10 th International Mathematical Competition for University Students Cluj-Napoca, July 2003 Day 1 1. (a) Let a 1 , a 2 , . . . be a sequence of real numbers such that a 1 = 1 and a n+1 > 3 2 a n for all n. Prove that the sequence a n  3 2  n−1 has a finite limit or tends to infinity. (10 points) (b) Prove that for all α > 1 there exists a sequence a 1 , a 2 , . . . with the same properties such that lim a n  3 2  n−1 = α. (10 points) Solution. (a) Let b n = a n  3 2  n−1 . Then a n+1 > 3 2 a n is equivalent to b n+1 > b n , thus the sequence (b n ) is strictly increasing. Each increasing sequence has a finite limit or tends to infinity. (b) For all α > 1 there exists a sequence 1 = b 1 < b 2 < . . . which converges to α. Choos ing a n =  3 2  n−1 b n , we obtain the required sequence (a n ). 2. Let a 1 , a 2 . . . , a 51 be non-zero elements of a field. We simultaneously replace each element with the sum of the 50 remaining ones. In this way we get a sequence b 1 . . . , b 51 . If this new sequence is a permutation of the original one, what can be the characteristic of the field? (The characteristic of a field is p, if p is the smallest positive integer such that x + x + . . . + x    p = 0 for any element x of the field. If there exists no such p, the characteristic is 0.) (20 points) Solution. Let S = a 1 + a 2 + · · · + a 51 . Then b 1 + b 2 + · · · + b 51 = 50S. Since b 1 , b 2 , · · · , b 51 is a permutation of a 1 , a 2 , · · · , a 51 , we get 50S = S, so 49S = 0. Assume that the characteristic of the field is not equal to 7. Then 49S = 0 implies that S = 0. Therefore b i = −a i for i = 1, 2, · · · , 51. On the other hand, b i = a ϕ(i) , where ϕ ∈ S 51 . Therefore, if the characteristic is not 2, the sequence a 1 , a 2 , · · · , a 51 can be partitioned into pairs {a i , a ϕ(i) } of additive inverses. But this is impossible, since 51 is an odd number. It follows that the characteristic of the field is 7 or 2. The characteristic can be either 2 or 7. For the case of 7, x 1 = . . . = x 51 = 1 is a possible choice. For the case of 2, any elements can be chosen such that S = 0, since then b i = −a i = a i . 3. Let A be an n × n real matrix such that 3A 3 = A 2 + A + I (I is the identity matrix). Show that the sequence A k converges to an idempotent matrix. (A matrix B is called idempotent if B 2 = B.) (20 points) Solution. The minimal polynomial of A is a divisor of 3x 3 − x 2 − x −1. This polynomial has three different roots. This implies that A is diagonalizable: A = C −1 DC where D is a diagonal matrix. The eigenvalues of the matrices A and D are all roots of polynomial 3x 3 − x 2 − x − 1. One of the three roots is 1, the remaining two roots have smaller absolute value than 1. Hence, the diagonal elements of D k , which are the kth powers of the eigenvalues, tend to either 0 or 1 and the limit M = lim D k is idempotent. Then lim A k = C −1 MC is idempotent as well. 4. Determine the set of all pairs (a, b) of positive integers for which the set of positive integers can be decomposed into two sets A and B such that a · A = b · B. (20 points) Solution. Clearly a and b must be different since A and B are disjoint. 