Tổng hợp các đề thi Toán Olympiads từ 1981-1995 tổ chức Ba Lan và Úc

Tổng hợp các đề thi Toán Olympiads từ 1981-1995 tổ chức Ba Lan và Úc

National Library of Australia Card Number and lSSN Australian Mathematics Trust Enrichment Series lSSN 1326-0170 Polish and Austrian Mathematical Olympiads 1981-1995

ISBN 1 876420 02 2

ED J BARBEAU, Toronto CANADA

GEORGE BERZSENYl, Terra Haute USA

RoN DUNKLEY, Waterloo CANADA

WALTER E MIENTKA, Lincoln USA

NIKOLAY KONSTANTINOV, Moscow RUSSIA

ANDY L!U, Edmonton CANADA

JORDAN B TABOV, Sofia BULGARIA

JOHN WEBB, Cape Town SouTH AFRICA

The books in this series are selected for the motivating, interesting and stimulating sets of quality problems, with a lucid expository style

in their solutions Typically, the problems have occurred in either national or international contests at the secondary school level

They are intended to be sufficiently detailed at an elementary level for the mathematically inclined or interested to understand but, at the same time, be interesting and sometimes challenging to the undergraduate and the more advanced mathematician lt is believed that these mathematics competition problems are a positive influence on the learning and enrichment of mathematics

ENRlCHMENT SERlES

BOOKS lN THE SERlES

1 1 ALL THE BEST FROM THE AUSTRALIAN MATHEMATICS COMPETITION

JD Edwards, DJ King Et PJ O'Halloran

1 2 MATHEMATICAL TOOLCHEST

AW Plank Et NH Williams

3 TOURNAMENT OF TOWNS QUESTIONS AND SOLUTIONS 1984-1989

PJ Taylor

4 AUSTRALIAN MATHEMATICS COMPE11110N BOOK 2 1985-1991

P J O'Halloran, G Pollard Et P J Taylor

5 PROBLEM SoLVING VIA THE AMC

The traditions of National Mathematical Olympiads in many European countries dates back to about 1950 (in some cases further back) These Olympiads are used, inter alia, to select national teams to participate in the International Mathematical Olympiad

The traditions of the Polish and Austrian Mathematical Olympiads are particularly strong, and to a certain extent they are linked, since together they developed the Austrian-Polish Mathematical Olympiad, one of the world's strongest regional events

As the reader will determine, the problems in this book are quite exquis­ite, having been hand-picked from the problems of many years They are also noted for having multiple independent solutions, making the math­ematics so much richer There can be little more satisfying than finding

a different, independent solution to a known one Being mathematics, of course, the result is always the same after having taken a quite different route

The authors of this book have many decades of experience at this level

Of the two, I have only had the pleasure of personally knowing Dr Kuczma Dr Kuczma has one of the world's highest reputations in prob­lem creation Indeed, he has had no less than four of his problems posed

in International Mathematical Olympiads Further, he has had many more reach the final preselection stage He is also equally renowned as

a problem solver From mutual acquaintances and examination of the work in this book, Erich Windischbacher is held in no less regard The Australian Mathematics Trust aims to set a high standard of mate­rial and exposition in this Enrichment Series The contents of this series involve pedagogical material in problem solving and instructive problems which have not appeared before in English We are confident that this book achieves the high standards to which we have aimed

Mathematics Olympiads have a long tradition in Poland as well as in Austria, and they have many features in common in both these countries Academic supervision comes from the Mathematical Societies and from university centres Financial support is provided, in the greatest part, by the Ministries of Education (the exact official name of that institution,

in each country, has changed several times during the past dec.ades) The effective running of the competitions relies on people (high school teachers and university teachers) whose enthusiasm and devotedness is practically the sole motive for their activities

The organizational format is much the same in the two countries Con­testants are high school students, most of them attending the last or the last but one grade College students are not allowed to participate The final round of the Austrian MO and of the Polish MO is a two day written exam, with three problems to be solved each day -just like the IMO As regards earlier stages, there are some differences; but, anyhow, ea<::h elimination round consists of problem solving All the problems posed at our olympiads are essay type; all steps of the reasoning have

to be explained and justified by the solver - short answer questions or multiple choice questions are not used

