IMO 1994 SL N2

Origin: AUS | Category: Number Theory

Problem

Determine all pairs (m, n) of positive integers such that n3+1 mn−1 is an integer.

Solution

Since mn −1 and m3 are relatively prime, mn −1 divides n3 + 1 if and only if it divides m3(n3 + 1) = (m3n3 −1) + m3 + 1. Thus n3 + 1 mn −1 \inZ ⇔m3 + 1 mn −1 \inZ; hence we may assume that m \geqn. If m = n, then n3+1 n2−1 = n + n−1 is an integer, so m = n = 2. If n = 1, then m−1 \inZ, which happens only when m = 2 or m = 3. Now suppose m > n \geq2. Since m3 + 1 \equiv1 and

mn −1 \equiv−1 (mod n), we deduce n3+1 mn−1 = kn −1 for some integer k > 0. On the other hand, kn−1 < n3+1 n2−1 = n+ n−1 \leq2n−1 gives that k = 1, and therefore n3+1 = (mn−1)(n−1). This yields m = n2+1 n−1 = n+1+ n−1 \inN, so n \in{2, 3} and m = 5. The solutions with m < n are obtained by symmetry. There are 9 solutions in total: (1, 2), (1, 3), (2, 1), (3, 1), (2, 2), (2, 5), (3, 5), (5, 2), (5, 3).