Problem 6. For a permutation σ = (i 1 , i 2 , ., i n ) of (1, 2, ., n) define D(σ) = n  k=1 |i k − k|. Let Q(n, d) be the number of permutations σ of (1, 2, ., n) with d = D(σ). Prove that Q(n, d) is even for d ≥ 2n. Solution. Consider the n × n determinant ∆(x) =          1 x . . . x n−1 x 1 . . . x n−2 . . . . . . . . . . . . x n−1 x n−2 . . . 1          where the ij-th entry is x |i−j| . From the definition of the determinant we get ∆(x) =  (i 1 , .,i n )∈S n (−1) inv(i 1 , .,i n ) x D(i 1 , .,i n ) where S n is the set of all permutations of (1, 2, ., n) and inv(i 1 , ., i n ) denotes the number of inversions in the sequence (i 1 , ., i n ). So Q(n, d) has the same parity as the coefficient of x d in ∆(x). It remains to evaluate ∆(x). In order to eliminate the entries below the diagonal, subtract the (n−1)-th row, multiplied by x, from the n-th row. Then subtract the (n − 2)-th row, multiplied by x, from the (n − 1)-th and so on. Finally, subtract the first row, multiplied by x, from the second row. ∆(x) =            1 x . . . x n−2 x n−1 x 1 . . . x n−3 x n−2 . . . . . . . . . . . . . . . x n−2 x n−3 . . . 1 x x n−1 x n−2 . . . x 1            = . . . =            1 x . . . x n−2 x n−1 0 1 − x 2 . . . x n−3 − x n−1 x n−2 − x n . . . . . . . . . . . . . . . 0 0 . . . 1 − x 2 x − x 3 0 0 . . . 0 1 − x 2            = (1 − x 2 ) n−1 . For d ≥ 2n, the coefficient of x d is 0 so Q(n, d) is even. 4 IMC2008, Blagoevgrad, Bulgaria Day 1, July 27, 2008 Problem 1. Find all continuous functions f : R → R such that f (x) − f (y) is rational for all reals x and y such that x − y is rational. Solution. We prove that f (x) = ax + b where a ∈ Q and b ∈ R. These functions obviously satify the conditions. Suppose that a function f(x) fulfills the required properties. For an arbitrary rational q, consider the function g q (x) = f(x+q)−f(x). This is a continuous function which attains only rational values, therefore g q is constant. Set a = f(1) − f (0) and b = f(0). Let n be an arbitrary positive integer and let r = f(1/n) − f(0). Since f(x + 1/n) − f(x) = f(1/n) − f (0) = r for all x, we have f(k/n) − f(0) = (f(1/n) − f(0)) + (f(2/n) − f(1/n)) + . . . + (f(k/n) − f((k − 1)/n) = kr and f(−k/n) − f(0) = −(f(0) − f (−1/n)) − (f (−1/n) − f (−2/n)) − . . . − (f(−(k − 1)/n) − f(−k/n) = −kr for k ≥ 1. In the case k = n we get a = f(1) − f(0) = nr, so r = a/n. Hence, f (k/n) − f (0) = kr = ak/n and then f(k/n) = a · k/n + b for all integers k and n > 0. So, we have f(x) = ax+ b for all rational x. Since the function f is continous and the rational numbers form a dense subset of R, the same holds for all real x. Problem 2. Denote by V the real vector space of all real polynomials in one variable, and let P : V → R be a linear map. Suppose that for all f, g ∈ V with P (fg) = 0 we have P (f ) = 0 or P (g) = 0. Prove that there exist real numbers x 0 , c such that P (f ) = c f(x 0 ) for all f ∈ V . Solution. We can assume that P = 0. Let f ∈ V be such that P (f) = 0. Then P (f 2 ) = 0, and therefore P (f 2 ) = aP (f ) for some non-zero real a. Then 0 = P (f 2 − af) = P (f (f − a)) implies P (f − a) = 0, so we get P (a) = 0. By rescaling, we can assume that P (1) = 1. Now P (X + b) = 0 for b = −P (X). Replacing P by ˆ P given as ˆ P (f(X)) = P (f (X + b)) we can assume that P (X) = 0. Now we are going to prove that P (X k ) = 0 for all k ≥ 1. Suppose this is true for all k < n. We know that P (X n + e) = 0 for e = −P (X n ). From the induction hypothesis we get P  (X + e)(X + 1) n−1  = P (X n + e) = 0, and therefore P (X + e) = 0 (since P (X + 1) = 1 = 0). Hence e = 0 and P (X n ) = 0, which completes the inductive step. From P (1) = 1 and P (X k ) = 0 for k ≥ 1 we immediately get P (f ) = f(0) for all f ∈ V . 