In the late seventies, a bilateral agreement on cultural exchange was concluded between the Polish and the Austrian Ministers of Education This resulted, in particular, in frequent visits of scientists and teachers, from one country to the other, and has led to exchange of experience­for instance, in the org�nization of math olympiads (remember that, in those years, Austria and Poland pertained to distinct political zon� of Europe) It is also in that time that the Austrian-Polish Mathematics Competition was launched.1

The authors of the present book are just two of those "enthusiasts of the Olympic idea in mathematics", for many years involved in the running of the national mathematics olympiads in Poland and in Austria It is quite

a time ago that we first met Soon the idea occurred to us to present, in book form, a selection of our countries' olympiad problems

As a guideline for the selection, we have decided to take the diversity of methods of solution Accordingly, each problem in this book is presented

1 A compilation of all the problems posed at the first sixteen rounds of that compe­ tition, with complete solutions, has appeared in book form: ME Kuczma, Problems

144 problems of the Austrian-Polish Mathematics Competition 1978-93, published

1994 by: The Academic Distribution Center, 1216 Walker Rd., Freeland, Maryland

21053, USA

with at least two solutions, and sometimes more than two; this feature of the book we consider important enough to be reflected in the sub-title

It is obvious that various ways of approach to any problem provide a better understanding of its nature, reveal several aspects of the relevant topics and teach various techniques

Now, it can always be questioned whether a different solution is a really

different one In some rare cases, it can be justly considered as such

In many cases, it cannot - and this is evident at first glance And in most other cases - also not; the "second" method can use other sym­bols, language, terminology, it may look quite unlike the "first" one, and still be, in fact, the same For instance: is the Law of Cosines any­thing else than operating with vectors and their inner products? Is the examination of divisibility of polynomials via manipulation in real do­main anything essentially different from complex roots and factorization technique? Combinatorial arguments, when disguised in the language

of polynomials (in fact, the generating functions of the quantities un­der consideration), do they really differ from the analogous arguments presented in pure form, without disguise?

This list can be continued, of course Viewed from a certain level of professionalism, all or almost all approaches to a particular olympiad­style problem are just like dressing the same idea in a robe of one or another colour What can be, however, immediately recognized by a mathematician, need by no means be evident to a young student who just makes the first steps in off-curricular areas of mathematics

Indeed, we think that - besides getting acquainted with various tools and tricks supplied by various methods - the reader's own discovery

of the intrinsic uniformity hidden behind apparently distinct ways of ap­proach is the true profit she or he can have from studying those solutions, and is the best we can offer her or him

Most of our solutions have been elaborated in detail The intention was

to make them accessible to a rather wide audience; some readers will find them unnecessarily lengthy, perhaps We are sure that the readers' invention will often go further; no doubt, they will find yet other ways

of resolving this or that problem, possibly more elegant or more general than the presented ones So much the better! Satisfaction from a good job done is the solver's true reward

(Another kind of satisfaction comes from detecting the authors' errors and mistakes; these are also very instructive!)

There is one more thing we must mention here There should be no surprise if a problem turns out to be identical or very closely related to a question that had appeared at some other competition or in the problem section of some journal It is no secret that problems "circulate" and

are being "borrowed" from one competition to another There is also nothing paradoxical in the fact that very similar ideas occur to people who independently devise olympiad stuff, in distant parts of the globe Although we have tried to avoid the use of problems of which we knew

to have been used elsewhere, we can by no means be sure

We are presenting 64 problems (a beautifully round number) from the two national olympiads, half from the Austrian, half from the Polish 2 They are arranged more or less thematically; the rough rules are (ilasy to spot Such rules can never be quite univalent; a problem may be difficult

to classify; it can pertain to more than one thematic area, sometimes depending on the solution method

The arrangement has nothing to do with the level of difficulty; quite challenging problems often follow or are followed by trivially simple ones The reader should not know "what to expect next"

We will be happy to receive any feedback from the readers: comments, communication about mistakes, any suggestions We wish all the readers joy, fun and pleasure in tackling the problems

