IMO 2001 SL N2

Origin: COL | Category: Number Theory

Problem

Consider the system x + y = z + u, 2xy = zu. Find the greatest value of the real constant m such that m \leqx/y for every positive integer solution x, y, z, u of the system with x \geqy.

Solution

We shall find the general solution to the system. Squaring both sides of the first equation and subtracting twice the second equation we obtain (x −y)2 = z2 + u2. Thus (z, u, x −y) is a Pythagorean triple. Then it is well known that there are positive integers t, a, b such that z = t(a2 −b2), u = 2tab (or vice versa), and x −y = t(a2 + b2). Using that x + y = z + u we come to the general solution: x = t(a2 + ab), y = t(ab −b2); z = t(a2 −b2), u = 2tab. Putting a/b = k we obtain x y = k2 + k k −1 = 3 + (k −1) + k −1 \geq3 + 2 \sqrt 2, with equality for k −1 = \sqrt 2. On the other hand, k can be arbitrarily close to 1 + \sqrt 2, and so x/y can be arbitrarily close to 3 + 2 \sqrt 2. Hence m = 3 + 2 \sqrt 2. Remark. There are several other techniques for solving the given system. The exact lower bound of m itself can be obtained as follows: by the system  x y −6 x y + 1 =  z−u y \geq0, so x/y \geq3 + 2 \sqrt 2.