1 is a Cauchy sequence in H. (This is the crucial observation.) Indeed, for m > n, the norm y m − y n  may be computed by the above remark as y m − y n  2 = d 2 2       1 m − 1 n , . . . , 1 m − 1 n , 1 m , . . . , 1 m  ⊤      2 R m = d 2 2  n(m − n) 2 m 2 n 2 + m − n m 2  = d 2 2 (m − n)(m − n + n) m 2 n = d 2 2 m − n mn = d 2 2  1 n − 1 m  → 0, m, n → ∞. By completeness of H, it follows that there exists a limit y = lim n→∞ y n ∈ H. We claim that y sastisfies all conditions of the problem. For m > n > p, with n, p fixed, we compute x n − y m  2 = d 2 2       − 1 m , . . . ,− 1 m , 1 − 1 m ,− 1 m , . . . ,− 1 m  ⊤      2 R m = d 2 2  m − 1 m 2 + (m − 1) 2 m 2  = d 2 2 m − 1 m → d 2 2 , m → ∞, showing that x n − y = d/ √ 2, as well as x n − y m , x p − y m  = d 2 2   − 1 m , . . . ,− 1 m , . . . , 1 − 1 m , . . . ,− 1 m  ⊤ ,  − 1 m , . . . , 1 − 1 m , . . . ,− 1 m , . . . ,− 1 m  ⊤  R m = d 2 2  m − 2 m 2 − 2 m  1 − 1 m  = − d 2 2m → 0, m → ∞, showing that x n − y, x p − y = 0, so that  √ 2 d (x n − y) : n ∈ N  is indeed an orthonormal system of vectors. This completes the proof in the case when T = S, which we can always take if S is countable. If it is not, let x ′ , x ′′ be any two distinct points in S \ T . Then applying the above procedure to the set T ′ = {x ′ , x ′′ , x 1 , x 2 , . . . , x n , . . .} it follows that lim n→∞ x ′ + x ′′ + x 1 + x 2 + ··· + x n n + 2 = lim n→∞ x 1 + x 2 + ··· + x n n = y satisfies that  √ 2 d (x ′ − y), √ 2 d (x ′′ − y)  ∪  √ 2 d (x n − y) : n ∈ N  is still an orthonormal system. This it true for any distinct x ′ , x ′′ ∈ S \ T ; it follows that the entire system  √ 2 d (x − y) : x ∈ S  is an orthonormal system of vectors in H, as required. 4 IMC2008, Blagoevgrad, Bulgaria Day 2, July 28, 2008 Problem 1. Let n, k be positive integers and suppose that the polynomial x 2k − x k + 1 divides x 2n + x n + 1. Prove that x 2k + x k + 1 divides x 2n + x n + 1. Solution. Let f(x) = x 2n + x n + 1, g(x) = x 2k − x k + 1, h(x) = x 2k + x k + 1. The complex number x 1 = cos( π 3k ) + i sin( π 3k ) is a root of g(x). Let α = πn 3k . Since g(x) divides f(x), f(x 1 ) = g(x 1 ) = 0. So, 0 = x 2n 1 + x n 1 + 1 = (cos(2α) + i sin(2α)) + (cos α + i sin α) + 1 = 0, and (2 cos α + 1)(cos α + i sin α) = 0. Hence 2 cos α + 1 = 0, i.e. α = ± 2π 3 + 2πc, where c ∈ Z. Let x 2 be a root of the polynomial h(x). Since h(x) = x 3k −1 x k −1 , the roots of the polynomial h(x) are distinct and they are x 2 = cos 2πs 3k + i sin 2πs 3k , where s = 3a ± 1, a ∈ Z. It is enough to prove that f(x 2 ) = 0. We have f (x 2 ) = x 2n 2 + x n 2 + 1 = (cos(4sα) + sin(4sα)) + (cos(2sα) + sin(2sα)) + 1 = (2 cos(2sα) + 1)(cos(2sα) + i sin(2sα)) = 0 (since 2 cos(2sα) + 1 = 2 cos(2s(± 2π 3 + 2πc)) + 1 = 2 cos( 4πs 3 ) + 1 = 2 cos( 4π 3 (3a ± 1)) + 1 = 0). Problem 2. Two different ellipses are given. One focus of the first ellipse coincides with one focus of the second ellipse. Prove that the ellipses have at most two points in common. Solution. It is well known that an ellipse might be defined by a focus (a point) and a directrix (a straight line), as a locus of points such that the distance to the focus divided by the distance