1 10 th International Mathematical Competition for University Students Cluj-Napoca, July 2003 Day 2 1. Let A and B be n × n real matrices such that AB + A + B = 0. Prove that AB = BA. Solution. Since (A + I)(B + I) = AB + A + B + I = I (I is the identity matrix), matrices A + I and B + I are inverses of each other. Then (A + I)(B + I) = (B + I)(A + I) and AB + BA. 2. Evaluate the limit lim x→0+  2x x sin m t t n dt (m, n ∈ N). Solution. We use the fact that sin t t is decreasing in the interval (0, π) and lim t→0+0 sin t t = 1. For all x ∈ (0, π 2 ) and t ∈ [x, 2x] we have sin 2x 2 x < sin t t < 1, thus  sin 2x 2x  m  2x x t m t n <  2x x sin m t t n dt <  2x x t m t n dt,  2x x t m t n dt = x m−n+1  2 1 u m−n du. The factor  sin 2x 2x  m tends to 1. If m − n + 1 < 0, the limit of x m−n+1 is infinity; if m − n + 1 > 0 then 0. If m − n + 1 = 0 then x m−n+1  2 1 u m−n du = ln 2. Hence, lim x→0+0 2x  x sin m t t n dt =      0, m ≥ n ln 2, n − m = 1 +∞, n − m > 1. . 3. Let A be a closed subset of R n and let B b e the set of all those points b ∈ R n for which there exists exactly one point a 0 ∈ A such that |a 0 − b| = inf a∈A |a − b|. Prove that B is dense in R n ; that is, the closure of B is R n . Solution. Let b 0 /∈ A (otherwise b 0 ∈ A ⊂ B),  = inf a∈A |a−b 0 |. The intersection of the ball of radius  + 1 with centre b 0 with set A is compact and there exists a 0 ∈ A: |a 0 − b 0 | = . 1 Denote by B r (a) = {x ∈ R n : |x − a| ≤ r} and ∂B r (a) = {x ∈ R n : |x − a| = r} the ball and the sphere of center a and radius r, respectively. If a 0 is not the unique nearest point then for any point a on the open line segment (a 0 , b 0 ) we have B |a−a 0 | (a) ⊂ B  (b 0 ) and ∂B |a−a 0 | (a)  ∂B  (b 0 ) = {a 0 }, therefore (a 0 , b 0 ) ⊂ B and b 0 is an accumulation point of set B. 4. Find all positive integers n for which there exists a family F of three-element subsets of S = {1, 2, . . . , n} satisfying the following two conditions: (i) for any two different eleme nts a, b ∈ S, there exists exactly one A ∈ F containing both a, b; (ii) if a, b, c, x, y, z are elements of S such that if {a, b, x}, {a, c, y}, {b, c, z} ∈ F, then {x, y, z} ∈ F. Solution. The condition (i) of the problem allows us to define a (well-defined) operation ∗ on the set S given by a ∗ b = c if and only if {a, b, c} ∈ F, where a = b. We note that this operation is still not defined completely (we need to define a ∗ a), but nevertheless let us investigate its features. At first, due to (i), for a = b the operation obviously satisfies the following three conditions: (a) a = a ∗ b = b; (b) a ∗ b = b ∗ a; (c) a ∗ (a ∗ b) = b. What does the condition (ii) give? It claims that (e’) x ∗ (a ∗ c) = x ∗ y = z = b ∗ c = (x ∗ a) ∗ c for any three different x, a, c, i.e. that the operation is asso c iative if the arguments are different. Now we can complete the definition of ∗. In order to save associativity for non- different arguments, i.e. to make b = a ∗ (a ∗ b) = (a ∗ a) ∗ b hold, we will add to S an extra element, call it 0, and define (d) a ∗ a = 0 and a ∗ 0 = 0 ∗ a = a. Now it is easy to check that, for any a, b, c ∈ S ∪ {0}, (a),(b),(c) and (d), still hold, and (e) a ∗ b ∗ c := (a ∗ b) ∗ c = a ∗ (b ∗ c). We have thus obtained that (S ∪ {0}, ∗) has the structure of a finite Abelian group, whose elements are all of order two. Since the order of every such group is a power of 2, we conclude that |S ∪ {0}| = n + 1 = 2 m and n = 2 m − 1 for some integer m ≥ 1. Given n = 2 m −1, according to what we have proven till now, we will construct a f amily of three-element subsets of S satisfying (i) and (ii). Let us define the operation ∗ in the following manner: if a = a 0 + 2a 1 + . . . + 2 m−1 a m−1 and b = b 0 + 2b 1 + . . . + 2 m−1 b m−1 , where a i , b i are either 0 or 1, we put a ∗ b = |a 0 − b 0 | + 2|a 1 − b 1 | + . . . + 2 m−1 |a m−1 − b m−1 |. 2 It is simple to check that this ∗ satisfies (a),(b),(c) and (e’). Therefore, if we include in F all possible triples a, b, a ∗ b, the condition (i) follows from (a),(b) and (c), whereas the condition (ii) follows from (e’) The answer is: n = 2 m − 1. 