_2Xn_1

Let = Then = Yn1 Yn2. i.e.. y,-, is an arithmeticsequence. If is a nonzero integer when n is in an infinite set 8, theyb's for n C S satisfy 1 1.Since an arithmetic sequence is unbounded unless the common differenceis 0, 7/n Yni = 0 for all n, which in turn implies that x1 = = m, anonzero integer.

Clearly, this condition is also sufficient.

Solution 23, Alternative 2An easy induction shows that

x1x2 x1x2xn_ (n 1)xi (n 2)x2 (x1 x2)n + (2x2 x1)

for n 3,4In this form we see that will be an integer for infinitely many valuesof n if and only if x1 = = in for some nonzero integer m.

Problem 24Solve the equation

3x =

Solution 24, Alternative 1It is clear that 2. We consider the following cases.

1. 2 x < 2. Setting x = 2cosa, 0 < a < ir, the equation becomes

8cos3 a 6cosa = + 1).or


from which it follows that cos 3a = cos

Then 3a = 2m7r, rn or 3a + = 2n7r, n E Z.

Since 0 a the solution in this case is

4ir 4irx=2cos0=2, andx=2cosT.

2. x> 2. Then x3 4x = x(x2 4) > 0 and

x2x2=(x2)(x4-1) >0

orx >

It follows that 3x > x>

Hence there are no solutions in this case.

Therefore, x = 2, x = 2cos4ir/5, and x = 2cos4ir/7.

Solution 24, Alternative 2For x> 2. there is a real number t > 1 such that

21x=tThe equation becomes

(t7 1)(t5 1) = 0,

which has no solutions for t> 1.Hence there are no solutions for x > 2.

For 2 x 2, please see the first solution.


problem 25 [AIME 1992]

For anysequence of real numbers A = {ai, a2, a3, }, define AA to be

Suppose that all of the terms of the sequence are 1, and that= a92 0.

Find al

SolutiOn 25Suppose that the first term of the sequence is d.

ThenAA={d,d+1,d+2,..}

with the term given by d + (n 1).

Hence

with the term given by

=a1 + (n 1)d+ 1)(n 2).

This shows that a quadratic polynomial in ii with leading coefficient1/2,

Since a19 = a92 = 0. we must have

= 19)(n 92),

soa1 =(119)(1 92)/2=819.

Problem 26 [Korean Mathematics Competition 2000]Find all real numbers x satisfying the equation

2X + 3X 4X + 6X 9X =

solution 26Setting = a and = b, the equation becomes

1-i-a2 -i-b2 a bab= 0.

Multiplying both sides of the last equation by 2 and completing theSquare5 gives

(1a)2+(ab)2+(b- 1)2

46 3. Solutions to Introductory Problem

Therefore 1 = = 3X and x = 0 is the only solution.

Problem 27 [China 1992]Prove that

16<

5olution5 to Introductory Problems 47

Uence there are 35solutions altogether: (0,33), (1.32), (33,0), and

(33, -33).

corn ient: More generally, we have

a3 + + c3 3abc

problem 29 [Korean Mathematics Competition 2001]

Let a, b, and c be positive real numbers such that a + b c 4 andab+bc+ca 4.Prove that at least two of the inequalities

abI2,

are true.

Solution 29We have

(a+b+c)2


Solution 30Let denote the desired sum. Then

s 1 (2n)! (2n)!k(n_k)!(n+k)!

n1 ,' 2n

/1 (272

(2n)! 2

(2n)! 2(n!)

Problem 31 [Romania 1983]Let 0 < a < 1. Solve

= aX

for positive numbers x.

Solution 31Taking loge yields

aX loge x =

Consider functions from

f(x)=aX. g(x)=Iogax, h(x)=xa.

Then both f and are decreasing and h is increasing. It follows thatf(x)g(x) = h(x) has unique solution x = a.

Problem 32What is the coefficient of x2 when

is expanded?


SolUtbohl 32

Let= an.o + afl1X+

It is eaSY to see that = 1 and

Since== (1 + 1)x + +...) (1 + 2Thx)=

we have

+

+

= a1,2 + + 26 + + 22n) (22 + + +24(2272_2 1)

1)=2+ -3

22Th+2_3.2n+1 +2 (2n+1 1) (2m+1 2)

3 3

Problem 33Let in and n be distinct positive integers.Find the maximum value of x is a real number in theinterval (0,

Solution 33By symmetry, we can assume that rn > n. Let y =Since 0 < x < 1, < xTh and 0 < y


Therefore

(Is x It I =(mn)(\ mm J

Equality holds if and only if

(rnn)y1

orIn \\ m

Comment: For m = n + 1, we have

5n+1 1 and 1+y7 0 and x y. By the symmetry, we can assume that x >y > 0. Then

(1 +x)(1 +x2)(1 +x4) >1 +x7>1 +y7,

showing that there are no solutions in this case.

4. x,y


5. x = y. Then solving

1 1 x xy7 1 x + X7 X8

leads to x = 0, 1, 1, which implies that (x, y) = (0.0) or (1, 1).

Therefore, (x, y) = (0,0) and (1, 1) are the only solutions to thesystem.

Problem 36Solve the equation

1)x2 + (2x2 2)x = 2x+1 2

for real numbers x.

Solution 36Rearranging terms by powers of 2 yields

1)2(x2+x1) =0. (1)

Setting y = 1 and dividing by 2 on the both sides, (1) becomes

2Yx+2xy_(x+y) =0

or(2)

Since f(x) = 1 and x always have the same sign,

0.Hence if the terms on the left-hand side of (2) are nonzero, they musthave the same sign, which in turn implies that their sum is not equal to0.

Therefore (2) is true if and only if x = 0 or y = 0. which leads to solutionsx = 1,0, and 1.

Problem 37Let a be an irrational number and let n be an integer greater than 1.Prove that

(a+ Va2_ + (a- -is an irrational number.


SolUtbol1 37

Let .1.N= (a+ + (a

and let1)*,

Then N = b + 1/b. For the sake of contradiction, assume that N isrational. Then by using the identity

+ = + + - + 1)

repeatedly for m 1, 2 we obtain that btm + 1/btm is rational for allm E N.

In particular,

=a+ 1+a- Va2 - 1=2a

is rational, in contradiction with the hypothesis.

Therefore our assumption is wrong and N is irrational.

Problem 38Solve the system of equations

(x1 X2 x3)2 = x2(x4 + X5 x2)(x2x3+x4)2 = x3(x5+x1x3)(x3x4+x5)2 = x4(x1+x2x4)

= x5(x2+x3x5)(x5x1-f-x2)2 = xi(x3+x4x1)

for real numbers x1, x2, x3, x4, S5.

38Let 5k+5 = Adding the five equations gives

5 5

+ 2XkXk+2) = + 2XkXk+2).k=1

It follow5 that

= 0.


Multiplying both sides by 2 and completing the squares yields

Xk+1)2 = 0,

from which x1 = x2 = X3 = x4 = Therefore the solutions to thesystem are

(xi,x2.x3,x4,xs) = (a, a, a. a, a)

for a C R.

Problem 39Let x. y, and z be complex numbers such that x+y+z = 2, x2+y2+z23, and xyz = 4.Evaluate

1 1 1+ +xy+z1 yz+xl zx+yl

Solution 39Let S be the desired value. Note that

xy+zl =xy+1x--y=(xl)(y1).

Likewise,yz+xl (y 1)(x 1)

andzx+y 1 = (z 1)(x 1).

Hence

1 1 1

S = (x_1)(y_1)+(y_1)(z_1)+(z_1)(x_1) x+y+z3 1- (x-l)(y-l)(z--l)

1

xyz(xy+yz+zx)+x+y+z-- 11

5(xy+yz+zx)

But

2(xy + yz + zx) = (x + y + z)2 (x2 + + z2) = 1.

Therefore S = 2/9.