to directrix is equal to a given number e < 1. So, if a point X belongs to both ellipses with the same focus F and directrices l 1 , l 2 , then e 1 · l 1 X = F X = e 2 · l 2 X (here we denote by l 1 X, l 2 X distances between the corresponding line and the point X). The equation e 1 · l 1 X = e 2 · l 2 X defines two lines, whose equations are linear combinations with coefficients e 1 ,±e 2 of the normalized equations of lines l 1 , l 2 but of those two only one is relevant, since X and F should lie on the same side of each directrix. So, we have that all possible points lie on one line. The intersection of a line and an ellipse consists of at most two points. Problem 3. Let n be a positive integer. Prove that 2 n−1 divides  0≤k<n/2  n 2k + 1  5 k . Solution. As is known, the Fibonacci numbers F n can be expressed as F n = 1 √ 5  1+ √ 5 2  n −  1− √ 5 2  n  . Expanding this expression, we obtain that F n = 1 2 n−1   n 1  +  n 3  5 + . +  n l  5 l−1 2  , where l is the greatest odd number such that l ≤ n and s = l−1 2 ≤ n 2 . So, F n = 1 2 n−1 s  k=0  n 2k+1  5 k , which implies that 2 n−1 divides  0≤k<n/2  n 2k+1  5 k . Problem 4. Let Z[x] be the ring of polynomials with integer coefficients, and let f(x), g(x) ∈ Z[x] be nonconstant polynomials such that g(x) divides f (x) in Z[x]. Prove that if the polynomial f (x)−2008 has at least 81 distinct integer roots, then the degree of g(x) is greater than 5. Solution. Let f(x) = g(x)h(x) where h(x) is a polynomial with integer coefficients. Let a 1 , . . . , a 81 be distinct integer roots of the polynomial f (x)−2008. Then f(a i ) = g(a i )h(a i ) = 2008 for i = 1, . . . , 81, Hence, g(a 1 ), . . . , g(a 81 ) are integer divisors of 2008. Since 2008 = 2 3 ·251 (2, 251 are primes) then 2008 has exactly 16 distinct integer divisors (including the negative divisors as well). By the pigeonhole principle, there are at least 6 equal numbers among g(a 1 ), . . . , g(a 81 ) (because 81 > 16 · 5). For example, g(a 1 ) = g(a 2 ) = . . . = g(a 6 ) = c. So g(x) − c is 1 a nonconstant polynomial which has at least 6 distinct roots (namely a 1 , . . . , a 6 ). Then the degree of the polynomial g(x) − c is at least 6. Problem 5. Let n be a positive integer, and consider the matrix A = (a ij ) 1≤i,j≤n , where a ij =  1 if i + j is a prime number, 0 otherwise. Prove that | det A| = k 2 for some integer k. Solution. Call a square matrix of type (B), if it is of the form          0 b 12 0 . . . b 1,2k−2 0 b 21 0 b 23 . . . 0 b 2,2k−1 0 b 32 0 . . . b 3,2k−2 0 . . . . . . . . . . . . . . . . . . b 2k−2,1 0 b 2k−2,3 . . . 0 b 2k−2,2k−1 0 b 2k−1,2 0 . . . b 2k−1,2k−2 0          . Note that every matrix of this form has determinant zero, because it has k columns spanning a vector space of dimension at most k − 1. Call a square matrix of type (C), if it is of the form C ′ =            0 c 11 0 c 12 . . . 0 c 1,k c 11 0 c 12 0 . . . c 1,k 0 0 c 21 0 c 22 . . . 0 c 2,k c 21 0 c 22 0 . . . c 2,k 0 . . . . . . . . . . . . . . . . . . . . . 0 c k,1 0 c k,2 . . . 0 c k,k c k,1 0 c k,2 0 . . . c k,k 0            By permutations of rows and columns, we see that | det C ′ | =     det  C 0 0 C      = | det C| 2 , where C denotes the k × k-matrix with coefficients c i,j . Therefore, the determinant of any matrix of type (C) is a perfect square (up to a sign). Now let X ′ be the matrix obtained from A by replacing the first row by  1 0 0 . . . 