22 FileGocHOMOMoi

Forexample, if ∠ABC = 2∠ADC,D lies either on the internal angle bisector of angle ABC, or on the external angle bisector of angle ABC we cannot write ∠ADC = 12∠ABC todetermine which line

Chapter 1 Introduction 1

1.1 Preface 1

1.2 Glossary 2

Chapter 2 Examples for practice 14

2.1 Arithmetic problems 15

2.1.1 Junior problems 15

2.1.2 Senior problems 36

2.1.3 Undergraduate problems 45

2.1.4 Olympiad problems 47

2.2 Algebraic problems 60

2.2.1 Junior problems 60

2.2.2 Senior problems 77

2.2.3 Undergraduate problems 86

2.2.4 Olympiad problems 88

2.3 Geometric problems 101

2.3.1 Junior problems 101

2.3.2 Senior problems 119

2.3.3 Olympiad problems 138

2.4 Analysic problems 153

2.4.1 Junior problems 153

2.4.2 Senior problems 156

2.4.3 Undergraduate problems 159

2.4.4 Olympiad problems 170

2.5 Problems of Other Topics 174

2.5.1 Junior problems 174

2.5.2 Senior problems 180

2.5.3 Undergraduate problems 184

2.5.4 Olympiad problems 185

Chapter 2 Exercises for training 196

3.1 Exercises 196

3.1.1 Exercises from Mathematic Reflextion 196

3.1.2 Exercises from OP from Around the World 210

3.2 Problems of Hanoi Open Mathematical Olympiad 215

3.2.1 Hanoi Open Mathematical Olympiad 2006 215

3.2.2 Hanoi Open Mathematical Olympiad 2007 216

3.2.3 Hanoi Open Mathematical Olympiad 2008 219

3.2.4 Hanoi Open Mathematical Olympiad 2009 221

3.2.5 Hanoi Open Mathematical Olympiad 2010 Senior Section 224

3.3 Singapore Open Mathematical Olympiad 2009 226

3.3.1 Junior Section 226

3.3.2 Senior Section 229

Bibliography 234

i

To guide students who are interested in mathematics and have the potential to enter the world of Olympiad mathematics, so that their mathematical ability can be promoted efficiently and comprehen- sively, it is important to improve their mathematical thinking and technical ability in solving mathematical problems.

Technical ability in solving mathematical problems does not only involve producing accurate and skilled computations and proofs, the standard methods available, but also the more unconventional, creative techniques.

It is clear that the usual syllabus in mathematical educations cannot satisfy the above requirements, hence the mathematical olympiad training books must be self-contained basically.

The book is based on the lecture notes used by the editor in the last 25 years for Olympiad training courses inBAC GIANG SPECIALIZING UPPER SECONDARY SCHOOL Its scope and depth significantly exceeds that of the usual syllabus, and introduces many concepts and methods of modern mathematics.

The core of each lecture are the concepts, theories and methods of solving mathematical problems Examples are then used to explain and enrich the lectures, and indicate their applications And from that, a number of questions are included for the reader to try Detailed solutions are provided in the book.

The examples given are not very complicated so that the readers can understand them more easily However, the practice questions include many from actual competitions which students can use to test themselves These are taken from a range of countries, e.g China, Russia, the USA and Singapore The questions are for students to practise, and test students’ ability to apply their knowledge in solving real competition questions Each section can be used for training courses with a few hours per week The test questions are not considered part of the lectures, since students can complete them on their own.

Acknowledgments

My great thanks to Doctor of Science, Professor Nguyen Van Mau, and Doctor Associate Professor Nguyen Vu Luong for their strong support I would also like to thank my colleagues, MA Bach Dang Khoa, MA Tran Thi Ha Phuong, MA Nguyen Danh Hao and MA Tran Anh Duc for their careful reading

of my manuscript, and their helpful suggestions This book would be not written today without their efficient assistance.

Nguyen Van Tien

1

Arithmetic mean-geometric mean (AM-GM) inequality If a1, a2, , an are

n nonnegative numbers, then their arithmetic mean is defined as 1

1n

n

∑

k=1

ak⩾ 1 1n

1.2 Glossary 3

Binomial coefficient

Cnk= n!

k!(n − k)!,the coefficient of xk in the expansion of (x + 1)n

Brocard angle ( See Brocard points.)

Brocard points Given a triangle ABC, there exists a unique point P such that

∠ABP = ∠BCP = ∠CAP and a unique point Q such that ∠BAQ = ∠CBQ = ∠ACQ.The points P and Q are the Brocard points of triangle ABC Moreover, ∠ABP and

∠BAQ are equal; their value φ is the Brocard angle of triangle ABC

Cauchy-Schwarz inequality For any real numbers a1, a2, , an,and b1, b2, , bn

with equality if and only if ai and bi are proportional, i= 1,2, ,n

Centrally symmetric A geometric figure is centrally symmetric (centrosymmetric)about a point O if, whenever P is in the figure and O is the midpoint of a segment P Q,then Q is also in the figure

Centroid of a triangle Point of intersection of the medians

Centroid of a tetrahedron Point of the intersection of the segments connectingthe midpoints of the opposite edges, which is the same as the point of intersection of thesegments connecting each vertex with the centroid of the opposite face

Ceva’s theorem and its trigonometric form Let AD, BE, CF be three cevians

of triangle ABC The following are equivalent:

(i) AD, BE, CF are concurrent;

sin ∠CADsin ∠DAB = 1.

Cevian A cevian of a triangle is any segment joining a vertex to a point on theopposite side

Chinese remainder theorem Let k be a positive integer Given integers a1, a2, , ak

and pairwise relatively prime positive integers n1, n2, , nk, there exists a unique integer

a such that 0⩽ a <∏k

i=1

ni and a≡ ai (modni) for i = 1,2, ,k

Circumcenter Center of the circumscribed circle or sphere

Circumcircle Circumscribed circle

Complex numbers in planar geometry If we introduce a Cartesian coordinatesystem in the Euclidean plane, we can assign a complex number to each point in the plane

by assigning α + βi to the point (α,β) for all reals α and β Suppose that A,B, ,F arepoints and a, b, , f are the corresponding complex numbers Then:

ˆ a +(c − b) corresponds to the translation of A under the vector ÐBC;→

ˆ given an angle θ, b+eiθ(a−b) corresponds to the image of A under a rotation through

(a − c) equals ∠ABC (directed and modulo 2π).

Using these facts, one can translate much of the language of geometry in the Euclideanplane into language about complex numbers

Congruence For integers a, b, and n with n≠ 1,a ≡ b (modn) (or ”a is congruent to

b modulo n”) means that a − b is divisible by n

Concave up (down) function A function f(x) is concave up (down) on [a,b] ⊆ R

if f(x) lies under (above) the line connecting (a1, f(a1)) and (b1, f(b1)) for all

a⩽ a1 < x < b1 ⩽ b

A function g(x) is concave up (down) on the Euclidean plane if it is concave up (down)

on each line in the plane, where we identify the line naturally with R

Concave up and down functions are also called convex and concave, respectively

Convex hull Given a nonempty set of points S in Euclidean space, there exists

a convex set T such that every convex set containing S also contains T We call T the

1.2 Glossary 5

convex hull of S

Cyclic polygon Polygon that can be inscribed in a circle

De Moivre’s formula For any angle a and for any integer n,

(cos a + isin a)n= cos na + isin na

Derangement A derangement of n items a1, , an is a permutation(b1, b2, , bn) ofthese items such that bi ≠ ai for all i According to a formula of Euler’s, there are exactly

