Local Mathematical Competitions

Question

Given a, b distinct positive integers, show that the system of equations xy + zw = a xz + yw = b has only finitely many solutions in integers x, y, z, w.

Accepted Answer

By adding and subtracting the equations we get (x + w)(y + z) = a + b and (x - w)(y - z) = a - b, hence by multiplying, 0 < |(x^2 - w^2)(y^2 - z^2)| = |a^2 - b^2|. Therefore 0 < |x^2 - w^2| |a^2 - b^2| and 0 < |y^2 - z^2| |a^2 - b^2|. But an equation 0 < |A^2 - B^2| |C| has only finitely many integer solutions A, B for fixed C.