0  , and let Y be the matrix obtained from A by replacing the entry a 11 by 0. By multi-linearity of the determinant, det(A) = det(X ′ ) + det(Y ). Note that X ′ can be written as X ′ =  1 0 v X  for some (n − 1) × (n − 1)-matrix X and some column vector v. Then det(A) = det(X) + det(Y ). Now consider two cases. If n is odd, then X is of type (C), and Y is of type (B). Therefore, | det(A)| = | det(X)| is a perfect square. If n is even, then X is of type (B), and Y is of type (C); hence | det(A)| = | det(Y )| is a perfect square. The set of primes can be replaced by any subset of {2} ∪ {3, 5, 7, 9, 11, . . .}. 2 Problem 6. Let H be an infinite-dimensional real Hilbert space, let d > 0, and suppose that S is a set of points (not necessarily countable) in H such that the distance between any two distinct points in S is equal to d. Show that there is a point y ∈ H such that  √ 2 d (x − y) : x ∈ S  is an orthonormal system of vectors in H. Solution. It is clear that, if B is an orthonormal system in a Hilbert space H, then {(d/ √ 2)e : e ∈ B} is a set of points in H, any two of which are at distance d apart. We need to show that every set S of equidistant points is a translate of such a set. We begin by noting that, if x 1 , x 2 , x 3 , x 4 ∈ S are four distinct points, then x 2 − x 1 , x 2 − x 1  = d 2 , x 2 − x 1 , x 3 − x 1  = 1 2  x 2 − x 1  2 + x 3 − x 1  2 − x 2 − x 3  2  = 1 2 d 2 , x 2 − x 1 , x 4 − x 3  = x 2 − x 1 , x 4 − x 1  − x 2 − x 1 , x 3 − x 1  = 1 2 d 2 − 1 2 d 2 = 0. This shows that scalar products among vectors which are finite linear combinations of the form λ 1 x 1 + λ 2 x 2 + ··· + λ n x n , where x 1 , x 2 , . . . , x n are distinct points in S and λ 1 , λ 2 , . . . , λ n are integers with λ 1 +λ 2 +···+ λ n = 0, are universal across all such sets S in all Hilbert spaces H; in particular, we may conveniently evaluate them using examples of our choosing, such as the canonical example above in R n . In fact this property trivially follows also when coefficients λ i are rational, and hence by continuity any real numbers with sum 0. If S = {x 1 , x 2 , . . . , x n } is a finite set, we form x = 1 n (x 1 + x 2 + ··· + x n ) , pick a non-zero vector z ∈ [Span(x 1 − x, x 2 − x, . . . , x n − x)] ⊥ and seek y in the form y = x + λz for a suitable λ ∈ R. We find that x 1 − y, x 2 − y = x 1 − x − λz, x 2 − x − λz = x 1 − x, x 2 − x + λ 2 z 2 . x 1 − x, x 2 − x may be computed by our remark above as x 1 − x, x 2 − x = d 2 2   1 n − 1, 1 n , 1 n , . . . , 1 n  ⊤ ,  1 n , 1 n − 1, 1 n , . . . , 1 n  ⊤  R n = d 2 2  2 n  1 n − 1  + n − 2 n 2  = − d 2 2n . So the choice λ = d √ 2nz will make all vectors √ 2 d (x i − y) orthogonal to each other; it is easily checked as above that they will also be of length one. Let now S be an infinite set. Pick an infinite sequence T = {x 1 , x 2 , . . . , x n , . . .} of distinct points in S. We claim that the sequence y n = 1 n (x 1 + x 2 + ··· + x n ) 3 Problem 3. Let p be a polynomial with integer coefficients and let a 1 < a 2 < . . . < a k be integers. a) Prove that there exists a ∈ Z such that p(a i ) divides p(a) for all i = 1, 2, . . . , k. b) Does there exist an a ∈ Z such that the product p(a 1 ) · p(a 2 ) · . . . · p(a k ) divides p(a)? Solution. The theorem is obvious if p(a i ) = 0 for some i, so assume that all p(a i ) are nonzero and pairwise different. There exist numbers s, t such that s|p(a 1 ), t|p(a 2 ), st = lcm(p(a 1 ), p(a 2 )) and gcd(s, t) = 1. As s, t are relatively prime