5oiutions to Introductory Problems 55

problem 40 [USSR 1990]

Mr Fat is going to pick three non-zero real numbers and Mr. Taf is going8rrange the three numbers as the coefficients of a quadratic equation

x2+ x+ =0.

Mr. Fat wins the game if and only if the resulting equation has twodistinct rational solutions.

Who has a winning strategy?

Solution 40Mr. Fat has the winning strategy. A set of three distinct rational nonzeronumbers a, b, and c, such that a + b + c = 0, will do the trick. Let A, B,and C be any arrangement of a, b, and c, and let f(x) = Ax2 + Bx + C.Then

f(1) = A + B + C = a + b + c =0.

which implies that 1 is a solution.Since the product of the two solutions is C/A, the other solution is C/A.and it is different from 1.

Problem 41 [USAMO 1978]Given that the real numbers a, b, c, d, and e satisfy simultaneously therelations

a+b+c+d+e=Sanda2+b2+c2+d2+c2= 16,

determine the maximum and the minimum value of a.

Solution 41, Alternative 1Since the total of b.c. d. and c is S a, their average is x = (S a)/4.Let

b=x+b1, c=x+c1, d=x+d1, e=x+c1.Then b1 + d1 + c1 = 0 and

or

(8a)2(1)

0 5a2 16a = a(5a 16).Therefore 0 a 16/5, where a = 0 if and only if b = c = d = c = 2nd a 16/5 if and only if cd 6/5

56 3. Solutions to Introductory Problem5

Solution 41, Alternative 2By the EtMS-AM inequality, (1) follows from

and the rest of the solution is the same.

Problem 42Find the real zeros of the polynomial

Pa(X) = (x2 + 1)(x 1)2 ax2.

where a is a given real number.

Solution 42We have

(x2 + 1)(x2 2x + 1) ax2 0.

Dividing by x2 yields

I 1\I 1'\I x + I (x 2 + I a 0.xj\ xj

By setting y = x + 1/x, the last equation becomes

y2 a = 0.

It follows thatx + - = 1

xwhich in turn implies that, if a 0, then the polynomial Pa(X') has thereal zeros

1 + 2

In addition, if a 8, then Pa(X) also has the real zeros

1- fiPi -2

Problem 43Prove that

1 3 2n1 12 4 2n

for all positive integers n.


1

4

We use induction

For n = 1, the result is evident.

SuPPOSe the statement is true for some positive integer k, i.e.,

1 3 2ki 12k

Then3 2k1 2k+1 1 2k+1

2k

In order for the induction step to pass it suffices to prove that

1 2k+1 I2k+2 < y3k+4

This reduces to(2k+1\2 3k+1


Solution 44Write P(x) = (x S)XCQ(X) and

Q(x) = b0xm + + ... +

where bm 0. Since Q has a positive root, by Descartes' rule of signs,either there must exist some k for which bk > 0 bk4l, or bm > 0.If there exists a k for which bk > 0 bk+1, then

ak+1 = sbk + bk+1 S.

If bm > 0, then = sbm S.In either case, there is a k such that ak s, as desired.

Problem 45Let rn be a given real number, Find all complex numbers x such that

/ '.2 / '.2IX \ IX \ 2I1 +I1 =772 +712.\x+1J \X1JSolution 45Completing the square gives

/ x x \2 2x2= 2 +m2+rn,\,x+l xlj x 1

i.e./ 2x2 \2 2x2 2

I = +712 +772.x21

Setting y = 2x2/(x2 1), the above equation becomes

y (rn2 + m) = 0,

i.e.(ym1)(y+m)=0.

Thus2x2 2x2

=772or =712+1,x21 x21

which leads to solutions

-2andx 1.

solutions to Introductory Problems 59

problem 46

The sequence given by xo = a, = b, and

= + 2\js periodic.

prove that ab = 1.

solution 46by on both sides of the given recursive relation yields

= + 1

or 1) = 1.

Let = 1 for n C N. Since Yn+i = is a geometricsequence. If xn is periodic, then so is which implies that = 0 forall n E N. Therefore

ab = XjXl = Yi + 1 = 1.

Problem 47Let a, b, c, and d be real numbers such that

(a2 + b2 1)(c2 + d2 1) > (ac+ bd 1)2.

Prove that

a2 + b2> I and c2 + d2 > 1.

Solution 47For the sake of the contradiction, suppose that one of a2 + b2 or c2 + d2'S less than or equal to 1. Since (ac+bd 1)2 0, a2 +b2 landC2 + d2 1 must have the same sign. Thus both a2 b2 and c2 + d2 areless than 1. Let

x=1a2b2andy=1c2---d2.Thefl 0 < x, y 1. Multiplying by 4 on both sides of the given inequality

4xy > (2ac + 2bd 2)2 = (2 2ac 2bd)2= (a2+b2+x-i-c2+d2+y2ac2bd)2= [(ac)2 (x+y)2=x2+2xy+y2,


or 0> x2 2xy f y2 = (x y)2, which is irhpossible.Thus our assumption is wrong and both a2 + b2 and c2 + d2 arethan 1.

Problem 48Find all complex numbers 2 such that

(3z+1)(4z+ 1)(6z+ 1)(12z+ 1) =2.

Solution 48

Note that

8(3z + 1)6(4z + 1)4(6z 4- l)2(12z -1- 1) = 768,

i.e.(24z+8)(24z +6)(24z +4)(24z+2) = 768.

Setting u = 24z + 5 and w = u2 yields

(u+3)(u+ 1)(u 1)(u3).= 768.

i.e.

(u2 1)(u2 9) =

i.e. lOw 759 = 0,

i.e.

(w33)(w+23) =0.

Therefore the solutions to the given equation are

Z24

andz=24

Problem 49Let x1,x2, be the zeros different from 1 of the polynomialP(x) = 1, n 2.Prove that

1 1 1 ni+ 1-

1x1 1x2 2


5olutiofl Alternative 1

Q(x) = P(1 x) = (1

Then

Q(x) = (_1)nxnl + + + -and are the nonzero roots of the polynomial Q(x). as

(1ai =0.

Thus the desired sum is the sum of .the reciprocals of the roots of poly-nomial Q(x), that is,

1 1 1+

1 X1 I I

1 1 1=++4-a1 a2

= a2a3 + ala3 -- + a1a2

By the Vieta's Theorem, the ratio between

and

is equal to the additive inverse of the ratio between the coefficient of xand the constant term in Q(x), i.e., the desired value is equal to

(n)

as desired.

Solution 49, Alternative 2For any polynomial R(x) of degree ni, whose zeros are Xi. Xn_ 1'the following identity holds:

1 1 1 R'(x)+ +...+ =

XX1 R(x)

62 3. So'utions to Introductory Problems

For Xl1 2 +...+x+1,

x1R(1) = n and

R'(i)=(n 1)+(n2)+'+ = n(n 1)

It follows that1 1 1 R'(l) ni

+ +...+1x1 1x2 R(1) 2

Problem 50Let a and b be given real numbers.Solve the system of equations

a

y

s/f x2 + y2

for real numbers x and y.

Solution 50Let u = x,+ y and v = x y. Then

2 2 U+V UV

Adding the two equations and subtracting the two equations in the orig-inal system yields the new system

= (a+b)s/1uv=

Multiplying the above two equations yields

uv(1 uv) = (a2 b2)(1 uv).

hence uv = a2 b2. It follows that

(a+b)i/1a2+b2 (ab)V1a2+b2u= andv=

which in turn implies that

(a+bV'a2_b2 b+a'1a2b2-i-b2

whenever 0 < a2 b2 < 1.

SOLUTiONS TOADVANCED PROBLEMS

4. SOLUTIONS TO

ADVANCED PROBLEMS

problem 51

f2000\ f2000\ 1 2000\ (2000

Solution 51

Let 20001(x) = (1 + x)200 =

Let w = (1 + Then w3 = 1 and w + 1 = 0. Hence

((2000\ (2000\ (2000

= f(1) + wf(w) + w2f(w2)= 22000 + + + + w2)200= 22000 + w(w2)200 + w2(w)200

= 22000 + + = 22000 1.

