" Đề thi Olympic sinh viên thế giới năm 1999 " . Đây là một sân chơi lớn để sinh viên thế giới có dịp gặp gỡ, trao đổi, giao lưu và thể hiện khả năng học toán, làm toán của mình. Từ đó đến nay, các kỳ thi Olympic sinh viênthế giới đã liên tục được mở rộng quy mô rất lớn. Kỳ thi này là một sự kiện quan trọng đối với phong trào học toán của sinh viên thế giới trong trường đại. | 6th INTERNATIONAL COMPETITION FOR UNIVERSITY STUDENTS IN MATHEMATICS Keszthely 1999. Problems and solutions on the first day 1. a Show that for any m 2 N there exists a real m x m matrix A such that A3 A 1 where I is the m x m identity matrix. 6 points b Show that det A 0 for every real m x m matrix satisfying A3 A 1. 14 points Solution. a The diagonal matrix 0 A 0 A AI I . 0 A is a solution for equation A3 A I if and only if A3 A 1 because A3 A I A3 A 1 I. This equation being cubic has real solution. b It is easy to check that the polynomial p x x3 x 1 has a positive real root Ai because p 0 0 and two conjugated complex roots A2 and A3 one can check the discriminant of the polynomial which is 1A I Q i Cl FTio mimmiim one movimiim vf FTio nnlimAmio 1 3 j y 2 j iQg a i i the local iiiiiiiiiilh11 called iiia . i die ylyii yiiiia l . If a matrix A satisfies equation A3 A I then its eigenvalues can be only A1 A2 and A3. The multiplicity of A2 and A3 must be the same because A is a real matrix and its characteristic polynomial has only real coefficients. Denoting the multiplicity of A1 by a and the common multiplicity of A2 and A3 by det A A AfAf A A2 A3 f. Because A1 and A2A3 A212 are positive the product on the right side has only positive factors. 2. Does there exist a bijective map N N such that 1 A x An 1 2_i n2 20 points Solution 1. No. For let be a permutation of N and let N 2 N. We shall argue that 3N E I n N 1 In fact of the 2N numbers N 1 . 3N only N can be N so that at least N of them are N. Hence 3N 3N x iw x -nn i- N 1 v 7 n N 1 Solution 2. Let be a permutation of N. For any n 2 N the numbers 1 . n are distinct positive integers thus 1 . n 1 . n 2 1 . By this inequality 1 X n 1 n n2 X 1 2 n 1 1 n 1 2 1 X n 1 n n 1 2 2n 1 n2 n 1 2 11 E2n 1 v 1 1 2n n 1 n 1 1 6th INTERNATIONAL COMPETITION FOR UNIVERSITY STUDENTS IN MATHEMATICS Keszthely 1999. Problems and solutions on the second day 1. Suppose that in a not necessarily commutative ring R the