A-8020 Graz Austria

2Problems from the Austrian MO: 1, 4, 5, 7, 9-12; 16-19, 23, 28-30, 32-35, 37,

38, 43, 52, 53, 55-61

Problems from the Polish MO: 2, 3, 6, 8, 13-15, 20-22, 24-27, 31, 36, 39-42, 44-51,

54, 62-64

We are happy to see our book appearing as an AMT publication Our sincere thanks go to Professor Peter Taylor, the executive director

of the AMT, for his invitation to publish this book in the AMT Enrich­ment Series and for his help in typesetting/formatting; and to Dr Andrei Storozhev for producing the diagrams

MEK&EW

May, 1998

>0

th

1 Show that 2<3n) + 1 is not divisible by 17, for any integer n > 0

2 Let a, b, c be positive integers with the properties: a3 is divisible

by b, b3 is divisible by c, c3 is divisible by a Show that (a+b+c)13

is divisible by abc

ln/3J

3 For every integer n > 2 show that the number L ( - 1)k ( n) is

(The symbol L x J denotes the Greatest Integer Function )

4 Calculate the sum of all divisors of the form 2x · 3Y (with x, y > 0)

of the number N = 1988 - 1

5 Show that there do not exist four successive integers whose product

is 1993 less than a perfect square

6 Show that there are infinitely many positive integers n such that each one of the three numbers n - 1 , n , n + 1 can be represented

as the sum of two perfect squares

7 Show that the following system of simultaneous equations has no solution in integers:

x2- 3xy + 3y2 - z2 31 -x2 + 6yz + 2z2 44

x2 + xy + 8z2 100

8 Solve the following equation in integers x, y:

9 If x, y, z are integers, at least one of which is 1990, show that

x2 + y4 + z6 > xy2 + y2z3 + xz3

for n = 1, 2, 3 , Define Yn = x; + 2n+2 Show that Yn is the square of an odd integer, for every non-negative integer n

11 The sequence (an} is defined recursively by

ao = 1, a1 = 2, an a;__1 + 1

= an-2 for n = 2, 3, 4,

Show that each an is an integer

12 Find·all functions f mapping non-negative integers into non-negat­ive integers and such that f (! ( n)) + f ( n) = 2n + 6 for every integer

n 2: 0

13 Show: that ln.J3J is a power of 2 for infinitely many natural num­bers n

(The symbol l x J denotes the Greatest Integer Function )

14 Four numbers are randomly chosen from the set { 1, 2, , 3n} ( n

is a fixed integer greater than 1) Compute the probability that the sum of those four numbers is divisible by 3

15 For what natural numbers n is it possible to tile the n x n-chess­board with 2 x 2 and 3 x 3-squares?

16 A triangular prism is a pentahedron whose two parallel faces ( "top base" and "bottom base" ) are congruent triangles and the remain­ing three faces are parallelograms We are given four non-coplanar points in space How many distinct triangular prisms having the four given points as vertices are there?

1 7 Consider the infinite chessboard with squares coloured white and black, in the usual manner SupposeS is a set of 1976 squares such that every two squares in S can be connected by a path consisting

of consecutively adjacent squares (Two squares are adjacent if they have a common edge ) Show that there are at least 494 white squares inS Moreover, show that 494 is the exact bound

18 Consider an alphabet consisting of three symbols a, b, c How many n-character words with the following properties (1) and (2) can be composed?

(1) the word should begin and end with an a;

(2) neighbouring positions must be occupied by different symbols

19 Nine trucks follow one another, in a line, on a highway At the end of a day's ride it turned out that each driver disliked the style

of the driving of the one in front of him They wish to rearrange themselves so that, next day, no truck would follow the same truck that it followed on the first day How many such rearrangements are possible?

