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 < e^x (the derivative of f(x) = e^x - x - 1 is positive and f(0) = 0) and hence (1 + k/n)^n < e^k. So, there are no solutions for n > 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 < n (m = 2, n = 4).
If n < 0 and m > 0, then n^m = 1/m-n. So m divides 1 and hence m = 1. But m must be even for this 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.
So, the only four pairs are (2, 4), (4, 2), (-2, -4), (-4, -2). |