Find all positive integers a, b, m, n with m relatively prime to n such that (a^{2} + b^{2})^{m} = (ab)^{n}.

**Solution**

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

a^{2} + b^{2} ≥ 2ab > ab, so m < n.

Let d be the greatest common divisor of a and b. Set a=dA, b = dB. Then (A^{2} + B^{2})^{m} = (AB)^{n}d^{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 (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.

© John Scholes

jscholes@kalva.demon.co.uk

7 Jan 2001