20 We are considering paths (Po, Pt , Pn) of length n over lattice points in the plane (i.e., points (x, y) with integer coordinates); for each i, the points Pi-l and Pi are assumed to be adjacent on the lattice grid Let F(n) be the number of those paths that begin in

Po = (0, 0) and end in a point Pn lying on the line y = 0 Prove that F(n) = e:)

21 Determine all real polynomials P (x) of degree not exceeding 5, such that P(x) + 1 is divisible by (x - 1)3 and P(x)- 1 is divisible by (x + 1)3

22 Prove that the polynomial xn + 4 factors into the product of two polynomials of lower degrees with inte_ger coefficients if and only if

n is divisible by 4

23 Find all natural numbers n for which the polynomial

Pn (x) = x2n + (x + 1)2n + 1

is divisible by the trinomial T(x) = x2 + x + 1

24 For every positive integer k show that the polynomial

is divisible by the binomial x5 + 1

25 Find all pairs of real numbers a, b such that the polynomials

have two distinct common real roots

26 Let a, x, y, z be real numbers such that

cos x + cos y + cos z sin x + sin y + sin z

cos(x + y + z) sin(x + y +z)

Prove the equality: cos(y + z ) + cos(z + x) + cos(x + y) = a

27 If a, b, e are pairwise distinct real numbers, show that the value of the expression

a - b b - e e - a

1 + ab + 1 + be + 1 + ea

is never equal to zero

28 Solve the system of equations:

x + y + xy = 19, y + z + yz = 11, z + x + zx = 14

29 Solve the system of equations

Xl (Xl - 1) = X2 - 1 X2 (X2 - 1) X3 - 1

axi-l + bxi + exi+l � 0 for i = 1, , n, where by definition xo = Xn, Xn+l = x1

33 Let a, b, e be the sides of a triangle Show that

35 Let a, b be non-negative real numbers with a2 + b2 = 4 Show that

ab In -<v2-1 a+b + 2 -and determine when equality holds

36 The real numbers ai, bi, ci, di are such that 0 � ci � ai � bi � di and ai + bi = Ci + di for i = 1, 2, , n Prove the inequality

II ai + II bi � II Ci + II di

i=l i=l i=l i=l

37 Prove the following inequality for all integers n > 1:

40 Prove that the inequality

holds for any real numbers a17 a2, , ar Find conditions for equality

41 For a fixed integer n ;::: 1 find the least value of the sum

42 On a given segment AD, find points B and C so as to maximize the product of the lengths of the six segments AB, AC, AD, BC,

47 Four sequences of real numbers

satisfy the simultaneous recursions

for n = 0, 1, 2, Suppose there exist integers k, r ;.::: 1 such that

48 The sequences xo, x1 x2, and yo, Yl Y2, are defined by:

Xo =YO= 1,

Xn + 2 Xn+l =

Xn + 1'

y� + 2 Yn+l = -2 for n = 0, 1, 2,

Yn

Show that Yn = X2"-l for every integer n ;.::: 0

49 Two sequences of integers all a2, a3, and b1, b2, b3, .. are de­fined uniquely by the equality (2 + v'3 ) n = an + bn v'3 Compute lim (an/bn)·

51 A sequence of real numbers ao, a1, a2, • satisfies the recurrence

!ani = an-1 + an+1 for n = 1 , 2, 3, ... Show that an+9 =an for all n

52 Construct a right triangle ABC with a given hypotenuse c such that two of its medians are perpendicular

53 Let ABC be a triangle, AC =/: BC Assume that the internal bi­sector of angle AC B bisects also t_he angle formed by the altitude and the median emanating from vertex C Show that ABC is a right triangle

54 If ABCDEF is a convex hexagon with AB = BC, CD = DE,

EF = FA, prove that the altitudes (produced) of triangles BCD, DEF, FAB , emanating from vertices C, E, A , concur

55 Let ABCDEF be a regular hexagon with M and N points on diagonals CA and CE (respectively) such that AM = CN If M ,

N and B are collinear, prove that AM = AB

56 Let ABC be an acute triangle with altitudes BD and CE Points

F and G are the feet of perpendiculars BF and CG to line DE

Prove that EF = DG

57 Consider the right triangle ABC with LC = 90° Let A1 and B1

be two points on line AB (produced beyond A and B) such that

AA1 = AB = BB1 and let N be the foot of the perpendicular from

A1 to line B1C Show that the rectangle with sides B1C and CN

has area twice as large as the square with side AB

58 Let ABC DE be a convex pentagon inscribed in a circle The dis­tances from A to lines BC, CD, DE, and BE are a, b, c, and d,

respectively Express d in terms of a, b, c

59 Let ABC be an isosceles triangle with base AB Let U be its circumcentre and M be the centre of the excircle tangent to side

AB and to sides C A and C B produced Show that

2 ·CU < CM < 4·CU

60 The diagonals AC and BD of a convex quadrilateral ABCD inter­sect in E Let F1, F2 and F be the areas of triangles ABE, CD E

and quadrilateral ABCD , respectively Show that

When does equality hold?

