IMO 1984 SL 2

Origin: CAN

Problem

Prove: (a) There are infinitely many triples of positive integers m, n, p such that 4mn −m −n = p2 −1. (b) There are no positive integers m, n, p such that 4mn −m −n = p2.

Solution

(a) For m = t(t −1)/2 and n = t(t + 1)/2 we have 4mn −m −n = (t2 −1)2 −1. (b) Suppose that 4mn −m −n = p2, or equivalently, (4m −1)(4n −1) = 4p2 +1. The number 4m−1 has at least one prime divisor, say q, that is of the form 4k + 3. Then 4p2 \equiv−1 (mod q). However, by Fermat’s theorem we have 1 \equiv(2p)q−1 =
4p2 q−1 \equiv(−1) q−1 (mod q), which is impossible since (q −1)/2 = 2k + 1 is odd.