Directed angles A directed angle contains information about both the angle’smeasure and the angle’s orientation (clockwise or counterclockwise) If two directed anglessum to zero, then they have the same angle measure but opposite orientations One oftentakes directed angles modulo π or 2π Some important features of directed angles modulo

p follow:

ˆ If A, B, C, D are points such that ∠ABC and ∠ABD are welldefined,

then ∠ABC = ∠ABD if and only if B,C,D are collinear

ˆ If A, B, C, D are points such that ∠ABC and ∠ADC are welldefined,

then ∠ABC = ∠ADC if and only if A,B,C,D are concyclic

ˆ Because 2(θ) = 2(π2 +θ), but θ ≠ π2 +θ, one cannot divide directed angles by 2 Forexample, if ∠ABC = 2∠ADC,D lies either on the internal angle bisector of angle ABC,

or on the external angle bisector of angle ABC we cannot write ∠ADC = 12∠ABC todetermine which line D lies on

These features show that using directed angles modulo π allows one to deal with tiple possible configurations of a geometry problem at once, but at the expense of possiblylosing important information about a configuration

mul-Euler’s formula (for planar graphs)

If F, V, and E are the number of faces, vertices, and edges of a planar graph, then

F + V − E= 2 This is a special case of an invariant of topological surfaces called the Eulercharacteristic

Euler’s formula (in planar geometry) Let O and I be the circumcenter and ter, respectively, of a triangle with circumradius R and inradius r Then OI2= R2−2rR

incen-Euler line The orthocenter, centroid and circumcenter of any triangle are collinear.The centroid divides the distance from the orthocenter to the circumcenter in the ratio

of 2 : 1 The line on which these three points lie is called the Euler line of the triangle

Euler’s theorem Given relatively prime integers a and m with m⩾ 1

aφ(m) ≡ a(modm), where φ(m) is the number of positive integers less than or equal to

mand relatively prime to m Euler’s theorem is a generalization of Fermat’s little theorem

Excircles or escribed circles Given a triangle ABC, there are four circles gent to the lines AB, BC, CA One is the inscribed circle, which lies in the interior ofthe triangle One lies on the opposite side of line BC from A, and is called the excircle(escribed circle) opposite A, and similarly for the other two sides The excenter opposite

tan-A is the center of the excircle opposite A; it lies on the internal angle bisector of A andthe external angle bisectors of B and C

Excenters See excircles

Exradius The radius of the three excircles of a triangle

Fermat number A number of the form 22n for some positive integer n

Fermat’s little theorem If p is prime, then ap ≡ a(modp) for all integers a

Feuerbach circle The feet of the three altitudes of any triangle, the midpoints ofthe three sides, and the midpoints of segments from the three vertices to the orthocenter,all lie on the same circle, the Feuerbach circle or the nine-point circle of the triangle.Let R be the circumradius of the triangle The nine-point circle of the triangle hasradius R/2 and is centered at the midpoint of the segment joining the orthocenter andthe circumcenter of the triangle

Feuerbach’s theorem The nine-point circle of a triangle is tangent to the incircleand to the three excircles of the triangle

Fibonacci sequence The sequence F0, F1, defined recursively by F0 = 0,F1 = 1,and

1.2 Glossary 7

then we also refer to f as the generating function for the sequence

Graph A graph is a collection of vertices and edges, where the edges are distinctunordered pairs of distinct vertices We say that the two vertices in one of these unorderedpairs are adjacent and connected by that edge

The degree of a vertex is the number of edges which ontain it

A path is a sequence of vertices v1, v2, , vn such that vi is adjacent to vi+1 for each i

A graph is called connected if for any two vertices v and w, there exists a path from

v to w

A cycle of the graph is an ordered collection of vertices v1, v2, , vn such that v1 ≡ vn

and such that the (vi, vi+1) are distinct edges

A connected graph which contains no cycles is called a tree, and every tree contains

at least two leaves, vertices with degree 1

Harmonic conjugates Let A, C, B, D be four points on a line in that order

If the points C and D divide AB internally and externally in the same ratio, (i.e.,

AC ∶ CB = AD ∶ DB), then the points C and D are said to be harmonic conjugates

of each other with respect to the points A and B, and AB is said to be harmonicallydivided by the points C and D If C and D are harmonic with respect to A and B, then

A and B are harmonic with respect to C and D

Harmonic range The four points A, B, C, D are referred to as a harmonic range,denoted by (ABCD), if C and D are harmonic conjugates with respect to A and B

Helly’s theorem If n> d and C1, , Cn are convex subsets of Rd, each d+1 of whichhave nonempty intersection, then there is a point in common to all the sets

Heron’s formula The area of a triangle with sides a, b, c is equal to

√

s(s − a)(s − b)(s − c), where s = a + b + c2 ⋅H¨older’s inequality Let w1, , wn be positive real numbers whose sum is 1 Forany positive real numbers aij,

Homothetic triangles Two triangles ABC and DEF are homothetic if they haveparallel sides Suppose that AB ∥ DE, BC ∥ EF, and CA ∥ F D Then lines AD, BE, and

CF concur at a point X, as given by a special case of Desargues’ theorem Furthermore,

some homothety centered at X maps triangle ABC onto triangle DEF.

Incenter Center of inscribed circle

Incircle Inscribed circle

Inversion of center O and ratio r Given a point O in the plane and a real number

r> 0, the inversion through O with radius r maps every point P ≠ O to the point P′ onthe ray Ð→OP such that OP ⋅ OP′ = r2 We also refer to this map as inversion through ω,the circle with center O and radius r Key properties of inversion are:

1 Lines through O invert to themselves (though the individual points on the line arenot all fixed)

2 Lines not through O invert to circles through O and vice versa

3 Circles not through O invert to other circles not through O

4 A circle other than ω inverts to itself (as a whole, not point-by-point) if and only

if it is orthogonal to ω, that is, it intersects ω and the tangents to the circle and to ω ateither intersection point are perpendicular

Isogonal conjugate Let ABCbe a triangle and let P be a point in the plane whichdoes not lie on any of the lines AB, BC, and CA There exists a unique point Q in theplane such that ∠ABP = ∠QBC,∠BCP = ∠QCA, and ∠CAP = ∠QAB, where theangles in these equations are directed modulo π We call Q the isogonal conjugate of P With this definition, we see that P is also the isogonal conjugate of Q

Jensen’s inequality If f is concave up on an interval [a,b] and λ1, λ2, , λn arenonnegative numbers with sum equal to 1, then

λ1f(x1) + λ2f(x2) + ⋯ + λnf(xn) ⩾ f (λ1x1+λ2x2+ ⋯ +λnxn)for any x1, x2, , xn in the interval [a,b] If the function is concave down, the inequality

is reversed

Kummer’s Theorem Given nonnegative integers a and b and a prime p, pt∣Ca

a+b ifand only if t is less than or equal to the number of carries in the addition a + b in base p

Lattice point In the Cartesian plane, the lattice points are the points (x,y) forwhich x and y are both integers

Law of cosines In a triangle ABC, CA2 = AB2+BC2−2AB ⋅ BC cos ∠ABC, andanalogous equations hold for AB2 and BC2

Law of quadratic reciprocity If p, q are distinct odd primes, then

(p

q) (q

p) = (−1)(p−1)(q−1)

4

1.2 Glossary 9

where (pq) and (qp) are Legendre symbols

Law of sines In a triangle ABC with circumradius equal to R one has

sym-n) is defined to equal 0 if n ∣ m,1 if m is a quadratic residue modulo n, and −1 if

m is a quadratic nonresidue modulo n

Lucas’s theorem Let p be a prime; let a and b be two positive integers such that

a= akpk+ak−1pk−1+ ⋯ +a1p + a0,

b= bkpk+bk−1pk−1+ ⋯ +b1p + b0,where 0⩽ ai, bi < p are integers for i = 0,1, ,k Then

n columns is an m × n matrix The object in the ith row and jth column of matrix A isdenoted ai,j If a matrix has the same number of rows as it has columns, then the matrix

is called a square matrix In a square n × n matrix A, the main diagonal consists of theelements a1,1, a2,2, , an,n

Menelaus’ theorem Given a triangle ABC, let F, G, H be points on lines BC, CA, AB,respectively Then F, G, H are collinear if and only if, using directed lengths,

Multiset Informally, a multiset is a set in which an element may appear more thanonce For instance, {1,2,3,2} and {2,2,2,3,1} are distinct multisets

Nine point circle (See Feuerbach circle.)