numbers, there exist m, n ∈ Z such that a 1 + sn = a 2 + tm =: b 2 . Obviously s|p(a 1 + sn) − p(a 1 ) and t|p(a 2 + tm) − p(a 2 ), so st|p(b 2 ). Similarly one obtains b 3 such that p(a 3 )|p(b 3 ) and p(b 2 )|p(b 3 ) thus also p(a 1 )|p(b 3 ) and p(a 2 )|p(b 3 ). Reasoning inductively we obtain the existence of a = b k as required. The polynomial p(x) = 2x 2 + 2 shows that the second part of the problem is not true, as p(0) = 2, p(1) = 4 but no value of p(a) is divisible by 8 for integer a. Remark. One can assume that the p(a i ) are nonzero and ask for a such that p(a) is a nonzero mul- tiple of all p(a i ). In the solution above, it can happen that p(a) = 0. But every number p(a + np(a 1 )p(a 2 ) . . . p(a k )) is also divisible by every p(a i ), since the polynomial is nonzero, there exists n such that p(a + np(a 1 )p(a 2 ) . . . p(a k )) satisfies the modified thesis. Problem 4. We say a triple (a 1 , a 2 , a 3 ) of nonnegative reals is better than another triple (b 1 , b 2 , b 3 ) if two out of the three following inequalities a 1 > b 1 , a 2 > b 2 , a 3 > b 3 are satisfied. We call a triple (x, y, z) special if x, y, z are nonnegative and x + y + z = 1. Find all natural numbers n for which there is a set S of n special triples such that for any given special triple we can find at least one better triple in S. Solution. The answer is n  4. Consider the following set of special triples:  0, 8 15 , 7 15  ,  2 5 , 0, 3 5  ,  3 5 , 2 5 , 0  ,  2 15 , 11 15 , 2 15  . We will prove that any special triple (x, y, z) is worse than one of these (triple a is worse than triple b if triple b is better than triple a). We suppose that some special triple (x, y, z) is actually not worse than the first three of the triples from the given set, derive some conditions on x, y, z and prove that, under these conditions, (x, y, z) is worse than the fourth triple from the set. Triple (x, y, z) is not worse than  0, 8 15 , 7 15  means that y  8 15 or z  7 15 . Triple (x, y, z) is not worse than  2 5 , 0, 3 5  — x  2 5 or z  3 5 . Triple (x, y, z) is not worse than  3 5 , 2 5 , 0  — x  3 5 or y  2 5 . Since x + y + z = 1, then it is impossible that all inequalities x  2 5 , y  2 5 and z  7 15 are true. Suppose that x < 2 5 , then y  2 5 and z  3 5 . Using x + y + z = 1 and x  0 we get x = 0, y = 2 5 , z = 3 5 . We obtain the triple  0, 2 5 , 3 5  which is worse than  2 15 , 11 15 , 2 15  . Suppose that y < 2 5 , then x  3 5 and z  7 15 and this is a contradiction to the admissibility of (x, y, z). Suppose that z < 7 15 , then x  2 5 and y  8 15 . We get (by admissibility, again) that z  1 15 and y  3 5 . The last inequalities imply that  2 15 , 11 15 , 2 15  is better than (x, y, z). We will prove that for any given set of three special triples one can find a special triple which is not worse than any triple from the set. Suppose we have a set S of three special triples (x 1 , y 1 , z 1 ), (x 2 , y 2 , z 2 ), (x 3 , y 3 , z 3 ). Denote a(S) = min(x 1 , x 2 , x 3 ), b(S) = min(y 1 , y 2 , y 3 ), c(S) = min(z 1 , z 2 , z 3 ). It is easy to check that S 1 :  x 1 − a 1 − a − b − c , y 1 − b 1 − a − b − c , z 1 − c 1 − a − b − c   x 2 − a 1 − a − b − c , y 2 − b 1 − a − b − c , z 2 − c 1 − a − b − c   x 3 − a 1 − a − b − c , y 3 − b 1 − a − b − c , z 3 − c 1 − a − b − c  2 is a set of three special triples also (we may suppose that a + b + c < 1, because otherwise all three triples are equal and our statement is trivial). If