19th APMC 1996 Problem 6 k, n are positive integers such that n > k > 1. Find all real solutions x 1 , x 2 , ... , x n to x i 3 (x i 2 + x i+1 2 + ... + x i+k-1 2 ) = x i-1 2 for i = 1, 2, ... , n. [Note that we take x 0 to mean x n , and x n+j to mean x j .]
All x i = 1/k 1/3 or all x i = 0. Solution If some x i = 0. Then from the i equation, x i-1 = 0. Continuing, all x i = 0, which is evidently a solution. So assume no x i are zero. Then it is immediate from the i equation that x i > 0. Now suppose x i is the smallest. Then x i 2 ≤ x i+k 2 , so x i 2 + x i+1 2 + ... + x i+k-1 2 ≤ x i+1 2 + x i+2 2 + ... + x i+k 2 . Also x i 3 ≤ x i+1 3 , so x i 3 (x i 2 + x i+1 2 + ... + x i+k-1 2 ) ≤ x i+1 3 (x i+1 2 + x i+2 2 + ... + x i+k 2 ), so x i-1 2 ≤ x i 2 and hence x i-1 ≤ x i . So x i-1 is also the smallest. Continuing, we get that all x i are equal. Then any equation gives that x i = 1/k 1/3 . 19th APMC 1996 © John Scholes jscholes@kalva.demon.co.uk 28 Nov 2003 Last corrected/updated 28 Nov 03