Thus the desired value is22000 1

3

Problem 52Let x, y, z be positive real numbers such that x4 + y4 + z4 = 1Determine with proof the minimum value of

1 x8+

1 y8+

1 z8

52For 0 < u < 1, let 1(u) = u(1 u8). Let A be a positive real number.RY the AM-GM inequality,

A(f(u))8 =Au8(1 _u8)..(1 _u8) < [Au8 +8(1_u8)]9

66 4. Solutions to Advanced

Setting A = 8 in the above inequality yields

8(f(u))8 < (8)

or

It follows that

1 x8+

1 y8+

1 = x(1 x8) + y(1 y8)+

z(1 z8)

>8

with equality if and only if

1x = y = z =

Comment: This is a simple application of the result of problem 33 inthe previous chapter.

Problem 53 [Romania 1990]Find all real solutions to the equation

2X + 3X + 6X = x2.

Solution 53For x < 0, the function f(x) = + + x2 is increasing, so theequation 1(x) = 0 has the unique solution x = 1.Assume that there is a solution s 0. Then

+ 3S + 6S 3,so s and hence 1.But then s LsJ yields

2S 2LsJ = (1 + I + s,which in turn implies that

GS > 4S = (23)2

Solutions to Advanced Problems 67

28 + > s2, a contradiction.x 1 is the only solution to the equation.

problem 54j4et {an}n1 be a sequence such that al = 2 and

1= +

for all n E N.

Find an explicit formula for

Solution 54Solving the equation

x 12 x

leads to x Note that

Therefore,

2+1)

and

+i]1

Problem 55Let x, y, and z be positive real numbers. Prove that

x+ yy+f(y+z)(y+x)

1.

55Note that

_____________

.J(x+y)(x+z)

68 4. Solutions to Advanced Problems

In fact, squaring both sides of the above inequality yields

x2+yz

which is evident by the AM-GM inequality. Thus

x x

-Likewise,

Y v'V

-and

z f(n') and f(f(n)) = 3nfor all ri.

Evaluate f(2001).

57, Alternative 1We prove the following lemma.

Lemma For ri=0,1,2,...,:1. and

2.


Proof We use induction.

For n = 0, note that f(1) 1, otherwise 3 = f(f(1)) = 1(1) = 1, whichis impossible. Since f N N, f(1) > 1. Since f(n + 1) > 1(n),f is increasing. Thus 1 < 1(1) < f(f(1)) = 3 or 1(1) = 2. HenceI (2) = 1 (1 (1)) = 3.

Suppose that for some positive integer ri 1,

fand

f (f(3n+1)) = 3n+2,as desired. This completes the induction. 0

There are 1 integers m such that < in < 2 . and there are3fl

1 integers m' such that

f2

3fl +in,

for 0 m 3n Therefore

+m) =f(f (3fl +m)) =

Hence

1(2001) = 1(2 . 36 + 543) = 3 (36 + 543) = 3816.

Solution 57, Alternative 2For integer n, let fl(3) = a1a2 denote the base 3 representation of

Using similar inductions as in the first solution, we can prove that

I ifa1 =1.= 1a2 . if al = 2.

Since 2001(3) = 2202010, f(2001)(3) = 12020100 or

1 (2001) = 1 32 +2 34 +2 36 + 37 = 3816.

Advanced Problems 71

problem 58 [China 1999]

Let F be the set of all polynomials f(x) with integers coefficients suchf(x) = 1 has at least one integer root.

For each integer k> 1, find the least integer greater than 1 for whichthere exists an f E F such that 1(x) = mk has exactly k distinct integerroots.

Solution 58Suppose that fk E F satisfies the condition that 1k (x) = has exactlyk distinct integer roots, and let a be an integer such that fk(a) = 1. Let

be the polynomial in F such that

gk(X) fk(x+a)

for all x.

Now = fk(a) = 1, so the constant term of is 1. Now gk(x) = mkhas exactly k distinct integer roots r1, r2 rk, so we can write

gk(x) . mk = (x r1)(x r2) ... (x

where qk(x) is an integer polynomial.Note that r1r2 . . Tk divides the constant term of gk(x) mk, whichequals 1- mk.Since rnk > 1, 1 172k cannot be 0,

1 mk! rir2 - -

Now are distinct integers, and none of them isO. so

rlr2..rkl hence

mk Ik/21! + 1.Thi5 value of mk is attained by

gk(x) = 1)(x+1)(x-2)(x+2)(x+ (_1)klk/2]) + Lk/2i! [k/2]! + 1.

Thus,

mk = Lk/2Th [k/2]! + 1.


Problem 59Let x1 = 2 and

= xn +

1 1 1 11

22 Xi X2 2

Solution 59Since x1 = 2 and

1 = 1),

is increasing.

Then 1 0.Hence

1 1 _1 1 1 1) 1

or

1 1 1 1+...+=1-x1 x2 xn+11

Thus it suffices to prove that, for m N,

1 1 11


Since xk4-1 is an integer, the lower bound of (2) implies that

xk+1 and 1 22k 1,from which it follows that

xk+2 1 = Xk+1(Xk+1 1) 22k (22k i) 22k+1

desired.This finishes the induction and we are done.

problem 60 [Iran 1997]Suppose that f : R+ is a decreasing function such that for allz,y E R+,

f(x + y) + f(f(x) + 1(y)) = f(f(x + f(y)) + f(y + 1(x))).

Prove that f(f(x)) = x.

Solution 60Setting y = x gives

f(2x) + f(2f(x)) = f(2f(x + f(x))).

Replacing x with f(x) yields

f(2f(x)) + f(2f(f(x))) = 1(21(1(x) + f(f(x)))).

Subtracting these two equations gives

1(21(1(x))) - f(2x) = f(2f(f(x) + f(f(x)))) - f(2f(x + 1(x))).

If 1(1(x)) > x, the left hand side of this equation is negative, so

f(f(x) +1(1(x))> f(x+ f(x))

and

f(x)+f(f(x))

74 Solutions to Advanced

Problem 61 [Nordic Contest 1998]Find all functions f : Q i Q such that

f(x+y)+f(x -y)2f(x)+2f(y)

for all x,y EQ.

Solution 61The only such functions are 1(x) = kx2 for rational k. Any such functionworks, since

f(x+y)+f(xy)=k(x+y)2+k(x--y)2=kx2+2kxy+ky2+kx2 2kxy+ky2= 2kx2 + 2ky2

= 21(x) + 21(y).

Now suppose f is any function satisfying

f(x+ y)+ f(xy) =2f(x)+2f(y).

Then letting x = y = 0 gives 21(0) = 41(0), so 1(0) = 0.We will prove by induction that f(nz) = n2f(z) for any positive integerri and any rational number z.The claim holds for ri = 0 and ri = 1; let ri 2 and suppose the claimholds for n 1 and n 2.

Then letting x = (n 1)z, y z in the given equation we obtain

f(riz) + f((n 2)z) = f((n 1)z + z) + f((n 1)z z)=2f((n1)z)+2f(z)

so

f(nz) = 2f((ri 1)z) + 2f(z) f((ri 2)z)= 2(n 1)21(z) + 2f(z) (n 2)21(z)(2n24n+2+2n2+4n4)f(z)= ri2f(z)

and the claim holds by induction.Letting x = 0 in the given equation gives

f(y) + f(y) = 2f(0) + 2f(y)

so f(y) = 1(y) for all rational y; thus f(nz) = n2f(z) for all integersn.


Now let Ic = f(1): then for any rational number x = p/q,

q2f(x) = f(qx) = 1(p) = p2f(1) = kp2

5

1(x) = kp2/q2 = kx2.

Thus the functions f(x) = kx2. k E Q. are the only solutions.

Problem 62 [Korean Mathematics Competition 2000]Let


Indeed

reduces to< 3 v'l + 4a

which is equivalent to

6V1 + 4a < 6 + 8a,

or 3a < 4a2, which is evident.