Orbit Suppose that S is a collection of functions on a set T , such that S is closedunder composition and each f ∈ S has an inverse T can be partitioned into its orbitsunder S, sets of elements such that a and b are in the same set if and only if f(a) = b forsome f ∈ S

Order Given a nonzero element g of a finite field, there exists a smallest positiveinteger d, named the order of g, such that gd= 1.

Orthocenter of a triangle Point of intersection of the altitudes

Pascal’s theorem If ABCDEF is a hexagon inscribed in a conic in the projectiveplane, such that each pair of opposite sides intersects at most one point, then the threeintersection points formed in this manner are collinear (If the hexagon is inscribed in aconic in the afine plane, then either the above result holds, or else each pair of oppositesides is parallel.)

This theorem the dual to Brianchon’s theorem

Pell’s equations If D is a prime congruent to 3 modulo 4, then the Diophantineequation

inte-is, φ(ab) = φ(a)φ(b) for all a,b relatively prime

Periodic Function f(x) is periodic with period T > 0 if

f(x + T) = f(x) for all x

Permutation Let S be a set A permutation of S is a one-to-one function π ∶ S → Sthat maps S onto S If S = {x1, x2, , xn} is a finite set, then we may denote a permuta-tion π of S by {y1, y2, , yn}, where yk= π(xk)

Pick’s theorem Given a non self-intersecting polygon P in the coordinate planewhose vertices are at lattice points, let B denote the number of lattice points on itsboundary and let I denote the number of lattice points in its interior The area of P isgiven by the formula

1.2 Glossary 11

line p′ that is perpendicular to rayÐ→OP and satisfies

d(O,P)d(O,p′) = R2,where d(A,B) denote the distance between the objects A and B

If q is a line that does not pass through O, then the pole of q is the point Q that haspolar q The pole-polar transformation with respect to the circle C is also called recipro-cation in the circle C

Polynomial in x of degree n Function of the form

to C is defined to be the product of the signed distances P X and P Y

The power of a point theorem states that this quantity is a constant; i.e., it does notdepend on which line was drawn More precisely,

P X ⋅ P Y = PO2−r2

no matter which line is drawn

Power mean inequality Let a1, a2, , an be any positive numbers for which

a1+a2+ ⋯ +an= 1 For positive numbers x1, x2, , xn we define

M−∞⩽ Ms⩽ Mt⩽ M+∞ for all s⩽ t

Primitive element For each prime p, a field F with p elements contains an element

g, called a primitive element of F , with the following property: for any nonzero element

h of F , there exists an integer k such that gk= h

Projective plane Let K be a field The projective plane over K is the set of alence classes of K3∖{(0,0)}, under equivalence by scalar multiplication (that is, where(a,b,c) and (d,e,f) are equivalent if and only if (a,b,c) = (dκ,eκ,fκ) for some κ ∈ K).The elements of K are called points, and the equivalence class containing (a,b,c) isoften denoted [a,b,c] or [a ∶ b ∶ c] Also, given (α,β,γ) ∈ K3∖{(0,0)}, the set of solutions[x,y,z] to

equiv-αx + βy + γz= 0

is called a line in the projective plane over K Any two distinct points (resp lines) aresaid to ”intersect in” or ”lie on” a unique line (resp point).

Ptolemy’s theorem In a convex cyclic quadrilateral ABCD,

Root of an equation Solution to the equation

Root of unity Solution to the equation zn−1= 0

Root Mean Square-Arithmetic Mean Inequality For positive numbers x1, x2, , xn,

Simson line For any point P on the circumcircle of ∆ABC, the feet of the diculars from P to the sides of line called the Simson line of P with respect to ∆ABC alllie on a line called The Simson line of P with respect to ∆ABC

perpen-Solid triangle inequality Given four points A, B, C, P in three-dimensional spacewhich are not coplanar, we have

∠AP B + ∠BP C > ∠APC

Stewart’s theorem In a triangle ABC with cevian AD, write

a= BC,b = CA,c = AB,m = BD,n = DC, and d = AD

Then

d2a + man= c2n + b2m

This formula can be used to express the lengths of the altitudes and angle bisectors

of a triangle in terms of its side lengths

1.2 Glossary 13

Thue-Morse sequence The sequence t0, t1, , defined by t0 = 0 and the recursiverelations

t2k= tk, t2k+1= 1 − t2k for all k⩾ 1

The binary representation of n contains an odd number of 1′s if and only if tn is odd

Triangular number A number of the form n(n + 1)

2 , where n is some positive teger

in-Trigonometric identities

sin2x + cos2x= 1,

1 + cot2x= sin12x = csc2x,tan2x + 1= cos12

x = sec2x;

addition and subtraction formulas:

sin(a ± b) = sin acos b ± cos asin b,cos(a ± b) = cos acos b ∓ sin asin b,tan(a ± b) =1 ∓ tan atan btan a ± tan b ;

double-angle formulas:

sin 2a= 2sin acos a = 1 + tan2tan a2

a,cos 2a= cos2a − sin2a= 2cos2a − 1= 1 − 2sin2a= 1 − tan1 + tan22aatan 2a= 1 − tan2tan a2

a;

triple-angle formulas:

sin 3a= 3sin a − 4sin3a,cos 3a= 4cos3a − 3 cos a,tan 3a= 3tan a − tan1 − 3tan2 3a

a ;

half-angle formulas:

sin2a

2 =1 − cos a2

cos2 a

2 = 1 + cos a2 ;sum-to-product formulas:

sin a + sin b= 2sina + b2 cosa − b

2 ,cos a + cos b= 2cosa + b2 cosa − b

2 ,tan a + tan b= sincos a cos b(a + b);

difference-to-product formulas:

sin a − sin b= 2cosa + b2 sin a − b

2 ,cos a − cos b= −2sina + b2 sina − b

2 ,tan a − tan b=cos a cos bsin(a − b);

product-to-sum formulas:

2 sin a cos b= sin(a + b) + sin(a − b),

2 cos a cos b= cos(a + b) + cos(a − b),

2 sin a sin b= −cos(a + b) + cos(a − b)

Wilson’s theorem If n> 1 be a positive integer, then

(n − 1)! = −1 (modn)

if and only if n is prime

Zeckendorf representation Let F0, F1, be the Fibonacci numbers 1, 2, Eachnonnegative integer n can be written uniquely as a sum of nonconsecutive positive Fi-bonacci numbers; that is, each nonnegative integer n can be written uniquely in the form

n=∑∞

k=0

αkFk

where ak∈ {0,1} and (ak, ak+1) ≠ (1,1) for each k

This expression for n is called its Zeckendorf representation

Example 2.2 Find all nine-digit numbers aaaabbbb that can be written as a sum of fifthpowers of two positive integers.