5. (a) Show that for each function f : Q × Q → R there exists a function g : Q → R such that f(x, y) ≤ g(x) + g(y) for all x, y ∈ Q. (b) Find a function f : R × R → R for which there is no function g : R → R such that f(x, y) ≤ g(x) + g(y) for all x, y ∈ R. Solution. a) Let ϕ : Q → N be a bijection. Define g(x) = max{|f(s, t)| : s, t ∈ Q, ϕ(s) ≤ ϕ(x), ϕ(t) ≤ ϕ(x)}. We have f(x, y) ≤ max{g(x), g(y)} ≤ g(x) + g (y). b) We shall show that the function defined by f(x, y) = 1 |x−y| for x = y and f(x, x) = 0 satisfies the problem. If, by contradiction there exists a function g as above, it results, that g(y) ≥ 1 |x−y| − f(x) for x, y ∈ R, x = y; one obtains that for each x ∈ R, lim y→x g(y) = ∞. We show, that there exists no function g having an infinite limit at each point of a bounded and closed interval [a, b]. For each k ∈ N + denote A k = {x ∈ [a, b] : |g(x)| ≤ k}. We have obviously [a, b] = ∪ ∞ k=1 A k . The set [a, b] is uncountable, so at least one of the sets A k is infinite (in fact uncountable). This set A k being infinite, there exists a sequence in A k having distinct terms. This sequence will contain a convergent subsequence (x n ) n∈N convergent to a point x ∈ [a, b]. But lim y→x g(y) = ∞ implies that g(x n ) → ∞, a contradiction because |g(x n )| ≤ k, ∀n ∈ N. Second solution for part (b). Let S be the set of all sequences of real numbers. The cardinality of S is |S| = |R| ℵ 0 = 2 ℵ 2 0 = 2 ℵ 0 = |R|. Thus, there exists a bijection h : R → S. Now define the function f in the following way. For any real x and positive integer n, let f(x, n) be the nth element of sequence h(x). If y is not a positive integer then let f(x, y) = 0. We prove that this function has the required property. Let g be an arbitrary R → R function. We show that there exist real numbers x, y such that f(x, y) > g (x) + g(y). Consider the sequence (n + g(n)) ∞ n=1 . This sequence is an element of S, thus (n + g(n)) ∞ n=1 = h(x) for a certain real x. Then for an arbitrary positive integer n, f(x, n) is the nth element, f(x, n) = n + g(n). Choosing n such that n > g(x), we obtain f(x, n) = n + g(n) > g(x) + g(n). 6. Let (a n ) n∈N be the sequence define d by a 0 = 1, a n+1 = 1 n + 1 n  k=0 a k n − k + 2 . Find the limit lim n→∞ n  k=0 a k 2 k , 3 if it exists. Solution. Consider the generating function f(x) =  ∞ n=0 a n x n . By induction 0 < a n ≤ 1, thus this series is absolutely convergent for |x| < 1, f(0) = 1 and the function is positive in the interval [0, 1). The goal is to compute f( 1 2 ). By the recurrence formula, f  (x) = ∞  n=0 (n + 1)a n+1 x n = ∞  n=0 n  k=0 a k n − k + 2 x n = = ∞  k=0 a k x k ∞  n=k x n−k n − k + 2 = f (x) ∞  m=0 x m m + 2 . Then ln f(x) = ln f(x) − ln f(0) =  x 0 f  f = ∞  m=0 x m+1 (m + 1)(m + 2) = = ∞  m=0  x m+1 (m + 1) − x m+1 (m + 2)  = 1 +  1 − 1 x  ∞  m=0 x m+1 (m + 1) = 1 +  1 − 1 x  ln 1 1 − x , ln f  1 2  = 1 − ln 2, and thus f( 1 2 ) = e 2 . 4 Let {a, b} be a solution and consider the sets A, B such that a · A = b · B. Denoting d = (a, b) the greatest common divisor of a and b, we have a = d·a 1 , b = d·b 1 , (a 1 , b 1 ) = 1 and a 1 ·A = b 1 ·B. Thus {a 1 , b 1 } is a solution and it is enough to determine the solutions {a, b} with (a, b) = 1. If 1 ∈ A then a ∈ a · A = b · B, thus b must be a divisor of a. Similarly, if 1 ∈ B, then a is a divisor of b. Therefore, in all solutions, one of numbers a,b is a divisor of the other one. Now we prove that if n ≥ 2, then (1, n) is a solution. For each positive integer k, let f(k) be the largest non-negative integer