Problem 63 [Thurnament of Towns 1997]Let a, b, and c be positive real numbers such that abc = 1.Prove that

1 1 1+ +

Advanced Problems 77

Likewise,(b2c+b2a+ 1) 3b

(c2a+c2b+ 1) 3c.

Therefore we only need to prove that

2(a+b+c)+3 3(a+b+c),

i.e.3 a + b + c,

which is evident from AM-GM inequality and abc = 1.

Solution 63, Alternative 2Let a = b = C = Then a1b1c1 = 1. Note that

(ai 0,

which implies that a1b1(a1+bi).

Therefore,

1 1

a+b+1 =

1

aibi(ai + b1) +a1b1c1

a1b1c1

aibi(ai + b1 + c1)

Cl

a1

b1

a1+b1+c1Adding the three inequalities yields the desired result.


Problem 64 [AIME 1988]Find all functions f, defined on the set of ordered pairs of positive inte..gers, satisfying the following properties:

f(x,x)x, f(x,y)=f(y.x). (x+y)f(x.y)=yf(x,x-4-y).

Solution 64We claim that f(x. y) = lcm(x. y). the least common multiple of x andy. It is clear that

lcm(x, x) = x

andlcm(x, y) = lcm(y, x).

Note thatlcm(x.y)= y

gcd(x.y)and

gcd(x.y) =gcd(x,x+y),

where gcd (u. v) denotes the greatest common divisor of u and v. Then

(x+y)lcm(x,y) =

x(x+y)gcd(x,x+y)

= ylcm(x,x+y).

Now we prove that there is only one function satisfying the conditions ofthe problem.For the sake of contradiction, assume that there is another functiong(x, y) also satisfying the given conditions.Let S be the set of all pairs of positive integers (x, y) such that f(x, y)g(x, y), and let (in, ii) be such a pair with minimal sum m+ri. It is clearthat in n, otherwise

I (in, ii) = f(m. m) = in = g(in. in) = g(ni. n).

By symmetry (f(x, y) = f(y, x)), we can assume that n in > 0.Note that

nf(rn, n in) = [in + (n m)]f(rn. ii in)= (m ni)f (in, rn + (n in))= (71. in)f(m., n)


or nrnf(rn.rirn)= f(rnn).

[,ikewiSe,

g(rn, n in)= ii

. g(rn. n).

Since f(rn. ii) g(m. n), f(rn. n in) g(rn. n in).

Thus (in. n rn) E S.

But (in. n in) has a smaller sum in + (n in) = n. a contradiction.Therefore our assumption is wrong and f(x, y) lcm(x. y) is the onlysolution.

problem 65 [Romania 1990]Consider ii complex numbers zk. such that IzkI 1. k = 1.2Prove that there exist c1.e2, . .. . E (1. 1} such that, for any in ii,

I+ e2z2 + + emzrnl 2.

Solution 65Call a finite sequence of complex numbers each with absolute value notexceeding 1 a green sequence.

Call a green sequence happy if it has a friend sequenceof is and is, satisfying the condition of the problem.We will prove by induction on ii. that all green sequences are happy.For n = 2. this claim is obviously true.

Suppose this claim is true when n equals some number in. For the caseof ii = rn + 1. think of the zk as points in the complex plane.For each k. let ek be the line through the origin and the point corre-sponding to zk. Among the lines some two are within 600 ofeach other; suppose they are and with the leftover one being 4.The fact that and are within 600 of each other implies that thereexists some number c9 {1, 1} such that z' = + e9z9 has absolutevalue at most 1.

the sequence z'. z4. z5 is a k-term green sequence. so,by the induction hypothesis, it must be happy: let c'. C4. C5 Ck+1be its friend.Let Ca 1.

Then the sequence is the friend of Induction is nowcomp'ete


Problem 66 [ARML 1997]Find a triple of rational (a, b, c) such that

Solution 66Let x = 1 and y = Then y3 = 2 and x = Note that

1 =y3 1 =(y 1)(y2+y+ 1).

and

2

3 3

3 3x =yl= =(y+l)3

or

(1)y+1

On the other hand,

3=y3+1=(y+1)(y2y+1)

from which it follows that

1 2 L1(2)y+l 3

Combining (1) and (2), we obtain

x=

Consequently,(4 2 1

is a desired triple.

4Solutions to Advanced Problems 81

problem 67 [Romania 1984)Find the minimum of

(X2

+ + +

where Xl.x2 are real numbers in the interval 1).

Solution 67Since logs x is a decreasing function of x when 0 < a < and. since(x 1/2)2 0 implies x2 x 1/4, we have

= xk+1 =It follows that

+ (X3

+ +

2 (logx2 + logx3 + + + logxilogx2 2n

by the AM-GM inequality.Equalities hold if and only if

Problem 68 [AIME 1984]Determine x2 + y2 z2 + w2 if

22_12 +2232+2252 +22_72 =1,

42_ 12 + 42_32 + 42_52 + 42_72 = 1,

z2 w262_12+62_32+ 62 52+62 72 1,

82_12 +8232+8252+8272=1.


Solution 68The claim that the given system of equations is satisfied byand w2 is equivalent to claiming that

z2 w2t_12+t_32+i_52+t_72_l (1)

is satisfied by t = 4, 16. 36, and 64.Multiplying to clear fractions, we find that for all values of t for which itis defined (i.e., t 1, 9,25, and 49), (1) is equivalent to the polynomialequation

P(t) = 0,

where

P(t) = (t 1)(t 9)(t 25)(t 49)x2(t 9)(t 25)(t 49) y2(t 1)(t 25)(t 49)z2(i l)(t 9)(t 49) w2(t 1)(t 9)(t 25).

Since deg P(t) = 4, P(t) = 0 has exactly four zeros t = 4, 16, 36, and 64,i.e.,

P(t) = (t 4)(t 16)(t 36)(t 64).

Comparing the coefficients of t3 in the two expressions of P(t) yields

I +9+25+49+x2+y2+z2+w2 =4+16+36+64,

from which it follows that

+ y2 + z2 + w2 = 36.

Problem 69 [Balkan 1997]Find all functions f : R IR such that

f(xf(x) + 1(y)) = (f(x))2 + y

for all x,y ER.

Solution 69Let f(0) = a. Setting x = 0 in the given condition yields

I (f (y)) = a2 +

for all y E R.

Since the range of a2 +y consists of all real numbers, f must be surjective.

4 Solutions to Advanced Problems 83

Thus there exists b C R such that f(b) = 0.getting x = b in the given condition yields

f(f(y)) = f(bf(b) + 1(y)) = (1(b))2 + y =

for all y C IL It follows that, for all x, y

(f(x))2 + y = f(xf(x) +f [1 (1 (x))f(x) + f(y)] = I [f(x)f(f (x)) + yJ

= f(f(x))2 + y = x2 + y,

that is,(f(x))2=x2. (1)

It is clear that f(x) = x is a function satisfying the given condition.Suppose that f(x) x. Then there exists some nonzero real number csuch that f(c) = c. Setting x = cf(c) + f(y) in (1) yields

[f(cf(c) + 1(y))]2 = [cf(c) + 1(y)]2 = fc2 + 1(y)]2,

for all y C R, and, setting x = c in the given condition yields

f(cf(c)+f(y)) = (f(c))2+y =c2+y,

for all y C IL

that (f(y))2 = y2.It follows that

fc2 +1(y)]2 = (c2 +y)2,

or

f(y)=y,for all y C R, a function which satisfies the given condition.

Therefore the only functions to satisfy the given condition are 1(x) = xor 1(x) = x, for x C R.

Problem 70The numbers 1000, ,2999 have been written on a board.Each time, one is allowed to erase two numbers, say, a and b, and replace

them by the number min(a, b)

After 1999 such operations, one obtains exactly one number c on theboard.

Prove that c < 1.


