IMO 1995 SL N8

Origin: GER | Category: Number Theory

Problem

Let p be an odd prime. Determine positive integers x and y for which x \leqy and \sqrt2p −\sqrtx −\sqrty is nonnegative and as small as possible.

Solution

Let u = \sqrt2p −\sqrtx −\sqrty and v = u(2\sqrt2p −u) = 2p −(\sqrt2p −u)2 = 2p −x −y −\sqrt4xy for x, y \inN, x \leqy. Obviously u \geq0 if and only if v \geq0, and u, v attain minimum positive values simultaneously. Note that v ̸= 0. Otherwise u = 0 too, so y = (\sqrt2p−\sqrtx)2 = 2p−x−2\sqrt2px, which implies that 2px is a square, and consequently x is divisible by 2p, which is impossible. Now let z be the smallest integer greater than \sqrt4xy. We have z2−1 \geq4xy, z \leq2p −x −y, and z \leqp because \sqrt4xy \leq(\sqrtx + \sqrty)2 < 2p. It follows that v = 2p −x −y −

4xy \geqz −

z2 −1 = z + \sqrt z2 −1 \geq p +

p2 −1 . Equality holds if and only if z = x + y = p and 4xy = p2 −1, which is satisfied only when x = p−1 and y = p+1 2 . Hence for these values of x, y, both u and v attain positive minima.