61 Let P1P2 be a fixed chord (not a diameter) of a circle k The tangents to k at P1 and P2 intersect at Ao Let P be a variable point on the minor arc P1P2 The tangent to k at P intersects lines

AoP1 and AoP2 at A1 and A2, respectively Determine the position

of P for which the area of triangle A0A1A2 is a maximum

62 Let P be a point inside a parallelepiped whose edges have lengths

a, b and c Show that there is a vertex whose distance from P does not exceed �v'a2 + b2 + c2

63 Do there exist two cubes such that each face of one of them meets each face of the other one (possibly at an edge or a corner)?

64 Let A1 , A2, Aa, A4 be points on the sphere circumscribed about the regular tetrahedron with edge 1 such that AiAj < 1 fori =1-j

Prove that these four points lie on one side of a certain great circle

of the sphere

Problem 1

Show that 2W> + 1 is not divisible by 17, for any integer n > 0

Problem 1, Solution 1

Assume, to the contrary, that 23" + 1 is divisible by 17 for a certain

n � 1 (we write just abe for aW>) Let r be the remainder left by 23"-1

23" = -1 (mod 17) forces 2 3"-1 = -1 (mod 17)

Descending, we conclude inductively that

So 23n + 1 is never congruent to 0 (mod 17)

Problem 2

Let a, b, c be positive integers with the properties: a3 is divisible by

b, b3 is divisible by c, c3 is divisible by a Show that (a+ b + c)13 is divisible by abc

k, m � 0 integers, k + m = 13 (2) The conditions of the problem imply that a9 is divisible by c and b9 is divisible by a Any number of the form (2) can now be represented as the product of four factors (separated by multiplication dots in the listing below):

m = 13

akbm = a · a3· a9·1;

akbm = a ·b · a9·(ak-10bm-1); akbm = a ·b ·b3 ·(ak-1bm-4); akbm = b9·b ·b3·1;

in each case the first factor is divisible by a, the second by b, the third

by c (and the fourth is an integer), and so the product abc is a factor of akbm That does the job

Problem 2, Solution 2

Let p be any prime divisor of the product abc Write

where a, (3, 7 � 0 and u, v, w � 1 are integers non-divisible by p

Since a3 is divisible by b, the exponents a and (3 satisfy 3 a � (3 Likewise,

3 (3 � 7 and 3 7 � a; and hence 9a � 7, 9 (3 � a, 9 7 � (3

Let r = min( a, f3, 7) We obtain

The numbers a, b, c are divisible by pr Thus (a+ b + c)13 is divisible

by p13r, hence by p01+/3+-r, in view of inequality (4) On the other hand, according to (3), o: + (3 + 'Y is the exact power in which p enters the prime factorization of abc Since p was an arbitrarily chosen prime factor of abc,

we conclude that (a+ b + c)13 is divisible by abc

For any fixed non-negative integer n, the fundamental Binomial Identity

is valid for every integer j, if we agree that

( ; ) = 0 for j < 0 and for j > n

Consider the following three sums:

Problem 3, Solution 2

Notation (3) together with the convention (2) is preserved from Solution

1 For any complex number z we have by the Binomial Theorem

(1 +w)n =An- Bnw2- Cnw for w3 = -1 {10) Setting w = -1 we hence obtain

Bn = An + Cn for n = 1, 2, 3,

Now consider the complex number

a= !+ !v'3i = cos(7r/3) + i sin(7r/3),

(11)

also satisfying a3 = -1 , and moreover, a2 = a -1 For w = a equalities (1 0) and (11) yield

(1 + a)n =An- Bn(a- 1 )- Cna =�An- (!An+ Cn)v'3i {12)

On the complex plane, the numbers 0, 1 and a represent the vertices of

an equilateral triangle, which is completed to a parallelogram (rhombus)

by the vertex 1 + a Thus 1 + a = J3 (cos( 71"/6) + i sin( 71"/6)), and by de Moivre's Theorem