Solution 70By symmetry, we may assume a b. Then

1. a=

We have 11 1

from which it follows that the sum of the reciprocals of all the numberson the board is nondecreasing (i.e.. the sum is a monovariant).At the beginning this sum is

1 1 1 1s-++...+ 20002 jij1jfJ

Rearranging terms in S yields

1 1 11 i\ /1->c 1000 2999) 2998)

> i,

or c < 1, as desired.

Problem 71 [Bulgaria 1998]Let a1, a2 be real numbers, not all zero.Prove that the equation

has at most one nonzero real root.

Solution 71Notice that = + a2x is concave. Hence

45olutions to Advanced Problems 85

is concave.

Since f'(x) exists. there can be at most one point on the curve y = f(x)with derivative 0.

Suppose there is more than one nonzero root.

Since x = 0 is also a root, we have three real roots x1 < x2 < x3. Ap-plying the Mean-Value theorem to 1(x) on intervals [xl.X2] and [x2,x3],we can find two distinct points on the curve with derivative 0, a contra-diction.

Therefore, our assumption is wrong and there can be at most one nonzeroreal root for the equation f(x) = n.

Problem 72 [Turkey 1998]Let be the sequence of real numbers defined by a1 = t and

=

for n 1.For how many distinct values of t do we have a1998 = 0?

Solution 72, Alternative 1Let 1(x) = 4x(1 x). Observe that

= {0. 1}, f1(1) = {1/2}. 1]) = [0,1],

and [0,1).Let = {x E R: = 0}; then

==

We claim that for alln 1, c [0. 1]. 1 E and

= + 1.

For n 1. we have

A1 = {x E RIf(x) = 0} = fO. 1},

and the claims hold.NOW suppose n 1 and C [0. 1]. 1 e and = 2n_1 + 1. Then

8o c [0. 1].


Since f(O) = f(1) = 0, we have = 0 for all n 1, so I eNow we have

==

aE

= {x:f(x)=l}I+ > I{x:f(x)=a}I

= 1+ 2aE

aE[O,1)

== 1+2(2n_1+1_1)=

Thus the claim holds by induction.Finally, a1998 = 0 if and only if I 1997(t) = 0, so there are 21996 + 1 suchvalues of t.

Solution 72, Alternative 2As in the previous solution, observe that if f(x) E [0, 1] then x e [0, 1],so if a1998 = 0 we must have t E [0. 1].Now choose U e ir/2[ such that sin U =

Observe that for any e

f(sin2 (1 = 4sin2 = sin2

since = sin2 0. it follows that

a2 = sin2 20. a3 = sin2 40 a1ggg = sin2 219970.

Therefore

a1998 = 0 sin 219970 = 0 0 =

for some k e 7Z.

Thus the values of t which give a1998 = 0 are

sin2 (kir/21997),

k Z, giving 21996 + 1 such values of t.

to Advanced Problems

problem 73 [IMO 1997 short list]

(a) Do there exist functions 1 : R p R and g : R such that

f(g(x))=x2 and

for all x e

(b) Do there exist functions f R i R and g R R such that

f(g(x)) = x2 and g(f(x)) =

for all x E R?

Solution 73

(a) The conditions imply that f(x3) = f(g(f(x))) = [1(x)]2, whence

XE fi, 0, 1} ; x3 = x : f(x) = ; f(x) E {O, 1}.

Thus, there exist different a, b E fI, 0, 1} such that 1(a) = 1(b).

But then a3 = g(f(a)) = g(f(b)) = b3, a contradiction.

Therefore, the desired functions f and p do not exist.

(b) Let

I if lxi 1p(x) = - In IxI if 0 < xi < I

1 0 ifx=0.Note that g is even and al = bi whenever g(a) = g(b); thus, weare allowed to define f as an even function such that

1(x) = where y is such that p(+y)= x.

We claim that the functions f, p described above satisfy the condi-tions of the problem.

It is clear from the definition of f that f(p(x)) = x2.

Now let y =Then g(y) = x and

p(f(x)) = 9(y2)(y2)Ifl(Y2) = y4Ifl Y = if i

= (yInY)4 if 0< y < 1( 0 ify=0

= [9(y)]4= x4.


Problem 74 [Weichao Wu]LetO< a1 a2 k.Prove that

a1a2 Solution 74, Alternative 1We define two new sequences. For i = 1.2 n, Let

/= and = .

Then/ , ak

= ak = (a1

or

= (ak

Therefore

nak = a2 + + + + ... +

Applying the AM-GM inequality yields

(bib2.bnafl*=

from which the desired result follows.

Solution 74, Alternative 2We define two new sequences. For = 1.2 n, Let

and

Then(1)

Note that, for cy(x y)(y + c) 0

x x+c>---. x>yandc>0:yy+c,

45olutions to Advanced Problems

Setting x = y = and c = ak the above inequality implies thataj/bi for i = 1. 2, . , n. Thus,

Using (1) and the AM-GM inequality yields

) =F +... +

. )

or

aia2 It is clear that the desired result follows from (2) and (3).

Problem 75Given eight non-zero real numbers a1, a2 a8, prove that at least oneof the following six numbers: a1a3 + a2a4, aja5 + a2as, a1a7 + a2a8,a3a5 + a4a5, a3a7 + a4a8, + a6a8 is non-negative.

Solution 75 [Moscow Olympiad 1978]First, we introduce some basic knowledge of vector operations.

Let u = [a. b] and v = Im, nJ be two vectors.Define their dot product u v = am + bn.It is easy to check that

(i) vv = m2+n2 = 1v12, that is, the dot product of vector with itselfis the square of the magnitude of v and v v 0 with equality ifand only if v = [0.0];

(ii) u v = v u;

(iii) u (v + w) = u v + w. where w is a vector;

(iv) (cu) . v = c(u . v), where c is a scalar.

When vectors u and v are placed tail-by-tail at the origin 0. let A and.8 be the tips of u and v. respectively. Then = v u.Let LAOB =0.Applying the law of cosines to triangle AOB yields

= AB2= 0A2-I-0B2-20A.OBcosO= juj2 + v]2 21u!JvJ cosO.

90 4. Solutions to Advanced Problem

It follows that

(v u) . (v u) = u u + v v 2IuIlvIcosO,

orcosO=

IuHvI

Consequently, if 0 0 900, u v 0.Consider vectors v1 = [ai, a2], v2 = [a3, a4], v3 = [a5, a6], and v4[a7, as].

Note that the numbers a1a3+a2a4. a1a5+a2a6, a1a7+a2a8, a3a5+a4a6,a3a7 + a4a8, a5a7 + a6a8 are all the dot products of distinct vectors VjandSince there are four vectors, when placed tail-by-tail at the origin, atleast two of them form a non-obtuse angle, which in turn implies thedesired result.

Problem 76 [IMO 1996 short list]Let a, b and c be positive real numbers such that abc = 1.Prove that

ab bc caa5+b5+ab+b5+c5+bc+c5+a5+ca 1.

Solution 76We have

a5 + b5 a2b2(a + b),

because(a3 b3)(a2 b2) 0,

with equality if and only if a = b. Hence

ab ab

a2b2(a+b)+ab1

ab(a+b)+1abc

ab(a+b+c)C

a+b+cLikewise,

bc ab5 + c5 + bc a + b + c

45olutions to Advanced Problems 91

andb

c5 + a5 + ca a + b + cthe last three inequalities leads to the desired result.

Equality holds if and only if a = b = c = 1.

Comment: Please compare the solution to this problem with thesecond solution of problem 13 in this chapter.

Problem 77 [Czech-Slovak match 1997]Find all functions f R * R such that the equality

f(f(x)+y) =f(x2 y) +4f(x)y

holds for all pairs of real numbers (x, y).

Solution 77

Clearly. f(x) = x2 satisfies the functional equation.Now assume that there is a nonzero value a such that f(a) a2.

Let y =x2 1(x) in the functional equation to find that

(f(x)+x2)=

(f(x)+x2)

2f(x)(x2 1(x) or 1(x) = x2.In both cases. f(0) = 0.

