IMO 1998 SL 14

Origin: GBR

Problem

Determine all pairs (x, y) of positive integers such that x2y+ x + y is divisible by xy2 + y + 7.

Solution

If x2y + x + y is divisible by xy2 + y + 7, then so is the number y(x2y + x + y) −x(xy2 + y + 7) = y2 −7x. If y2 −7x \geq0, then since y2 −7x < xy2 +y+7, it follows that y2 −7x = 0. Hence (x, y) = (7t2, 7t) for some t \inN. It is easy to check that these pairs really are solutions. If y2 −7x < 0, then 7x −y2 > 0 is divisible by xy2 + y + 7. But then xy2 + y + 7 \leq7x −y2 < 7x, from which we obtain y \leq2. For y = 1, we are led to x + 8 | 7x −1, and hence x + 8 | 7(x + 8) −(7x −1) = 57. Thus the only possibilities are x = 11 and x = 49, and the obtained pairs (11, 1), (49, 1) are indeed solutions. For y = 2, we have 4x + 9 | 7x −4, so that 7(4x + 9) −4(7x −4) = 79 is divisible by 4x + 9. We do not get any new solutions in this case. Therefore all required pairs (x, y) are (7t2, 7t) (t \inN), (11, 1), and (49, 1).