there is a special triple (x, y, z) which is not worse than any triple from S 1 , then the triple ((1 − a − b − c)x + a, (1 − a − b − c)y + b, (1 − a − b − c)z + c) is special and not worse than any triple from S. We also have a(S 1 ) = b(S 1 ) = c(S 1 ) = 0, so we may suppose that the same holds for our starting set S. Suppose that one element of S has two entries equal to 0. Note that one of the two remaining triples from S is not worse than the other. This triple is also not worse than all triples from S because any special triple is not worse than itself and the triple with two zeroes. So we have a = b = c = 0 but we may suppose that all triples from S contain at most one zero. By transposing triples and elements in triples (elements in all triples must be transposed simultaneously) we may achieve the following situation x 1 = y 2 = z 3 = 0 and x 2  x 3 . If z 2  z 1 , then the second triple (x 2 , 0, z 2 ) is not worse than the other two triples from S. So we may assume that z 1  z 2 . If y 1  y 3 , then the first triple is not worse than the second and the third and we assume y 3  y 1 . Consider the three pairs of numbers x 2 , y 1 ; z 1 , x 3 ; y 3 , z 2 . The sum of all these numbers is three and consequently the sum of the numbers in one of the pairs is less than or equal to one. If it is the first pair then the triple (x 2 , 1 − x 2 , 0) is not worse than all triples from S, for the second we may take (1 − z 1 , 0, z 1 ) and for the third — (0, y 3 , 1 − y 3 ). So we found a desirable special triple for any given S. Problem 5. Does there exist a finite group G with a normal subgroup H such that |Aut H| > |Aut G|? Solution. Yes. Let H be the commutative group H = F 3 2 , where F 2 ∼ = Z/2Z is the field with two elements. The group of automorphisms of H is the general linear group GL 3 F 2 ; it has (8 − 1) · (8 − 2) · (8 − 4) = 7 · 6 · 4 = 168 elements. One of them is the shift operator φ : (x 1 , x 2 , x 3 ) → (x 2 , x 3 , x 1 ). Now let T = {a 0 , a 1 , a 2 } be a group of order 3 (written multiplicatively); it acts on H by τ(a) = φ. Let G be the semidirect product G = H ⋊ τ T . In other words, G is the group of 24 elements G = {ba i : b ∈ H, i ∈ (Z/3Z)}, ab = φ(b)a. G has one element e of order 1 and seven elements b, b ∈ H, b = e of order 2. If g = ba, we find that g 2 = baba = bφ(b)a 2 = e, and that g 3 = bφ(b)a 2 ba = bφ(b)aφ(b)a 2 = bφ(b)φ 2 (b)a 3 = ψ(b), where the homomorphism ψ : H → H is defined as ψ : (x 1 , x 2 , x 3 ) → (x 1 + x 2 + x 3 )(1, 1, 1). It is clear that g 3 = ψ(b) = e for 4 elements b ∈ H, while g 6 = ψ 2 (b) = e for all b ∈ H. We see that G has 8 elements of order 3, namely ba and ba 2 with b ∈ Ker ψ, and 8 elements of order 6, namely ba and ba 2 with b ∈ Ker ψ. That accounts for orders of all elements of G. Let b 0 ∈ H \Kerψ be arbitrary; it is easy to see that G is generated by b 0 and a. As every automorphism of G is fully determined by its action on b 0 and a, it follows that G has no more than 7 · 8 = 56 automorphisms. Remark. G and H can be equivalently presented as subgroups of S 6 , namely as H = (12), (34), (56) and G = (135)(246), (12). 3 . polynomial f (x) 2008. Then f(a i ) = g(a i )h(a i ) = 2008 for i = 1, . . . , 81, Hence, g(a 1 ), . . . , g(a 81 ) are integer divisors of 2008. Since 2008 = 2. orthonormal system of vectors in H, as required. 4 IMC2008, Blagoevgrad, Bulgaria Day 2, July 28, 2008 Problem 1. Let n, k be positive integers and suppose