Setting x = a, it follows from above that either f(a) = 0 or 1(a) = 0 orf(a)=a2.

The latter is false, so f(a) = 0.Now, let x = 0 and then x = a in the functional equation to find that

1(u) = f(y), 1(y) = f(a2

and so

1(n) = f(y) = f(a2 + y);that is, the function is periodic with nonzero period a2.Let y = a2 in the original functional equation to obtain

f(f(x)) = 1(1(x) + a2) = f(x2 a2) + 4a2f(x) = f(x2) + 4a2f(x).

However, putting y = 0 in the functional equation gives 1(1(x)) = 1(x2)for all x.


Thus, 4a2f(x) = 0 for all x. Since a is nonzero, 1(x) = 0 for all x.Therefore, either 1(x) = x2 or f(x) = 0

Problem 78 [KvantjSolve the system of equations:

3x yx+ =3x2 + y2x+3y =0.x2 + y2

Solution 78, Alternative 1Multiplying the second equation by i and adding it to the first equationyields

(3xy)(x+3y)i =3,

or i(xyi)

x2+y2 x2+y2Let z = x + yi. Then

!_ xyiz x2 + y2

Thus the last equation becomes

3iz + =

z

orz23z+(3i) =0.

Hence3v'34-4i 3(1+2i)

2 2

that is, (x,y) = (2,1) or (x,y) = (1,i).

Solution 78, Alternative 2Multiplying the first equation by y, the second by x, and adding up yields

2xy +(3x y)y (x + 3y)x

=x2 + y2

or 2xy 1 = 3y. It follows that y 0 and

3y + I


this into the second equation of the given system gives

[(3y+ 1)2+

(3Y+ 1) = o.

or

4y4 3y2 1 = 0.

it follows that y2 1 and that the solutions to the system are (2, 1) and(1,i).

problem 79 [China 1995]

Mr. Fat and Mr. Taf play a game with a polynomial of degree at least 4:

+ ... +_x + i.

They fill in real numbers to empty spaces in turn.

If the resulting polynomial has no real root. Mr. Fat wins; otherwise, Mr.Taf wins.

If Mr. Fat goes first, who has a winning strategy?

Solution 79

Mr. Taf has a winning strategy.

We say a blank space is odd (even) if it is the coefficient of an odd (even)power of x.

First Mr. Taf will fill in arbitrary real numbers into one of the remainingeven spaces, if there are any.

Since there are only n 1 even spaces, there will be at least one oddSpace left after 2n 3 plays, that is, the given polynomial becomes

p(x) = q(x) +_x2t_1,

where s and 2t I are distinct positive integers and q(x) is a fixedPolynomial.

We claim that there is a real number a such that

p(x) = q(x) + +_x2t1

Will always have a real root regardless of the coefficient of

Then Mr. Taf can simply fill in a in front of x8 and win the game.

4. Solutions to Advanced

Now we prove our claim. Let b be the coefficient of x2t_1 in p(x). Notethat

+ p(l)

= + 2s_2t+la + b) + [q(1) + (1)8a

= + q(_1)) + +

Since s 2t 1 2s2t+1 + (_1)8 o.Thus

1+ q(1)

a = 2s2t+1 + (_1)8is well defined such that a is independent of b and

+p(l) = 0.

It follows that either p(l) = p(2) = 0 or p(l.) and p(2) have differentsigns, which implies that there is a real root of p(x) in between 1 and2.

In either case, p(x) has a real root regardless of the coefficient ofas claimed.

Our proof is thus complete.

Problem 80 [IMO 1997 short list}Find all positive integers k for which the following statement is true: ifF(x) is a polynomial with integer coefficients satisfying the condition

for c=0,1

then F(O) = F(1) = ... = F(k + 1).Solution 80The statement is true if and only if /c 4.We start by proving that it does hold for each Ic 4.Consider any polynomial F(x) with integer coefficients satisfying theinequality 0 F(c) Ic for each c e {0. 1 k + 1}.Note first that F(k + 1) = F(0), since F(k + 1) F(0) is a multiple ofIc + 1 not exceeding Ic in absolute value.Hence

F(x) F(0) = x(x Ic 1)G(x).


where G(x) is a polynomial with integer coefficients.

consequently,

k jF(c) F(O)j = c(k + I c)IC(c)I

for each c E 2 k}.

The equality c(k + 1 c) > k holds for each cC {2, 3 k 1}, as it isequivalent to (c 1)(k c) > 0.

Note that the set f2. 3 k 1} is not empty if k 3, and for any c inthis set. (1) implies that < 1.Since C(c) is an integer, C(c) 0.

Thus

F(x)F(0)=x(x2)(x3)''(xk+1)(xk1)H(x). (2)

where H(x) is a polynomial with integer coefficients.To complete the proof of our claim, it remains to show that 11(1) =H(k) = 0.Note that for c = 1 and c = k, (2) implies that

Ic IF(c) F(0)I = (Ic 2)! . Ic IH(c)I.

Fork4,(k2)!>1.Hence H(c) = 0.We established that the statement in the question holds for any k 4.But the proof also provides information for the smaller values of Ic aswell.

More exactly, if F(x) satisfies the given condition then 0 and k + 1 areroots of F(x) and F(0) for any Ic and if Ic 3 then 2 must also bea root of F(x) F(0).

Taking this into account, it is not hard to find the following counterex-amples:

F(x)=x(2x) fork=1.F(x) = x(3 x) for Ic = 2.F(x)=x(4x)(x2)2 for k=3.


Problem 81 [Korean Mathematics Competition 2001JThe Fibonacci sequence is given by

(n.EN).

Prove thati'3 r'3L 2n+2 2n2

9

for all n 2.

Solution 81Note that

= = =

whence = 0 (1)

for all n 2.Setting a = b = and c = in the algebraic identity

abbcca)

givesrv,n3 r,3 n3 n htf2n

Applying (1) twice gives

= 2 2

The desired result follows from

r'3 r' i r' r' r'2 \ tlr2n+2r2nr2n_2 r2n)

4. Solutions to Advanced Problems 97

problem 82 [Romania 1998]Find all functions u : R * R for which there exists a strictly monotonicfunction f : R * R such that

f(x + y) = f(x)u(y) + f(y)

for all x,y ER.

Solution 82The solutions are u(x) = ax. a E R.To see that these work, take f(x) = x for a = 1.

1, take 1(x) =a 1; then

f(x + y) = 1 = (ax 1 = + f(y)

for all x, y E R.Now suppose u R * R, f R R are functions for which f is strictlymonotonic and f(x + y) = f(x)u(y) + f(y) for all x, y E R.We must show that u is of the form u(x) = ax for some a E R+. First,letting y = 0, we obtain f(x) = f(x)u(0) + f(0) for all x E R.Thus, n(O) 1 would imply f(x) = f(O)/(1 u(O)) for all x, whichwould contradict the fact that f is strictly monotonic, so we must haveu(0) = 1 and f(0) = 0.Then f(x) 0 for all x 0.Next, we have

f(x)u(y) + f(y) = f(x + y) = f(x) + f(y)u(x),

or

f(x)(u(y) 1) = f(y)(u(x) 1)

for all x,y ER. That is,

u(x) I u(y) 1f(x) 1(y)

for all xy 0.

It follows that there exists C E R such that

u(x) I=f(x)

for all x 0.Thus, u(x) = 1+Cf(x) for x 0: sinceu(0) = 1, f(0) = 0, this equationalso holds for x = 0.


