### 21st Putnam 1960

Problem B1

Find all pairs of unequal integers m, n such that mn = nm.

Solution

Answer: (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 = nk. But for x > 1 we have 1 + x < ex (the derivative of f(x) = ex - x - 1 is positive and f(0) = 0) and hence (1 + k/n)n < ek. So there are no solutions for n > 2. If n = 1, then nm = 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 = 2h. 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 < n.

If n < 0 and m > 0, then nm = 1/m-n. So m divides 1 and hence m = 1. But m must be even for to make nm 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.