21st Putnam 1960 Problem B1 Find all pairs of unequal integers m, n such that m n = n m . Solution
: (2, 4), (4, 2), (-2, -4), (-4, -2). Suppose first that m and n are both positive. Assume m > n. Then we can put m = n + k with k > 0. Hence (1 + k/n) n = n k . But for x > 1 we have 1 + x  2. If n = 1, then n m = 1 and hence m = 1, contradicting the fact that m and n are unequal. If n = 2, then m must be a power of 2. Suppose m = 2 h . Then we find h = 1 or 2. h = 1 is invalid (because m and n are unequal), so m = 4. There is also the corresponding solution with m  0, then n m = 1/m -n . So m divides 1 and hence m = 1. But m must be even for to make n m positive. Contradiction. So there are no solutions of this type. If m and n are both negative, then -m, - n is a solution, so the only possibilities are (-2, -4) and (-4, -2) and it is readily checked that these are indeed solutions. 21st Putnam 1960 © John Scholes jscholes@kalva.demon.co.uk 15 Feb 2002