IMO 2003 SL N3

Origin: BUL | Category: Number Theory

Problem

Determine all pairs (a, b) of positive integers such that a2 2ab2 −b3 + 1 is a positive integer.

Solution

Let a and b be positive integers for which a2 2ab2−b3+1 = k is a positive integer. Since k > 0, it follows that 2ab2 \geqb3, so 2a \geqb. If 2a > b, then from 2ab2 −b3 + 1 > 0 we see that a2 > b2(2a −b) + 1 > b2, i.e. a > b. Therefore, if a \leqb, then a = b/2. We can rewrite the given equation as a quadratic equation in a, a2 − 2kb2a + k(b3 −1) = 0, which has two solutions, say a1 and a2, one of which is in N0. From a1 + a2 = 2kb2 and a1a2 = k(b3 −1) it follows that the other solution is also in N0. Suppose w.l.o.g. that a1 \geqa2. Then a1 \geqkb2 and 0 \leqa2 = k(b3 −1) a1 \leqk(b3 −1) kb2 < b.

By the above considerations we have either a2 = 0 or a2 = b/2. If a2 = 0, then b3 −1 = 0 and hence a1 = 2k, b = 1. If a2 = b/2, then b = 2t for some t, and k = b2/4, a1 = b4/2 −b/2. Therefore the only solutions are (a, b) \in{(2t, 1), (t, 2t), (8t4 −t, 2t) | t \inN}. It is easy to show that all of these pairs satisfy the given condition.