53rd Putnam 1992

Problem A3

Find all positive integers a, b, m, n with m relatively prime to n such that (a2 + b2)m = (ab)n.



Answer: a, b, m, n = 2r, 2r, 2r, 2r+1 for any positive integer r.

a2 + b2 ≥ 2ab > ab, so m < n.

Let d be the greatest common divisor of a and b. Set a=dA, b = dB. Then (A2 + B2)m = (AB)nd2(n-m). Hence if p divides A, then it must also divide B. But A and B are coprime, so A = 1. Similarly, B = 1. Now 2(n-m) > 1, so d divides (A2 + B2)m = 2m. So d is a power of 2. Hence a = b = 2r 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
7 Jan 2001