Solution

15

By Newton’s binomial formula,(5u+v)5 = v5(mod25) for any integers u,v, or any fifthpower n5 leaves remainders −7, −1, 0, 1, 7 modulus 25, when n is respectively congruent

to 3, 4, 0, 1, 2 modulus 5 Since aaaabbbbb= m5+n5 for integers m, n, and m5+n5 leavesremainders 0, ±1, ±2, ±6, ±7, ±8, ±14, then b∈ {0,1,3,4,7,9} for the last two digits of m5+n5

to be equal

Since 15 = 1 (mod11),25 = 32 = −1 (mod11),35 = 243 = 1 (mod11),45 = 1024 =

1 (mod11) and 55 = 3125 = 1 (mod11), then m5 +n5 leaves re- mainders −2, −1, 0, 1, 2modulus 11 Moreover, aaaabbbb0 is clearly a multiple of 11, and since no digit may becongruent to −1 modulus 11, then b ∈ {9,0,1,2} Together with the previous result, wemay conclude that the possible values for b are b = 9 with wlog m = 0 (mod5) and

n= 4 (mod5) and exactly one of m,n is a multiple of 11, or b = 0 with m + n a multiple

of 5, or b= 1 with m,n = 3 (mod5) and exactly one of m,n a multiple of 11

If b= 1, clearlym,n cannot finish in the same digit or otherwise m5+n5 would finish

in 8, hence wlog m ends in 3 and n ends in 8 Since 885 > 999999999, m = 33 must be amultiple of 11, and n= 10u + 8 for some digit u Now, 85 ends in 68 and 333 ends in 93,but (10u + 8)5−85 is a multiple of 100, so m5+n5 formed in this way ends in 61, not in

11 No solution is possible in this case

If b= 9, either m is a multiple of 10 and n5 finishes in 99999, or m is odd and finishes

in 5 and n5 must finish in 4 In the first case, a digit u must exist such that(10u+9)5 mustend in 99999, or since 95 = 59049,50 ⋅ 94u= 328050u, hence 50u must end in 950, yielding

u= −1 (mod20), impossible for a digit In the second case, m = 10u + 5 and n = 10v + 4,and m5+n5 ends in the same two digits as 55+45+5510u + 4450v, which are also the samethree digits as

125 + 24 + 250u + 800v= 800v + 250u + 149

Clearly u must be odd, and 250u ends in 750 or 250, and 800v must finish respectively

in 100 (impossible) or in 600, hence v= 2 or v = 7 But 745 > 999999999, or n = 24 not amultiple of 11, yielding m= 55 Now, 555+245 = 511246999, and no solution exists in thiscase

Finally, if b = 0, then m + n is a multiple of 5 with multiplicity 2 = a = 1, because

if m + n is a multiple of 53 , either m + n = 250 and either m5 or n5 exceeds 1010 , or

m + n = 125, and wlog m5 ends in 0 and n5 ends in 5, absurd Note that 55 divides

m5+n5 = (m + n)5−5mn(m + n)3 +5m2n2(m + n), yielding either a = 1 and m2n2 is amultiple of 54 , or a = 2 and m2n2 is a multiple of 52 In either case, either m or n is amultiple of 5, hence so is the other, and clearly one ends in 0 iff the other one ends in 0,hence either m= 10u +5 and n = 10v +5 or m = 10u and n = 10v In the first case, the lastfour digits of m5+n5 are also the last four digits of 531000(u2+v2) + 5510(u + v) + 2 ⋅ 55,hence the last four digits of 5000(u2+v2−1) + 1250(u + v + 1) Now, u + v + 1 must be amultiple of 4 for the last three digits to be 0, hence u2+v2−1 is even, and 1250(u +v +1)must end in 0000, yielding u + v + 1= 8, since u +v +1 = 16 produces either u or v = 8, and

855> 999999999 Trying all possible alternatives with u + v = 7, we find that u = 6 results

in m5 > 999999999,u = 5 and v = 2 results in m5+n5 = 513050000, and u = 4 and v = 3results in 237050000, so no solution exists in this case Finally, when m= 10u and n = 10v,the problem is equivalent to finding all digits u, v such that u5+v5 = aaaa Combining thenumbers whose fifth powers do not exceed 104, we find that this result is only true when

65+15= 7777

We conclude that the only number of the form aaaabbbbb that is the sum of two fifth

1) if x∈ I0 , it satisfies{x} = {nx},

2) if x∈ Ir, r⩾ 1, only those x in the subinterval Jr⊂ Ir, Jr= [k + n − 1r , k + r + 1

n )

Example 2.4 Find all quadruples (x,y,z,w) of integers satisfying the system of tions

equa-x + y + z + w= xy + yz + zx + w2−w= xyz − w3= −1

First solution

Note that

(x + y)(y + z)(z + x) = (x + y + z)(xy + yz + zx) − xyz = 2

This implies that (x + y,y + z,z + x) equals some permutation of (1,1,2),(−1,−1,2)

or (1,−1,−2) It follows that (x,y,z) equals some permutation of (0,1,1),(1,1,−2) or(0,1,−2)

The first case implies that w = −1 − (x + y + z) = −3 and w3 = 1 + xyz = 1, which is acontradiction The second case implies that

w= −1 − (x + y + z) = −1,w2−w= −1 − (xy + yz + zx) = 2and w3 = xyz + 1 = −1, from where it follows that w = −1 The third case implies that

w= −1−(x+y+z) = −2 and w3 = xyz+1 = 1, which is a contradiction It follows that all ble quadruples (x,y,z,w) are (1,1,−2,−1) with all posible permutations among x,y,z (⊠)

ˆ x= −w − 3; then y + z = −1 − w − x = 2, and

yz= −1 + w − w2+2(w + 3) = −w2+3w + 5,leading to w3−1= xyz = (w + 3)(w2−3w − 5) = w3−14w − 15, or w = −1, for x = −2 and

yz= 1 Note that (y − z)2 = (y + z)2−4yz= 0, or y = z = 1 because y + z = 2

ˆx= −w − 2; then y + z = −1 − w − x = 1, and yz = −1 + w − w2+(w + 2) = 1 + 2w − w2,leading to

w3−1= xyz = (w + 2)(w2−2w − 1) = w3−5w − 2;

no solution in integers exists in this case because no integer w satisfies 5w= −1

ˆ x= −w; then y + z = −1 − w − x = −1, and yz = −1 + w − w2−w= −w2−1, leading to

w3−1= xyz = w3+w, or w= −1, for x = 1 and yz = −2

Note that (y − z)2 = (y + z)2−4yz = 1 + 8 = 32 for y − z = ±3, with solutions y = 1 and

z= −2, or y = −2 and z = 1

ˆ x= 1 − w; then y + z = −1 − w − x = −2, and

yz= −1 + w − w2+2(1 − w) = 1 − w − w2,leading to w3−1= xyz = (w −1)(w2+w − 1) = w3−2w + 1, or w= 1, for x = 0 and yz = −1.Note that(y −z)2 = (y +z)2−4yz= 8 is not a perfect square, hence no solution in integersexist in this case

Restoring generality, all solutions are w= −1, and (x,y,z) a permutation of (1,1,−2).(⊠)

Example 2.5 Find the sequences of integers (an)n⩾0 and (bn)n⩾0 such that

(2 +√5)n= an+bn

1 +√52for each n⩾ 0

where a = (1 + 2 + ⋯ + k)2−1= (k − 1)(k + 2)(k4 2+k + 2) Since x = a7n4, y = a5n3 forany positive integer n , then x3 = ay4 Hence

a perfect squares There are infinitely many such integers since the relation tn+k = u2 isequivalent to the Pell’s equation(2n+2k +1)2−2u2 = 1 The fundamental solution to thisPell equation is (3,2), hence all these integers are given by the sequence (ns), where