For each n, Kn is an integer; in particular, K3 = 0 Hence A3 = 0; and for n ;::: 4 we have q ;::: 2, so An = 3q-l K n is divisible by 3

Problem 4

Calculate the sum of all divisors of the form 2z · 3Y (with x, y > 0) of the number N = 1988 - 1

Problem 4, Solution 1

The only trouble is to determine the highest powers of 2 and 3 that divide

N This can be done using the Binomial Theorem:

1 988 ( 20- 1)88

c:) - c18) 20 + (8:) 202 - c:) 203 + + (::) 2088 1- 88 · 20 + (terms divisible by 26)

(we used the fact that (8:) = ! · 88 · 87 = 22 · 3 · 1 1 · 29); and

Consequently, the sum we are about to evaluate equals

s =

(x,y): x,y>O 2"' ·3Y dividing N

L 2"'· 3Y xE{l,2,3,4,5}

yE{l,2}

5 2

= L::2"'·L::3y x=l y=l

In particular, 4>(64) = 32 and 4>(27) = 18 Thus

(mod 64) and 1918 = 1 (mod 27)

Raising the first of these relations to third power and the second one to fifth power, we get

(mod 64) and 1990 = 1 (mod 27);

or which is exactly the same

-(mod 64) and 192 1988 = 1 (mod 27)

Consequently, in view of (1) and ( 2),

(mod 64) and 1988 "¢ 1 (mod 27) (3) (if 1988 were 1 (mod 64), the product 198 · 1988 would be 33 rather than

1 (mod 64); and the second relation of (3) is justified similarly)

On the other hand, equation (1) shows that 198 = 1 (mod 3 2) Besides,

19 = 1 (mod 9) If we raise the first relation to power 11 and the second

Show that there do not exist four successive integers whose product is

1993 less than a perfect square

Problem 5, Solution 1

Assume that the equation

x(x + 1) (x + 2)(x + 3) + 1993 = y2 (1)

is fulfilled for some integers x and y Examine equation (1) modulo 5 Either the product x(x + 1)(x + 2)(x + 3) is divisible by 5 or its four factors leave remainders 1, 2, 3, and 4, in which case the product equals

(y - z)(y + z) = 1992 (2)

We see from (2) that z cannot be - 1 ; hence z ;::: 0 We may also assume (see (1)) that y ;::: 0 So the second factor in equation (2) is non-negative; consequently, both factors must be positive, the second one greater than the first Both factors are integers of the same parity; their product

is even, so they both are even In view of the prime decomposition

1992 = 23 · 3 · 83, the prime factor 83 must enter y + z and we conclude that the pair (y - z, y + z) must be one of the following:

(2, 996), (4, 498), (6, 332), (12, 166)

Accordingly, z equals 497, 247, 163, or 77, which means that the product

(x + 1)(x + 2) equals 498, 248, 164, or· 78 However, it is easily verified that no one of these four numbers is equal to the product of two successive integers Contradiction ends the proof

Problem 6, Solution 2

Now let nk = 2m% + 1, where mk = k (k + 1) It is enough to notice that

nk - 1 = m% + m%, nk = (k2 + 2k)2 + (k2 - 1)2,

nk + 1 = (mk + 1)2 + (mk - 1)2

Problem 6, Solut ion 3

Define the sequences a1 a2, a3, and b1 b2, b3, . recursively by

ao = 4, bo = 3,

and notice the equality a% + 2 = 2b% (easy proof by induction) Hence,

if we set nk = a% + 1, we are done because

Problem 7, Solut ion 1

Since the terms x2 and z2 appear in all the three equations, it is tempting

to apply the method of elimination so as to get rid of them If we multiply the first equation by a, the second by b, and the third by c, and add the resulting equations, we obtain an equation in which the coefficients of

x2 and z2 are a - b + c and -a + 2b + 8c, respectively Setting these expressions to be zero, we find that e.g a = 10, b = 9 and c = -1 do the job, producing the equation