If C = 0, then u(x) = 1 for all x, and we are done.Otherwise, observe

u(x+y) = 1+Cf(x+y)= 1 + Cf(x)u(y) + Cf(y)= u(y)+Cf(th)u(y)= u(x)u(y)

for all x,y cR.Thus u(nx) = u(x) for all n Z, x R.Since u(x) = 1 + Cf(x) for all x, u is strictly monotonic, and u(x) =1/u(x) for all x, so u(x) > 0 for all x as u(O) = 1.Let a = u(1) > 0; then u(n) = a for all n N, and

u(p/q) = (u(p))lk =

for allpe Z, q E N, so u(x) = ax for allxE Q.Since u is monotonic and the rationals are dense in we have u(x) = axfor all x E R.Thus all solutions are of the form u(x) = ax, a E

Problem 83 [China 1986]Let z1 , Z2, , be complex numbers such that

Prove that there exists a subset S of {zi, Z2,... , such that

zS

Solution 83, Alternative 1Let 2, and 3 be three rays from origin that form angles of 60, 180,and 300, respectively, with the positiveFor i = 1,2,3, let 7Z, denote the region between and (here 4 =including the ray Then

1= jZkj+ > ZkH !ZkIZkEl?..3

By the Pigeonhole Principle, at least one of the above sums is not lessthan 1/3.

4. Solutions to Advanced Problems

Say it's (otherwise, we apply a rotation, which does not effect themagnitude of a complex number), Let zk = xk + 1Yk. Then for Zk e 7Z3,Xk XkI ZkI/2.

>

as desired.

Solution 83; Alternative 2We prove a stronger statement: there is subset S of {zi, z2, .. . , z,.,} suchthat

zES

For I k n, let Zk Xk + iyk. Then

1 == >

XkO Xk


Solution 84The roots of P(x) are clearly integer roots of P(P(x)); we claim thereare no other integer roots.We prove our claim by contradiction. Suppose, on the contrary, thatP(P(k)) = 0 for some integer k such that P(k) 0.Let

P(x) =a(xri)(x r2)(xr3).. .(x rn),

where a.rl,r2,...,rfl are integers,

Since P(k) 0, we must have 1k 1 for all i.Since the are all distinct, at most two of 1k r2j. k r31, 1k equal1, so

Ia(kr2).(krn_i)I

and IP(k)l Also note that P(k) = for some i0, so P(k)!Now we consider the following two cases:

1. 1k! rn!. Then P(k)I rn)! > rn!, a contradic-tion.

2. 1k! < rn!, that is, 1 1k! 1. Let a. b, c be real numbers,a b. For x e [a, b}, the function

f(x) =x(cx)

reaches its minimum value at an endpoint x = a or x = b, or atboth endpoints.

Thus

k(k rn)I = k!!rn kI !k!(!Tn! 1k!) 1.It follows that

rn! P(k)! 1),which implies that Tn! K 2. Since n 5, this is only possible if

P(x) = (x + 2)(x + 1)x(x 1)(x 2).

But then it is impossible to have k and 1k! rn!, a contra-diction.

4 Solutions to Advanced Problems 101

Thus our assumption was incorrect, and the integer roots of P(P(x)) areexactlY all the integer roots of P(x).

problem 85 [Belarus 1999}Two real sequences x1, x2 and Y2 are defined in the following

way.

andyn

1 + +

for all n 1. Prove that 2 < XnYn 1.

Solution 85, Alternative 1Let = 1 and note that the recursion for is equivalent to

Zn+1 = Zn + +

Also note that Z2 = = x1; since the and satisfy the samerecursion, this means that Zn = Xn_1 for all n> 1.

Thus,Xn Xn

XnYn = = .2n Xn_1

Note that>

Thus 2Xn_i and XnYn > 2, which is the lower bound of the desiredinequality.

Since Xn5 are increasing for n> 1, we have

= 3>

which implies that

2Xn_i >

Thus 3Xn_i > Xn. which leads to the upper bound of the desired inequal-ity.

Solution 85, Alternative 2Setting = cot 0,, for 0 < 0,,

102 4. Solutions to Advanced Problen

300Since = 30, we have in general Similar calculation showsthat

2tan

2

1 tan

tan 0. tan2 is positive and XnYn > 2.

And since for n> 1 we have < 30. we also hare

tan2 f(k) k)

= (_1)Pkk!

p2

(modp).

It follows that

S(f) := f(o) + f(1) + ... + f(p -1) 0 (mod p) (1)


On the other hand, (ii) implies that S(f) j (mod p), where jthe number of those Ic E 1 p 1} for which 1(k) 1 (mod p)But (i) implies that 1 j < p 1.So S(f) 0 (mod p), which contradicts (1).Thus our original assumption was wrong. and our proof is complete.

Solution 89, Alternative 2Again, we approach the problem indirectly.Assume that d p 2, and let

1(x) = + + a1x + a0.

Then

pi pi p2 p2 pi p2

S(f) = 1(k) = = =k=0 k=O i=0 i=0 k=O i=0

pi

where =

(modp)foralli=0,1 p2.We use strong induction on I to prove our claim.The statement is true for i 0 as So = p.Now suppose that So S1 .. 0 (mod p) for some 1 i p2. Note that

p pi pI

k=Oj=0 j0

(I + (mod p)

Since 0 < i + 1

4Solutions to Advanced Problems 109

problem 90Let n be a given positive integer.

Consider the sequence a0, a1. with a0 = and

1ak = ak_I + n

for k = 1, 2, , n.

Prove that1

1 < 1.Ti

Solution 90, Alternative 1We prove a stronger statement: For k = 1. 2

71

(1)

We use induction to prove both inequalities.

We first prove the upper bound, For k = 1, it is easy to check that

1 1 2n+1 na1 = + =

2 4n 4n 2ni

Suppose that

2n_k+2+n(2n_k+2)2

It follows that

n+1 n+1 (ri.l-1)2ak+i_2k+l (2n_k+l)(2n_k+2)+n(2n_k+2)2

n+1 "ri+l 2nk+2\=

n+1 /1 1 2nk+2)2 2n_k+1) >0.

This complete the induction step. In particular, for n = k 1, we obtain

n+1 72+12n(n1)+1 n+2 n+2 ri

as desired.

Solution 90, Alternative 2Rewriting the given condition as

1 Ti 1 1

ak a

ak rt+ak_1

It is clear that aks are increasing.Thus

I

2


Thus (2) implies that1 1 1a0 n+1

or 11,n+2 n-f-2 nwhich is the desired lower bound.

Problem 91 [IMO 1996 short list]Let a1, a2,.. . , be nonnegative real numbers. not all zero.

(a) Prove that .. = 0 has precisely onepositive real root R.

(b) Let A = and B = jaj.

Prove that AA RB.


Solution 91

(a) Consider the function

a1 a2

Note that f decreases from oo to 0 as x increases from 0 to oc.

Hence there is'a unique real number R such that f(R) = 1, that is,there exists a unique positive real root R of the given polynomial.

(b) Let c3 = ad/A.

Then c3s are non-negative and c2 = 1.

Since ln x is a convex function on the interval (0, oo), by Jensen'sinequality,

Ecj ln =-In(f(R))=0.

ft follows that

c3(lnA+jlnR) 0j=1

or

c3 = a3 /A, we obtain the desired inequality.

Comment: Please compare the solution of (a) with that of the problem15 in the last chapter.

Problem 92Prove that there exists a polynomial P(x, y) with real coefficients suchthat P(x, y) 0 for all real numbers x and y, which cannot be writtenas the sum of squares of polynomials with real coefficients.

Solution 92We claim that

P(x,y) = (x2 +y2 1)x2y2 +

is a polynomial satisfying the given conditions.

First we prove that P(x, y) 0 for all real numbers x and y.


If x2 + y2 I 0, then it is clear that P(x,y) > 0; if x2 + y2 1


But, finally, comparing the coefficients of the term x2y2, we have

= -1,

which is impossible for real numbers

Thus our assumption is wrong, and our proof is complete.

Problem 93 [IMO 1996 short list]For each positive integer n, show that there exists a positive integer ksuch that

k = f(x)(x + 1)2n +

g with integer coefficients, and find the smallestsuch /c as a function of n.

Solution 93First we show that such a k exists.

Note that x + 1 divides 1 Then for some polynomial a(s) withinteger coefficients, we have

(1 + x)a(x) = 1 = 2 (1 +

or2=(1+x)a(x)+(1 +x272).