It is clear that for s big enough we have ns⩾ 1 and n(n + 1)2 > n + k, hence we get an

Example 2.7 Find all triples (x, y, z) of integers satisfying the system of equations

⎧⎪⎪

⎨⎪⎪

⎩

(x2+1) (y2+1) +10z2 = 2010(x + y)(xy − 1) + 14z = 1985

2.1 Arithmetic problems 21

and the system becomes

{ppq + 140k2+q2+10k2 = 2010= 1985 ⇔{ppq2+q2 = 2010 − 10k= 1985 − 140k2 (iii)Since(p−q)2 = 2010−10k2−2(1985−140k) = −10(k−14)2 then only k= 14 can providesolvability to (iii) And for k = 14, (iii) becomes

{ppq2+q2 = 50= 25 ⇔ p= q = 5

Hence,

{x + yxy = 5= 4 ⇔{xy = 4= 1 or {xy = 1= 4and triples(4,1,140),(1,4,140) are all integer solutions of the original system in integers.(⊠)

Example 2.8 Let n be a positive integer Find the least positive integer a such that thesystem

If n is prime with 10 it is prime with 5 and 2

Since ϕ(52) = 5⋅4 = 20 and ϕ(23) = 22⋅1= 4, where ϕ(n) is Euler’s totient function, then

n20= 1 (mod25) and n20 = (n4)5 = 15 = 1 (mod8), or n20 = 1 (mod200), and uniqueness

of this residue modulus 200 is guaranteed by the Chinese Remainder Theorem, hence thelast three digits of n20 are a01, where a is even The conclusion follows (⊠)Example 2.10 Find all integers n for which 9n + 16 and 16n + 9 are both perfect squares.Solution

If 9n + 16 and 16n + 9 are both perfect squares then n⩾ 0 and the number

pn= (9n + 16)(16n + 9) = (12n)2+(92+162)n + 122

is also a perfect square Since

(12n + 12)2⩽ (12n)2+(92+162)n + 122 < (12n + 15)2

it follows that if n> 0 then we must have pn= (12n +13)2 or pn= (12n +14)2 The formercondition gives n = 1 and the latter, n = 52 Therefore, n = 0,n = 1, and n = 52 are theonly integers n for which the expressions 9n+16 and 16n+9 simultaneously return perfect

Example 2.11 Find all n for which there are n consecutive integers whose sum of squares

is a prime

Solution

If n = 1 any integer square cannot be a prime, so n is at least 2 In cases n = 2 and

n= 3 we can find consecutive numbers 1,2 and 2,3,4 for which sum of squares is 5 and 29,

so prime numbers Suppose now n> 3 If a ∈ Z such that A = a2+(a+1)2+ ⋯ +(a+n−1)2

is prime, then

A= na2+2(1+⋯+(n−1))a+12+ ⋯ +(n−1)2 = n(a2+(n − 1)a + (n − 1)(2n − 1)6 ) = p⋅B.Case 1: n is prime

Because n > 3, we have 2 ∤ n,3 ∤ n, so 6 ∤ n, and because A is prime it follows

6∣ (n − 1)(2n − 1), B = 1 and A = n But A is a sum of n > 3 integer consecutive squares,

so A> n, contradiction

Case 2: n is not prime

If p prime, p∣ n,p > 3, we get

n= pm,A = p(ma2+m(pm − 1)a + m(pm − 1)(2pm − 1)6 ) = pB = p,

because A is prime But A is a sum of n > 3 integer consecutive squares, so A > n > p,contradiction We get n= 2a3b, a= 1,b = 1

If a= 2,n = 4m,A = 2(2ma2 + 2m(4m − 1)a + 3) = 2B = 2, because A is prime But forthe same above reason, A> n > 2, contradiction

If b= 2 we have the same approach If n = 2 ⋅ 3 = 6, we can find integers −1,0,1,2,3,4for which the sum of squares is 31, prime number Finally the answer is n∈ {2,3,6} (⊠)Example 2.12 Find all positive integers n for which(n−2)!+(n+2)! is a perfect square.Solution

Note the identity

(n − 2)! + (n + 2)! = (n − 2)! ⋅ (n2+n − 1)2

Thus, if (n − 2)! + (n + 2)! is a perfect square, then (n − 2)! must also be a perfectsquare For n = 2,3 we have that (n − 2)! = 1, so n = 2 and n = 3 are solutions In thecase of n⩾ 4, we note that the largest prime number smaller than or equal to n−2, which

we will denote by p, can only appear once in the prime factorization of (n − 2)!, implyingthat (n − 2)! is never a perfect square for n ⩾ 4 If it were to appear at least twice in thefactorization, then it is necessary to have n − 2 = 2p But, by Bertrand’s postulate, weknow that there is a prime number in between p and 2p, contradicting p’s maximality

2.1 Arithmetic problems 23

Example 2.13 Find all primes p and q such that both pq −555p and pq +555q are perfectsquares

First solution

Since pq − 555p= p(q −555) is a perfect square, p divides q −555 and q > 555 Therefore

Likewise q divides p + 555, so there exists an integer b⩾ 1 such that p + 555 = bq (β)From (α) and (β) it follows that

If p= 3 ⇒ 3∣(q − 555) ⇒ 3∣q ⇒ q = 3 contradiction since q > 555 Analogously for p = 5and p= 37 Thus we arrive to p∣(k − 1), hence k = ph + 1

But q − 555= pr and hence

p + 555= q(ph + 1) ⇒ p = pqh + pr ⇒ 1 = qh + r ⇒ h = 0,r = 1 ⇒ k = 1

Hence q = p + 555, if p ⩾ 3 we obtain that p + 555 is even, contradiction, so p = 2 and

Example 2.14 Find all pairs (x,y) for which x! + y! + 3 is a perfect cube

Solution

Let x!+y!+3= z3 with x, y⩾ 7 We have that z3 = 3 mod 7, so this gives a contradictionsince the cubic residues modulo 7 are only 0, 1, 6 Thus, we can suppose without loss ofgenerality that x⩽ 6

Now let us do the case work:

But f(y!) = f(243k(3k2+3k + 1)) = 5 + f(k) = 5 + log3k, so we can see that for k⩾ 1,

[243k(3k23+3k + 1)] = 81k(3k2+k + 1) > 5 + log3kand hence the equation has no solution Finally, we get that the solutions to originalequation are (x,y) = (2,5);(x,y) = (5,2);(x,y) = (3,6);(x,y) = (6,3) (⊠)Example 2.15 Prove that there are infinitely many pairs(p,q) of primes such that p6+q4

has two positive divisors whose difference is 4pq

Solution

Let p= 2 and let q be odd prime number Then

p6+q4= 64 + q4= (8 + 4q + q2)(8 − 4q + q2)and we have

8 + 4q + q2−8 − 4q + q2= 8q = 4pq (⊠)Example 2.16 Prove that there is no n for which

g(k + 1) = ((k + 1)2−(k + 1) + 1)((k + 1)2+(k + 1) + 1) = (k2+k + 1)(k2+3k + 3)

So multiplying g(1),g(2), ,g(n), we see that there are a sequence of squares in theproduct(32, 72, 132, ), but with a final factor of n2+n + 1 left over at the end It followsthat since n2+n + 1cannot be a square (since it lies between n2 and (n+1)2), the product

∏

k=1(k4+k2+1) = (n2+n + 1)n−1∏

k=1(k2+k + 1)2.Moreover, because the difference between (n + 1)2 and n2 is 2n + 1, n2+n + 1 cannot

be a perfect square for any n> 1 Thus, there is no n for which∏n

Clearly, for each non-negative integer n, these numbers form different penta- sequences

Example 2.18 Find all positive integers n for which √√

n +√

n + 2009 is an integer.First solution

Because 2009= 72⋅41, it follows that m= 7 and n = 16

Thus n= 16 is the only integer n for which √√

n +√

n + 2009 is an integer (⊠)

Second solution

Suppose there is a positive integer k such that √

n +√

n + 2009= k2.Taking the conjugate we obtain √

n + 2009 −√

n= 2009k2 ⋅Subtracting the above equations we get 2√

n= k2−2009

k2 ⋅From this equation we infer that √

n must be rational and hence integer, since n isinteger Thus, k2 is a divisor of 2009 However, the only square numbers that divide 2009are 12 and 72 It is clear that k = 1 If k = 7 we get n = 16 and it is easily verifed that

b2−a2 = 2009 = 72 ⋅ 41 giving the systems

{b − ab + a = 41= 49 or {b − ab + a = 2009= 1.

