53rd Putnam 1992 Problem A3 Find all positive integers a, b, m, n with m relatively prime to n such that (a 2 + b 2 ) m = (ab) n . Solution
: a, b, m, n = 2 r , 2 r , 2r, 2r+1 for any positive integer r. a 2 + b 2 ≥ 2ab > ab, so m  1, so d divides (A 2 + B 2 ) m = 2 m . So d is a power of 2. Hence a = b = 2 r for some r. So (2r+1)m = 2r n. But 2r+1 and 2r are coprime, so m = M·2r, n = N·(2r+1). Hence M = N. But m and n are coprime, so M = N = 1. 53rd Putnam 1992 © John Scholes jscholes@kalva.demon.co.uk 7 Jan 2001