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Number theory Difficulty 4.7 AIME Prove it Romania

Given aa, bb distinct positive integers, show that the system of equations
xy+zw=a xy + zw = a
xz+yw=b xz + yw = b
has only finitely many solutions in integers xx, yy, zz, ww.

Solution

By adding and subtracting the equations we get (x+w)(y+z)=a+b(x + w)(y + z) = a + b and (x−w)(y−z)=a−b(x - w)(y - z) = a - b, hence by multiplying, 0<∣(x2−w2)(y2−z2)∣=∣a2−b2∣0 < |(x^2 - w^2)(y^2 - z^2)| = |a^2 - b^2|. Therefore 0<∣x2−w2∣≤∣a2−b2∣0 < |x^2 - w^2| \le |a^2 - b^2| and 0<∣y2−z2∣≤∣a2−b2∣0 < |y^2 - z^2| \le |a^2 - b^2|. But an equation 0<∣A2−B2∣≤∣C∣0 < |A^2 - B^2| \le |C| has only finitely many integer solutions AA, BB for fixed CC.

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