First system gives the solution n= 16 Second system implies a = −1004, contradiction

Example 2.19 How many positive integers n less than 2012 are divisible by [√3

n].Solution

Let a= [√3

n] Then a ⩽ √3

n < a + 1, i.e a3 ⩽ n < (a + 1)3 Because [√3

n] divides n, itfollows that n= ab for some positive integer b We get a3 = ab < (a + 1)3, hence

a2 ⩽ b < (a + 1a )3 = a2+3a + 3 + 1

a ⩽ a2+3a + 4

All positive integers n with the property that [√3

n] divides n are given by

a3, a3+a, a3+2a, , a3+(3a + 3)a,a ∈ Z∗.Note that the largest a with a3< 2012 is a = 12 and we have 123+12k⩽ 2012 is equivalent

to 12k⩽ 284, i.e k ⩽ 23 +23, hence k= 23 The desired number is given by

11

∑

a=1(3a + 4) + 24 = 3 ⋅11 ⋅ 122 +44 + 24= 198 + 44 + 24 = 266 (⊠)Example 2.20 Find all n for which the number of diagonals of a convex n-gon is aperfect square

Solution

Since the number of diagonals of a convex n-gon is n(n − 3)

2 ,we seek to find positiveinteger solutions to the Diophantine equation

n(n − 3)

2 = m2

2.1 Arithmetic problems 27

that is

n2−3n − 2m2 = 0for n⩾ 3 and m ⩾ 0 Note that n2+m2 = 0 (mod3) which implies that both n and m aredivisible by 3 So, by letting n= 3N and m = 3M , the equation becomes

N2−N − 2M2 = 0that is

nk= 3Nk=3(Xk+1)

2 for k⩾ 0,

Example 2.21 For an even integer n consider a positive integer N having exactly n2

divisors greater than 1 Prove that N is the fourth power of an integer

First solution

The following results will be of use

Lemma 1 If an odd prime p divides x2+1 for some integer x, then p≡ 1 mod 4.Proof: Since p is odd, p is of the form 4y + 1 or 4y + 3; showing that p= 3mod 4 isimpossible will establish the claim Assuming to the contrary that p= 4y + 3, we have

x2 = −1modp;(x2)2 +1= x4 +2= xp−1= (−1)2 +1= −1modpwhich is impossible since (noting that p cannot divide x) xp−1 = 1 (by Fermat’s ”little”

Lemma 2 If 4x2+1= f1f2 fk, then for each factor fi, fi= 1mod4

Proof Since 4x2+1 is odd, it can only have odd divisors Each factor is either 1

or some product of (odd) prime factors of 4x2+1, each of which must be congruent to

1 mod 4 by lemma 1 It follows that every factor is congruent to 1 mod 4 (◻)Considering now the original problem, let n = 2m, then the condition that N has n2

divisors greater than 1 is equivalent to

τ(N) = (e1+1)(e2+1) (ek+1) = 4m2+1where τ is the number of divisors function and the prime factorization of N is

N = pe 1

1 pe2

2 pek

k

By lemma 2 each of the factors ei +1 is of the form 4yi +1; therefore each ei is a

Second solution

Let N be factored (into primes) as N = pe 1

1 ⋯pek

k

It is well known that N has (e1+1) × ⋯ × (ek+1) = n2+1 divisors (including 1) Since n

is even, n2+1≡ 1 (mod4), so ei+1 must be either 1 mod 4 or 3 mod 4 for 1⩽ i ⩽ k Wewill show that n2+1 has no factors of the form 3 mod 4; thus ei+1≡ 1 (mod4) so ei isdivisible by 4 for all i, implying that N is a fourth power

Proof: Assume for a contradiction that there exists some number p such that p= 4k+3and n2 + 1 is divisible by p WLOG let p be prime; otherwise it must have some primefactor also of the form 3 mod 4 (since p is 3 mod 4, its prime factors cannot all be 1 mod 4).Thus n2≡ −1 (mod p) Note that n must then be relatively prime to p; thus by Fermat’sLittle Theorem, np−1≡ 1 (modp) However,

np−1≡ n4k+2≡ (n2)2k+1

≡ (−1)2k + 1 ≡ −1 (mod p)

Thus 1≡ −1 (mod p), but since p ⩾ 3, we have the desired contradiction (⊠)Example 2.22 Let ABCD be a quadrilateral inscribed in a circle and circumscribedabout a circle such that the points of tangency form a quadrilateral A1B1C1D1 Prove that

A1C1⊥ B1D1

Solution

Let the tangency points of AB, BC, CD, and DA intersect the inner circle at point

A1, B1, C1, and D1, respectively, and let P be the intersection point of A1C1 and B1D1.Then, since ABCD is cyclic,

Example 2.23 Solve in prime numbers the equation: xy+yx= z

First solution

We first note that neither x and y equal nor are they both odd, for otherwise z wouldn’t

be prime Without loss of generality, suppose that x = 2 The equation then reduces to

2y+y2= z, or,

z= 2y−2 + y2+2= 2(2y−1−1) + (y + 1)(y + 2) − 3y

First, (y + 1)(y + 2) is divisible by three since all primes greater than 3 are either one

or two less than a multiple of three (if some prime was three less, then it would contradictthe prime property) Second, since y is an odd prime of the form 2k + 1 (for some speciff

c integer k), we rewrite the first term as

2(2k+1)−1−1= 4k − 1

2.1 Arithmetic problems 29

Since 4≡ 1mod3, it is easy to show that 4k≡ 1mod3 Thus, 4k−1 is divisible by three.We’ve shown that z is divisible by three for y > 3 It follows that the only solution inprime numbers is x= 2,y = 3,z = 17 or x = 3,y = 2,z = 17 (⊠)

First of all, we can observe that x, y can’t be both odd primes, or z would be even;

so, one of them must be equal to 2, say x We now have the equation 2y +y2 = z Thefollowing lemmas hold:

Lemma 1 If a is an odd number then 2a≡ 2 (mod 3)

Proof 2≠ 0 (mod 3) ⇒ 2k≠ 0 (mod 3) ⇒ (2k−1) (2k+1) ≡ 0 (mod 3)

Thus

Lemma 2 If p= 3 is a prime then p2= 1 (mod 3)

Proof If p is such a prime then(p,3) = 1 ∶ we can apply Fermat’s Theorem, so

Lemma 3 If p> 3 is a prime then 2p+p2= 0 (mod 3)

It follows from Lemma 3 that y must be equal to 3: the only one solution is (2,3,17),