for which n f(k) |k. Then let A = {k : f(k) is odd} and B = {k : f (k) is even}. This is a decomposition of all positive integers such that A = n · B. 5. Let g : [0, 1] → R be a continuous function and let f n : [0, 1] → R be a sequence of functions defined by f 0 (x) = g(x) and f n+1 (x) = 1 x  x 0 f n (t)dt (x ∈ (0, 1], n = 0, 1, 2, . . .). Determine lim n→∞ f n (x) for every x ∈ (0, 1]. (20 points) B. We shall prove in two different ways that lim n→∞ f n (x) = g(0) for every x ∈ (0, 1]. (The second one is more lengthy but it tells us how to calculate f n directly from g.) Proof I. First we prove our claim for non-decreasing g. In this case, by induction, one can easily see that 1. each f n is non-decrasing as well, and 2. g(x) = f 0 (x) ≥ f 1 (x) ≥ f 2 (x) ≥ . . . ≥ g(0) (x ∈ (0, 1]). Then (2) implies that there exists h(x) = lim n→∞ f n (x) (x ∈ (0, 1]). Clearly h is non-decreasing and g(0) ≤ h(x) ≤ f n (x) for any x ∈ (0, 1], n = 0, 1, 2, . . Therefore to show that h(x) = g(0) for any x ∈ (0, 1], it is enough to prove that h(1) cannot be greater than g(0). Supp ose that h(1) > g(0). Then there exists a 0 < δ < 1 such that h(1) > g(δ). Using the definition, (2) and (1) we get f n+1 (1) =  1 0 f n (t)dt ≤  δ 0 g(t)dt +  1 δ f n (t)dt ≤ δg(δ) + (1 − δ)f n (1). Hence f n (1) − f n+1 (1) ≥ δ(f n (1) − g(δ)) ≥ δ(h(1) − g(δ)) > 0, so f n (1) → −∞, which is a contradiction. Similarly, we can prove our claim for non-increasing continuous functions as well. Now suppose that g is an arbitrary continuous function on [0, 1]. Let M(x) = sup t∈[0,x] g(t), m(x) = inf t∈[0,x] g(t) (x ∈ [0, 1]) Then on [0, 1] m is non-increasing, M is non-decreasing, both are continuous, m(x) ≤ g(x) ≤ M(x) and M(0) = m(0) = g(0). Define the sequences of functions M n (x) and m n (x) in the same way as f n is defined but starting with M 0 = M and m 0 = m. Then one can easily see by induction that m n (x) ≤ f n (x) ≤ M n (x). By the first part of the proof, lim n m n (x) = m(0) = g(0) = M(0) = lim n M n (x) for any x ∈ (0, 1]. Therefore we must have lim n f n (x) = g(0). 2 Proof II. To make the notation clearer we shall denote the variable of f j by x j . By definition (and Fubini theorem) we get that f n+1 (x n+1 ) = 1 x n+1  x n+1 0 1 x n  x n 0 1 x n−1  x n−1 0 . . .  x 2 0 1 x 1  x 1 0 g(x 0 )dx 0 dx 1 . . . dx n = 1 x n+1  0≤x 0 ≤x 1 ≤ ≤x n ≤x n+1 g(x 0 ) dx 0 dx 1 . . . dx n x 1 . . . x n = 1 x n+1  x n+1 0 g(x 0 )   x 0 ≤x 1 ≤ ≤x n ≤x n+1 dx 1 . . . dx n x 1 . . . x n  dx 0 . Therefore with the notation h n (a, b) =  a≤x 1 ≤ ≤x n ≤b dx 1 . . . dx n x 1 . . . x n and x = x n+1 , t = x 0 we have f n+1 (x) = 1 x  x 0 g(t)h n (t, x)dt. Using that h n (a, b) is the same for any permutation of x 1 , . . . , x n and the fact that the integral is 0 on any hyperplanes (x i = x j ) we get that n! h n (a, b) =  a≤x 1 , ,x n ≤b dx 1 . . . dx n x 1 . . . x n =  b a . . .  b a dx 1 . . . dx n x 1 . . . x n =   b a dx x  n = (log(b/a)) n . Therefore f n+1 (x) = 1 x  x 0 g(t) (log(x/t)) n n! dt. Note that if g is constant then the definition gives f n = g. This implies on one hand that we must have 1 x  x 0 (log(x/t)) n n! dt = 1 and on the other hand that, by replacing g by g − g(0), we can suppose that g(0) = 0. Let x ∈ (0, 1] and ε > 0 be fixed. By continuity there exists a 0 < δ < x and an M such that |g(t)| < ε on [0, δ] and |g(t)| ≤ M on [0, 1] . Since lim n→∞ (log(x/δ)) n n! = 0 there exists an n 0 sucht that log(x/δ)) n /n! < ε whenever n ≥ n 0 . Then, for any n ≥ n 0 , we have |f n+1 (x)| ≤ 1 x  x 0 |g(t)| (log(x/t)) n n! dt ≤ 1 x  δ 0 ε (log(x/t)) n n! dt + 1 x  x δ |g(t)| (log(x/δ)) n n! dt ≤ 1 x  x 0 ε (log(x/t)) n n! dt + 1 x  x δ Mεdt ≤ ε + Mε. Therefore lim n f(x) = 0 = g(0). 