Raising both sides to the (2n)th power, we obtain

= (1 + + (1 +

where b(s) is a polynomial with integer coefficients.

This shows that a k satisfying the condition of the problem exists. Let/c0 be the minimum such k.

Let 2n = 2r . q, where r is a positive integer and q is an odd integer.We claim that =

First we prove that 2q divides k0. Let t = Note that + 1(5t + 1)Q(x), where

Q(x) = 5t(q_l) + ... + 1.

The roots of + 1 are

((2m1)ir\ . . f(2m1)ir\), m=1,2,...,t;


that is,R(x)=xt+l=(x_wi)(x_w2)..(x_wt).

Let f(s) and g(x) be polynomials with integer coefficients such that

k0 f(x)(x+1)272+g(x)(x272+1)

= f(x)(x + 1)2n + g(x)Q(x)(xt + 1).

It follows that

f(Wm)(Wm+1)Th=ko, (1)

Since r is positive, t is even. So

2=R(1) = +w2)(1+wt).

Since is a symmetric polynomial in .. ,with integer coefficients, it can be expressed as a polynomial with integercoefficients in the elementary symmetric functions in w1, w2,. .. ,and therefore

F = f(wl)f(w2) .

is an integer.

Taking the product over m = 1,2,... , t, (1) gives 2272F = ku or =It follows that 2q divides k0.

It now suffices to prove that k0

Note that Q(1) = 1.

It follows thatQ(x) =(x+1)c(x)+1,

where c(s) is a polynomial with integer coefficients.

Hence(x + 1)2n = Q(x)d(x) + 1, (2)

for some polynomial d(x) with integer coefficients.

Also observe that, for any fixed m,

Thus(1 +(iJm)(1 (1 +w2t_1) = R(1) = 2,

and writing


we find that for some polynomial h(x), independent of m. with integercoefficients such that

(1 + w)th(w) = 2.But then (x + 1)h(x) 2 is divisible by xt + 1 and hence we can write

(x + 1)h(x) = 2+ (xt + 1)u(x),

for some polynomial u(x) with integer coefficients.Raising both sides to the power q we obtain

(x + = 2q + (xt + 1)v(x), (3)

where v(x) is a polynomial with integer coefficients.

Using (2) and (3) we obtain

(x + + 1)v(x)= Q(x)d(x)(xt + 1)v(x) + (xt + 1)v(x)= Q(x)d(x)(xt -I- 1)v(x) + (x + 2q,

that is,2q = fi(x)(x + + g1(x) + 1),

where (x) and 91(x) are polynomials with integer coefficients.

Hence k0 < 2q, as desired.

Our proof is thus complete.

Problem 94 [USAMO 1998 proposal, Kiran Kedlaya]Let x be a positive real number.

(a) Prove that(ni)! _!

(b) Prove that

(ni)! 1


Solution 94

We use infinite telescoping sums to solve the problem.

(a) Equivalently, we have to show that

1

1

that

n!x

(ni)! n!

and this telescoping summation yields the desired result.

(b) Let

f( )- (n-i)!X

Then, by (a), 1(x)

In particular, 1(x) converges to 0 as x approaches oo, so we canwrite f as an infinite telescoping series

= + k -1) - f(x + k)]. (1)

On the other hand, the result in (a) gives

f(x-1)-f(x)=

(ni)! x1(x+1)(x+n)


Substituting the la.st equation to (1) gives

as desired.

Problem 95 [Romania 1996]Let n 3 be an integer, and let

XcS={1,2,,,.,n3}

be a set of 3n2 elements.Prove that one can find nine distinct numbers (i = 1,2,3) in Xsuch that the system

aix+biy+ciz = 0a2x+b2y+c2z = 0a3x+b3y+c3z = 0

has a solution (xo, Yo, zo) in nonzero integers.

Solution 95Label the elements of X in increasing order Xi


Note that cannot equal since X1 and X2 are disjoint, and thatai a2 implies that the triples (ai,bi,c1) and (a2,b2,c2) are identical,a contradiction.Hence the nine numbers chosen are indeed distinct.

problem 96 [Xuanguo Huang]Let n 3 be an integer and let x1, x2, , be positive real numbers.Suppose that

= 1.

Prove that

(1 1

Solution 96By symmetry, we may assume that x1 x2 We have thefollowing lemma.

Lemma

1 + 1 + x3

Prooft Since n 3, and, since

1

= 1,

we have1 1

or>

It follows that x2x3> 1. Thus

1+x2 1+x3 (1+x1)(1+x3)

- (1+x2)(1+x3)

0,


as desired. o

By the lemma, we have

>1+xi 1+x2

and, since

it follows by the Chebyshev Inequality

11 (1

By the Cauchy-Schwartz Inequality, we have

2

or

(2)

Multiplying by

on both sides of (2) and applying (1) gives

which in turn implies the desired inequality.


problem 97Let X1,X2,. . . be distinct real numbers.Define the polynomials

P(x) =

and /1 1Q(x)=P(x)( + +..,+\XXj XX2

Let Y1,Y2 Yn1 be the roots of Q.

Show thatminlx,,x31

For the sake of contradiction, assume that

d minlx2 = minx, = mins3 (1)i(j

Since P has no double roots, it shares none with Q.Then

(1

+1

+1

= Q(Yi) 0,\Y,X1 yjX2

or1 1 1

+ =0.YiX1 Y%X2In particular, setting i = 1 and i = 2 gives

(2)

We claim that there is a k such that sk(sk + d)

1224. Solutions to Advanced Problems

for. all k, which in turn implies that

which contradicts (2).

Let j be the number of ks such that sk(sk+d) < 0, that is, Sk < 0 < Sk+d.

A simple but critical fact is that 8k + d and have the same sign. Infact, suppose that

then

Then > 0 if and only if /c > i + 1, that is Sk + d> 0.

From (1), we obtain Sk + d Sk+3, and, since sk + d and 3k+3 have thesame sign, we obtain

1 >1Sk + d 8k+j

for all k = . , n j. Therefore,

or

(3)

Also note that

E!


Problem 98 [Romania 1998]Show that for any positive integer n, the polynomial

f(x)=(x2+x)2"+l

cannot be written as the product of two non-constant polynomials withinteger coefficients.

Solution 98Note that 1(x) = g(h(x)), where h(x) = x2 + x and 9(y) = y2" + 1.Since

and is even for 1 k 1, g is irreducible, by Eisenstein'scriterion.

Now let p be a non-constant factor of f, and let r be a root of p.Then g(h(r)) = f(r) = 0, so .s h(r) is a root of g.Since s = r2 -f- r C Q(r), we have Q(s) C Q(r), so

deg p deg(Q(r)/Q) deg(Q(s)/Q) = deg g =

Thus every factor of f has degree at least T'.Therefore, if f is reducible, we can write f(x) = A(x)B(x) where A andB have degree

2 2"5 +5 +x+1) (mod2).

Since + x + 1 is irreducible in Z2[x], by unique factorization we musthave

A(s) B(s) + x + 1)2"' + + 1 (mod 2).

Thus, if we write

A(s) = + + ao,B(s) = + .. + b0,

then ao, b0 are odd and all the other coefficientsare even. Since f is monic, we may assume without loss of generality


that = = 1; also, a0bo = f(O) = 1. but a0 > 0, > 0 as f hasno real roots, so a0 = b0 = 1.

Therefore,

([x2n+2"1] + [x2"1])(g(x)h(x))

(+

i=2"1

+ + +EQ (mod 4)

as + is even.But

+ [x2"'])(f(x))=

and is odd by Lucas's theorem, so

([x2"+2"'] + [x2"')) (1(x)) 2 (mod 4),

a contradiction.Hence f is irreducible.

Problem 99 [Iran 1998]Let Ii, 12,13 R R be functions such that

aifi + a212 + a3f3

is monotonic for all a1, a2, a3 E

Prove that t

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