Example 2.24 Let a and b be integers such that ∣b − a∣ is an odd prime Prove that

P(x) = (x − a)(x − b) − p is irreducible in Z[x] for any prime p

First solution

We want to show that

(x − a)(x − b) − p = x2−(a + b)x + ab − p = 0has no solutions in integers when∣b − a∣ is an odd prime and p is prime The discriminant

of this quadratic is

(a + b)2−4ab + 4p= (a − b)2+4p= q2+4pwhere q= ∣b − a∣, so the problem is equivalent to showing that q2+4p cannot be a square.Suppose to the contrary that for some integer y

Second solution

Assume for a contradiction that P(x) = (x−a)(x−b)−p is indeed reducible over Z[x]for some prime p Then, since P(x)is a quadratic, there must be some integers r and ssuch that P(x) = (x − r)(x − s) Thus, P(r) = 0, or (r − a)(r − b) = p However, since

∣(r − a) − (r − b)∣ = ∣a − b∣ is odd, their product, p, must be even, so p = 2

Let x= r − a and y = r − b, and without loss of generality let x < y Since xy = p > 0,either x, y are both positive or x, y are both negative Without loss of generality let themboth be positive Then since ∣x − y∣ = ∣a − b∣, which is an odd prime, xy ⩾ 1 ⋅ (1 + 3) > 2 = p,

Example 2.25 Let p and q be odd primes Prove that for any odd integer d> 0 there is

an integer r such that the numerator of the rational number

If p > q, positive integers a,b exist such that p = aq + b with b < q If b is even, take

r= 2b , and if b is odd, take r = q + b2 In either case, we have that b − 2r≡ 0 (modq), or

if n≡ r (modq), then p − n ≡ b − r ≡ r (modq)

2.1 Arithmetic problems 31

Since n and p − n cannot be equal because p is odd, all numbers n ∈ {1,2, ,p − 1}such that n= r (modq) may be grouped up in distinct pairs of the form (n,p − n) Foreach one of these pairs, their contribution to the total sum is

1

nd+ 1(p − n)d = nndd(p − n)+(pn)dd⋅But for odd d,

nd+(p − n)d= p(nd−1−nd−2(p − n) + nd−3(p − n)2− +(p − n)d−1) ,

and the numerator of this fraction is divisible by p, but not its denominator, since n and

p − n are both smaller than prime p Adding any number of such fractions, a commonfactor p will always appear in the numerator, but never in the denominator (⊠)Example 2.26 Find all integers that can be represented as a3+b3+c3−3abc for somepositive integers a, b, and c

Solution

Let us say an integer is nice if it can be represented as a3+b3+c3 −3abc for somepositive integers a, b, c Assume without loss of generality that b= a + x and c = a + x + y,for some nonnegative integers x, y Therefore,

a3+b3+c3−3abc= (3a + 2x + y)(x2+xy + y2)

For x = y = 0 it follows that 0 is nice Suppose that x,y are not both zero Since(3a + 2x + y)(x2+xy + y2) > 0 we have that any nonzero nice integer is nonnegative Let

us prove first that 1 and 2 are not nice We have that

(3a + 2x + y)(x2+xy + y2) > 3a(x2+xy + y2) ⩾ 3,from where it follows the claim Let us prove also that any nice integer divisible by 3 must

be divisible by 9 We have that

0≡ (3a + 2x + y)(x2+xy + y2) ≡ (y − x) ⋅ (x − y)2≡ (y − x)3 (mod3),

from where it follows that x≡ y (mod3) Therefore,

3a + 2x + y≡ x2+xy + y2≡ 0 (mod3),which implies the claim Let us prove that 9 is not nice From the previous result we havethat x≡ y (mod3), from where it follows that

(3a + 2x + y)(x2+xy + y2) = (3a + 3) ⋅ 3 > 9

Let us proceed to find which integers are nice Taking x = 0,y = 1 it follows that anypositive integer of the form 3a + 1 is nice Taking x= 1,y = 0 it follows that any positiveinteger of the form 3a + 2 is nice Taking x = y = 1 it follows that any positive integer

of the form 9(a + 1) is nice From these we conclude that all the nice integers are 0, anypositive integer greater than 3 of the form 3a + 1 or 3a + 2, and the integers greater than

Example 2.27 Find all positive integers n such that a divides n for all odd positiveintegers a not exceeding √

n

First solution

If 1 is the largest odd integer not exceeding √

n, the result is trivially true, and√

n< 3,

or n⩾ 8 Assume now that m ⩾ 1 is an integer such that 2m + 1 is the largest odd integernot exceeding√

n.Then, 2m + 3> √n ⩾ 2m+1, or 4m2+12m + 9> n ⩾ 4m2+4m + 1 Since2m + 1 and 2m − 1 are positive odd integers with difference 2, they are coprime, and ifboth divide n, then their product 4m2−1 must also divide n, which is larger than 4m2−1.Therefore, n ⩾ 2(4m2−1), and 4m2+12m + 9> 8m2−2, or 4m2−12m − 11< 0 Now, if

m⩾ 4, then 4m2−12m − 11= (m2−11) + 3(m − 4)m > 0, and necessarily m ⩽ 3 Assumethat m= 3 Then 2m+3 = 9 > √n, and n < 81, but n must be divisible by 3,5 and 7, whichare coprime Therefore, n must be divisible by 105, which is absurd, and m⩾ 2 Assumenow m= 2 Then, 2m + 3 = 7 > √n ⩾ 5 = 2m + 1, and 25 ⩽ n < 49, but n must be divisible

by 3 and 5, which are coprime Therefore, n must be divisible by 15, or n= 30,45 Assumenext that m= 1 Then, 9 ⩽ n < 25 and n must be divisible by 3, or n = 9,12,15,18,21,24.The integers that we are looking for are then 1 through 9, 12, 15, 18, 21, 24, 30 and 45 (⊠)

Second solution

For n⩽ 25 such integers can be computed as 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18, 21, and

24 In order to get some intuition, consider what happens for n ⩾ 25: up to n = 72 = 49,these are the integers divisible by 3 and 5, that is multiples of 15, which are 30 and 45

in this range After n= 49, up to n = 92= 81, these are the integers divisible by 3, 5, and

7, that is multiples of 105, a contradiction Recall Bertrand’s postulate, that is there isalways a prime between m and 2m, where m is any integer with m> 1 Using induction

we can see that every time we jump from n to 4n we get at least one more prime in therange [√n,2√n]

This prime is greater than 4 for n ⩾ 9, so the product of primes that must divide ngrows faster than n In summary, the only positive integers n such that a divides n for allodd positive integers a not exceeding √

n are

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18, 21, 24, 30, 45 (⊠)Example 2.28 Find all primes q1, q2, , q6 such that q2

1 = q2

2+ ⋯ +q2

6.Solution

Every square is 0 or 1 modulo 3 and clearly q1 ≠ 3 Suppose there are 0 ⩽ a ⩽ 5 primesbetween q2, , q6 not equal to 3 Then 1 = 1 ⋅ a + 0(5 − a) (mod3), from which a = 1 or

2 ⩾ (x + y2 )m−ngives inequality

xm+ym⩾ (xn+yn)(x + y2 )m−n (2.3)Applying (2.3) to positive real numbers a and b with a + b= 2 we obtain

am+bm⩾ (an+bn)(a + b2 )m−n= an+bn

Suppose that there is natural m< n such that for any positive real numbers a and bwith a + b= 2 holds inequality

am+bm⩾ an+bn⇔ am+(2 − a)m= an+(2 − a)n.Then we obtain inequality

2m = lim

a→2(am+(2 − a)m) = lim

a→2(an+(2 − a)n) = 2n

which contradict to inequality 2m < 2n⇐ m< n

Second solution

First, let us prove that for any nonnegative positive integer k and for any positiveinteger n, an+k +bn+k ⩾ an +bn, if a + b = 2 and a ⩾ 0,b ⩾ 0 To prove this, we usemathematical induction on k

