IMO 1986 SL 5

Origin: FRG

Problem

The set S = {2, 5, 13} has the property that for every a, b \inS, a ̸= b, the number ab −1 is a perfect square. Show that for every positive integer d not in S, the set S \cup{d} does not have the above property.

Solution

Suppose that for every a, b \in{2, 5, 13, d}, a ̸= b, the number ab −1 is a perfect square. In particular, for some integers x, y, z we have 2d −1 = x2, 5d −1 = y2, 13d −1 = z2. Since x is clearly odd, d = (x2 + 1)/2 is also odd because 4 ∤x2 + 1. It follows that y and z are even, say y = 2y1 and z = 2z1. Hence (z1 − y1)(z1 +y1) = (z2 −y2)/4 = 2d. But in this case one of the factors z1 −y1, z1 + y1 is odd and the other one is even, which is impossible.