10 · ( -3xy + 3y2) + 9 · 6yz - xy = 10 · 31 + 9 · 44 -100,

i.e.,

y ( -31x + 30y + 54z) = 606

This yields the possible values of IYI: 1, 2, 3, 6, 101, 202, 303, 606

In a similar manner we can eliminate the terms x2 and xy, multiplying the first, the second and the third equation of the system by suitable factors a, b, c; now we need that a - b + c and - 3a + c (the coefficients

of x2 and xy in the resulting equation) should be zero When we take

D = (24y)2 - 4 · 31· (3y2 - 507) = 4(51y2 + 15717);

then the roots z1, z2 are: (- 12y ± fDJ4 )/3l Thus 51y2 + 15717 ought

to be a square number in order that ZI, z2 be integers Yet, for the previously found values of IYI this expression takes values 15768, 15921,

16176, 17553, 535968, 2096721, 4697976, 18744753, no one of which is a perfect square So the system has no integer solutions

P roblem 7, Solution 2

An astonishingly simple proof results from examination of the two outer equations modulo 5 (the middle equation is not needed!) Multiplying the first equation by 8 and adding the third equation we get

9x2 - 23xy + 24y2 = 348,

which is

-x2 + 2xy - y2 = 3, i.e., (x - y)2 = 2 (mod 5)

Yet the square of an integer can only be 0, 1 or 4 (mod 5); the claim follows

(u + v + 4) (uv + 1)

1 · ,

1· , 5· ,

5

One of the factors must be equal to 5 or -5 and the other to 1 or -1

(respectively) This means that the sum u + v and the product uv have

to satisfy one of the four equation systems:

If this has to be an integer, the denominator s + 2 must be a divisor of

5, which means that s must be one of the numbers -7, -3, -1, 3 For each of these values of s , the corresponding value of q is computed from

(2) and we arrive at the four possible systems of equations for s = x + y,

(they correspond, in a certain order, to the four systems obtained in Solution 1) The numbers x and y must be the roots of the respective

quadratic trinomial

t2 + 7t - 10 j t2 + 3t - 10 j t2 - 3t + 2

Of these, only the second and the fourth have integer roots; these are, respectively, -5, 2 and 1, 2 So (x, y) is one of the pairs (-5, 2), (2, -5), (1, 2), (2, 1)

Problem 8, Solution 3

Use the symmetric forms s = x + y, q = xy The resulting relation (1)

can be viewed as a quadratic equation with the unknown s and parameter

One of these roots has to be equal to x + y, an integer Therefore D

must be the square of an integer: D = d2; d � 0 Then

with discriminant

D = x4 + 4(x - 1)(x2 + 1) = x4 + 4x3 - 4x2 + 4x - 4, (6) which must be a perfect square in order that equation (5) has an integer root y

Suppose x > 2 Then the following inequalities hold:

D - (x2 + 2x - 4)2 =

D - (x2 + 2x - 2)2 2

0(x-1) > 0,

-4(x - 1)(x - 2) < 0, showing that D is strictly comprised between the squares of two skip­consecutive integers x2 + 2x - 4 and x2 + 2x -2 Therefore D has to be the square of x2 + 2x -3 This, however, cannot be the case, since this last number is of different parity than D (see (6))

The only possibility that remains is that x = 2 Equation (5) then be­comes y2 + 4y - 5 = 0; equivalently; (y -1) (y + 5) = 0, and we get y = 1

or y = -5 So (2, 1) and (2, -5) are all pairs of integers (x,.y) with

x > 1, satisfying the equation Symmetry yields two other pairs (1, 2)

and ( -5, 2); and there are no more- as the argument shows

P roblem 8, Solut ion 5

Assume that the integers x, y satisfy the equation Its left side is the sum of two addends, one of which must be 2:: 1 and the other one � 0 Let e.g y2(x -1) 2:: 1, x2 (y -1) � 0 Then x 2:: 2, y # 0, y � 1

If y = 1, then of course x = 2 (just look at the equation)

Assume y < 0 for the sequel (remember that y = 0 has been excluded) Again let x + y = 8 and rewrite the equation in the form

