IMO 1988 SL 9

Origin: FRG

Problem

Let a and b be two positive integers such that ab+1 divides a2 + b2. Show that a2+b2 ab+1 is a perfect square.

Solution

Let us assume a2+b2 ab+1 = k \inN. We then have a2 −kab + b2 = k. Let us assume that k is not an integer square, which implies k \geq2. Now we observe the minimal pair (a, b) such that a2 −kab+b2 = k holds. We may assume w.l.o.g. that a \geqb. For a = b we get k = (2 −k)a2 \leq0; hence we must have a > b. Let us observe the quadratic equation x2 −kbx + b2 −k = 0, which has solutions a and a1. Since a + a1 = kb, it follows that a1 \inZ. Since a > kb implies k > a + b2 > kb and a = kb implies k = b2, it follows that a < kb and thus b2 > k. Since aa1 = b2 −k > 0 and a > 0, it follows that a1 \inN and a1 = b2−k a < a2−1 a < a . We have thus found an integer pair (a1, b) with 0 < a1 < a that satisfies the original equation. This is a contradiction of the initial assumption that (a, b) is minimal. Hence k must be an integer square.