13th Iberoamerican 1998 Problem B3 k is the positive root of the equation x 2 - 1998x - 1 = 0. Define the sequence x 0 , x 1 , x 2 , ... by x 0 = 1, x n+1 = [k x n ]. Find the remainder when x 1998 is divided by 1998.
Put p(x) = x 2 - 1998x - 1. Then p(1998) = -1, p(1999) = 1998, so 1998  k x n-1 - 1. Hence x n /k < x n-1 < x n /k + 1/k, so [x n /k] = x n-1 - 1. k = (1998k + 1)/k = 1998 + 1/k. Hence kx n = 1998x n + x /k. Hence x n+1 = [kx n ] = 1998x n + [x n /k] = 1998x n + x n-1 - 1. Hence x n+1 = x n-1 - 1 mod 1998. So x 1998 = 1 - 999 = 1000 mod 1998. 13th Ibero 1998 © John Scholes jscholes@kalva.demon.co.uk 24 Oct 2002