53rd Putnam 1992

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

Solution

Answer: a, b, m, n = 2^r, 2^r, 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)ⁿd^2(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 = 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