Expanding and regrouping,

Equality f (x) = 0 implies that both inequalities in (7) must turn into equalities Now, f(x) = ! (2) means that x = 2, while s2 = 9 means that

s = -3 Hence y = s-x = -5 Recalling the case of y = 1 (mentioned

at the beginning), we obtain the two solving pairs (x, y) with y� 1: (2, 1) and (2, -5) Interchanging the roles of x and y we get the other two pairs: (1, 2) and ( -5, 2) ; and these four pairs constitute the complete solution

Problem 9

If x, y, z are integers, at least one of which is 1990, show that

x2 + y4 + z6 > xy2 + y2z3 + xz3

Problem 9, Solution 1

This is in fact the Cauchy-Schwarz inequality for the triple of numbers

x, y2, z3; it can be settled (in the weak form) as follows, using the arithmetic mean-geometric mean inequality for pairs of numbers:

x2 + y4 x2 + z6 y4 + z6 x2 + y4 + z6 2 + 2 + 2

= lxly2 + lxllzl3 + Y21zl3

> xy2 + y2z3 + xz3 (because lxl ;:::: x and lzl ;:::: z )

Equality would require that x2 = y4 = z6 and either y = 0, xz ;:::: 0, or

y =/= 0, x, z ;:::: 0 In the first case we get x = y = z = 0, in contradiction

to the "1990" condition

Regarding the second case, we now have z3 = y2 = x > 0 Since x, y,

z have to be integers, z3 = y2 forces that z is itself a perfect square:

z = u2, with u being a positive integer Thus y = ±u3, x = u6 By assumption, one of the numbers x = u6, y = ±u3, z = u2 has to be

1990 And since 1990 is neither a square or cube or sixth power, equality cannot occur and the given inequality holds (in the strict form)

These expressions are non-negative Now, x, y, z are integers, one of them being equal to 1990 If x = 1990, then y2 =I x If y = 1990 or

z = 1990, then z3 =I y2 In each case one of the terms (y2 - x )2 and

(z3 - y2)2 is strictly positive, and so is the difference expressed by for­mulas (1)

showing that Yn is the square of an odd integer, as asserted

or, which is the same (in view of xo = 0),

So we have formula (1) of Solution 1 (without guessing), and it remains just to repeat the last paragraph of that solution

If one prefers (unwisely enough) to work out a recursive formula for the

YnS, that is also possible Squaring the equation that defines Xn+2 we obtain

Multiply this by 12 and insert expression (4) (and the analogous expres­ sion for 12zn+l):

-Yn+3 + 9yn+2 + 4Yn+l - 18 · 2n+3

36(Yn+l - 2n+3) - 2( -Yn+2 + 9Yn+l + 4yn - 18 · 2n+2)

The powers of 2 cancel out and we are left with

Apply the method described in Solution 3 The characteristic equation

is

q3 - 7q2 + 14q -8 = 0

Its coefficients sum up to 0, hence one of the roots is 1 and the equa­ tion factors into (q - 1) (q2 - 6q + 8) = 0 The roots of the quadratic factor are found e.g from the Viete's Formulas; they are 2 and 4 So we postulate

Yn = A· 4n + B · 2n + C (5) The initial terms xo = 0, x1 = 1 , x2 = 3 yield the initial terms of the sequence (Yn}: Yo = 4, Yl = 9, Y2 = 25 Setting these in (5) we obtain the system of linear equations for the constants A, B, C :

We proceed by induction The first t hree terms ao = 1 , a1 = 2 and a2 = 5

are integers Fix n � 3 and assume that the aks are integers for all k :::; n;

we will show that an+l i s a n integer also According t o the defining formula, an-1 = (a;_2 + 1)fan-3 i thus

The numbers an- 1 an-2, an-3 are integers, by the inductive assumption The last formula shows that an-1 and an-2 are coprime Now,

an+l = (a� + 1)/an_1 is an integer This completes the inductive step

Problem 1 1 , Solution 2

According to the definition,

Replacing n by n + 1 we obtain

Subtracting the first equation from the second one,

This shows that the sequence ((an+l + an-1)/an) is constant It begins with (a2 + ao)/a1 = (5 + 1)/2 = 3, and hence (an+l + an-dian = 3 for all n ; equivalently,

an+1 = 3an - an-1 for n = 1, 2, 3, Since ao = 1 and a1 = 2, this forces that all the ans are integ�rs