The case k = 0 is readily checked

Assume that the statemnt is true for k = l and let us prove, that it is true too for

Thus an+l+1+bn+l+1 ⩾ an+l +bn+l ⩾ an+bn or an+l+1+bn+l+1 ⩾ an+bn This completesthe induction and proves that the statemnt is true for all nonnegative integers The lastresult implies that the given relation is only satisfied for all integers m⩾ n (⊠)

Third solution

Let n= 0 or n = 1, we need to find all integers m such that am+bm ⩾ 2 Let f(x) =

xm+(2 − x)m with 0 < x < 2 Since g(x) = xm is convex in this interval we have that

f(x) = g(x) +g(2 −x) ⩾ 2g(1) = 2 by Jensen’s inequality, so all integers m satisfy To notethat g′′(x) = m(m−1)xm−2 and m(m−1) ⩾ 0 If n ⩾ 2 then m ⩾ n Let f(x) = xn+(2−x)n.Since f(x) = f(2 − x) it suffices consider 1 ⩽ x < 2 fixed Now let g(n) = xn+(2 − x)n,

we will prove that g(n) is non decreasing We have g′(n) = xn ln(x) + (2 − x)n ln(2 − x).Now suppose n ⩾ 2 fixed, that is to say h(x) = xn ln(x) + (2 − x)n ln(2 − x) We need

to prove that h(x) ⩾ 0 Since t(x) = xn ln(x) is convex if 1 ⩽ x < 2 we have that h(x) =

t(x) + t(2 − x) ⩾ 2t(1) = 0 by Jensen’s inequality and we are done To note that

t′′(x) = xn−2[n(n − 1) ln (x) + 2n − 1] > 0

If n⩽ −1 then m ⩽ n

The idea is the same, but in this case we consider 0< x ⩽ 1 (⊠)

2.1 Arithmetic problems 35

Example 2.31 Let a, b, c, d be integers such that a + b + c + d= 0

Prove that a5+b5+c5+d5 is divisible by 30

First Solution

Because x5 ≡ x (mod5),x3 ≡ x (mod3),x2 ≡ x (mod2) then x5 ≡ x3 ≡ x (mod3) and

x5 ≡ x (mod2) for any integer x Therefore,

a5+b5+c5+d5 ≡ a + b + c + d ≡ 0 (modk)where k= 2,3,5 Thus, a5+b5+c5+d5 ⋮30

For any natural m> 1 product of m consequtive integers allways divisible by m, then

x2−x= (x − 1)x divisible by 2,x3−x= (x − 1)x(x + 1) divisible by 3 and

In the following assume x − 1 > 0 Then there will always be factors of 2 and 3 among

x − 1, x, and x + 1, and in some cases a factor of 5 as well The only triples x − 1, x, x + 1for which there is not a factor of 5 are those of the form

We will prove a lemma first:

Lemma Let x be an integer, then 30∣(x5−x)

Proof We have x5−x= (x−1)x(x+1)(x2 + 1) Because 6∣(x−1)x(x+1) so 6∣(x5−x)

ˆ If one of p or q is even Since x2 = p2q + 1, x is odd, ie

x2 = 1 (mod8) ⇒ p2q + 1= 1 (mod8) ⇒ p2q= 0 (mod8)Because both p and q are primes, and 2 is the only even prime, p= q = 2

ˆ If both p, q> 2 in that case x > 2 We have

◇ For p= 3, we have solution q = 11

◇ If p≠ 3, then p2+2= 0 (mod3) which cannot be prime

So, all the solutions are: (p,q) ∈ {(2,2),(3,7),(3,11)} (⊠)

2.1.2 Senior problems

Example 2.33 Find the least odd positive integer n such that for each prime p, thenumber M =n24−1+np4+p8 is divisible by at least four primes

First solution

Let n= 2k + 1 with k nonnegative integer For k = 0,1,2,3 it is easy to see that when

p= 2 there are less than four prime divisors

M = p8+np4+n2−1

4 = (p4+n

2)2−14

2.1 Arithmetic problems 37

= (p2+2p + 2, 4p3+8p2+4p + 1)

= (p2+2p + 2, 4p3+8p2+4p + 1 − 4p3−8p2−4p)

= (p2+2p + 2, 1) = 1,and

(p2−2p + 2, p4+5) = (p2−2p + 2, 4p3−8p2+4p + 1) = (p2−2p + 2, 1) = 1

Thus p2+2p + 2, p2−2p + 2 and p4+5 are pairwise coprime As p4+5= 2 (mod 4) forall odd p, then 21 is the greatest power of 2 dividing p4+5 Since both p2 +2p + 2 and

p2−2p + 2 are odd, there is another prime different from 2 and from all the divisors of

p2+2p + 2 and p2−2p + 2 which divides p4+5, and so n= 9 is the least desired number.(⊠)

For p = 2 the result holds Assume that p is an odd prime Note that 2∣(p4+5) Toprove that another prime divides p4+5 it is enough to prove that 4∤ p4+5 This resultsfollows from the fact that 4∣p4+3 In order to prove that two primes divide(p2+2p+2)(p2−2p + 2) it is enough to prove that (p2+2p + 2, p2−2p + 2) = 1 Let (p2+2p + 2, p2−2p + 2) = d.Note that d is odd and that d∣4p This implies that d∣p If d = p then p∣(p2+2p + 2), which

is a contradiction Therefore, d= 1, as we wanted to prove This implies that k = 4 is theleast integer value, from where we conclude that n = 9 is the least odd positive integerthat satisfies the condition

Example 2.34 (a) Prove that for each positive integer n there is a unique positiveinteger an such that

(1 +√5)n= √an+√

an+4n.(b) When n is even, prove that an is divisible by 5 ⋅ 4n−1 and find the quotient.First solution

(a) Let 2α= 1 +√5 With this we have

2nαn= √an+√

(2.4) Squaring both sides leads to

n= xn+yn

√

5= (1 +√5)n.(b) If n is even, then we have an= x2

2.1 Arithmetic problems 39

Example 2.35 If a1, a2, , ak ∈ (0,1), and k,n are integers such that k > n ⩾ 1, provethat the following inequality holds

min{a1(1 − a2)n, a2(1 − a3)n, , ak(1 − a1)n} ⩽ (n + 1)nn n+1 ⋅Solution ( Method ”Reductio ad absurdam”)

Let’s suppose that the inequality doesn’t hold Therefore

a1(1 − a2)n> (n + 1)nn n+1

a2(1 − a3)n> (n + 1)nn n+1

ak(1 − a1)n> (n + 1)nn n+1⋅Multiplying these relations up, we get

a1⋅a2⋅ ⋅ ak⋅(1 − a1)n⋅(1 − a2)n⋅ ⋅(1 − ak)n> [(n + 1)nn n+1]

k

(∗)

But, for a∈ (0,1), we have a(1 − a)n⩽ (n + 1)nn n+1

Let’s prove this inequality

The equality holds for: na= 1 − a ⇔ a = n + 11 ∈ (0,1)

Using the proved inequality for a1, a2, , ak , we get:

a1(1 − a1)n⩽ (n + 1)nn n+1

a2(1 − a2)n⩽ (n + 1)nn n+1

ak(1 − ak)n⩽ (n + 1)nn n+1⋅