3 6. Let f(z) = a n z n + a n−1 z n−1 + . . . + a 1 z + a 0 be a polynomial with real coefficients. Prove that if all roots of f lie in the left half-plane {z ∈ C : Re z < 0} then a k a k+3 < a k+1 a k+2 holds for every k = 0, 1, . . . , n − 3. (20 points) Solution. The polynomial f is a product of linear and quadratic factors, f(z) =  i (k i z + l i ) ·  j (p j z 2 +q j z +r j ), with k i , l i , p j , q j , r j ∈ R. Since all roots are in the left half-plane, for each i, k i and l i are of the same sign, and for each j, p j , q j , r j are of the same sign, too. Hence, multiplying f by −1 if necessary, the roots of f don’t change and f becomes the polynomial with all positive coefficients. For the simplicity, we extend the sequence of coefficients by a n+1 = a n+2 = . . . = 0 and a −1 = a −2 = . . . = 0 and prove the same statement for −1 ≤ k ≤ n − 2 by induction. For n ≤ 2 the statement is obvious: a k+1 and a k+2 are positive and at least one of a k−1 and a k+3 is 0; hence, a k+1 a k+2 > a k a k+3 = 0. Now assume that n ≥ 3 and the statement is true for all smaller values of n . Take a divisor of f(z) which has the form z 2 + pz + q where p and q are positive real numbers. (Such a divisor can be obtained from a conjugate pair of roots or two real roots.) Then we can write f(z) = (z 2 + pz + q)(b n−2 z n−2 + . . . + b 1 z + b 0 ) = (z 2 + pz + q)g(x). (1) The roots polynomial g(z) are in the left half-plane, so we have b k+1 b k+2 < b k b k+3 for all −1 ≤ k ≤ n − 4. Defining b n−1 = b n = . . . = 0 and b −1 = b −2 = . . . = 0 as well, we also have b k+1 b k+2 ≤ b k b k+3 for all integer k. Now we prove a k+1 a k+2 > a k a k+3 . If k = −1 or k = n − 2 then this is obvious since a k+1 a k+2 is positive and a k a k+3 = 0. Thus, assume 0 ≤ k ≤ n − 3. By an easy computation, a k+1 a k+2 − a k a k+3 = = (qb k+1 + pb k + b k−1 )(qb k+2 + pb k+1 + b k ) − (qb k + pb k−1 + b k−2 )(qb k+3 + pb k+2 + b k+1 ) = = (b k−1 b k − b k−2 b k+1 ) + p(b 2 k − b k−2 b k+2 ) + q(b k−1 b k+2 − b k−2 b k+3 )+ +p 2 (b k b k+1 − b k−1 b k+2 ) + q 2 (b k+1 b k+2 − b k b k+3 ) + pq(b 2 k+1 − b k−1 b k+3 ). We prove that all the six terms are non-negative and at leas t one is positive. Term p 2 (b k b k+1 − b k−1 b k+2 ) is positive since 0 ≤ k ≤ n−3. Also terms b k−1 b k −b k−2 b k+1 and q 2 (b k+1 b k+2 −b k b k+3 ) are non-negative by the induction hypothesis. To check the sign of p(b 2 k − b k−2 b k+2 ) consider b k−1 (b 2 k − b k−2 b k+2 ) = b k−2 (b k b k+1 − b k−1 b k+2 ) + b k (b k−1 b k − b k−2 b k+1 ) ≥ 0. If b k−1 > 0 we can divide by it to obtain b 2 k −b k−2 b k+2 ≥ 0. Otherwise, if b k−1 = 0, either b k−2 = 0 or b k+2 = 0 and thus b 2 k − b k−2 b k+2 = b 2 k ≥ 0. Therefore, p(b 2 k − b k−2 b k+2 ) ≥ 0 for all k . Similarly, pq(b 2 k+1 − b k−1 b k+3 ) ≥ 0. The sign of q(b k−1 b k+2 − b k−2 b k+3 ) can be checked in a similar way. Consider b k+1 (b k−1 b k+2 − b k−2 b k+3 ) = b k−1 (b k+1 b k+2 − b k b k+3 ) + b k+3 (b k−1 b k − b k−2 b k+1 ) ≥ 0. If b k+1 > 0, we can divide by it. Otherwise either b k−2 = 0 or b k+3 = 0. In all cases, we obtain b k−1 b k+2 − b k−2 b k+3 ≥ 0. Now the signs of all terms are checked and the proof is complete. 4 . each element with the sum of the 50 remaining ones. In this way we get a sequence b 1 . . . , b 51 . If this new sequence is a permutation of the original. minimal polynomial of A is a divisor of 3x 3 − x 2 − x −1. This polynomial has three different roots. This implies that A is diagonalizable: A = C −1 DC where