IMO 1989 SL 25

Origin: KOR

Problem

Let a, b be integers that are not perfect squares. Prove that if x2 −ay2 −bz2 + abw2 = 0 has a nontrivial solution in integers, then so does x2 −ay2 −bz2 = 0.

Solution

We may assume w.l.o.g. that a > 0 (because a, b < 0 is impossible, and a, b ̸= 0 from the condition of the problem). Let (x0, y0, z0, w0) ̸= (0, 0, 0, 0) be a solution of x2 −ay2 −bz2 + abw2. Then x2 0 −ay2 0 = b(z2 0 −aw2 0). Multiplying both sides by (z2 0 −aw2 0), we get (x2 0 −ay2 0)(z2 0 −aw2 0) −b(z2 0 −aw2 0)2 = 0 ⇔(x0z0 −ay0w0)2 −a(y0z0 −x0w0)2 −b(z2 0 −aw2 0)2 = 0. Hence, for x1 = x0z0 −ay0w0, y1 = y0z0 −x0w0, z1 = z2 0 −aw2 0, we have x2 1 −ay2 1 −bz2 1 = 0. If (x1, y1, z1) is the trivial solution, then z1 = 0 implies z0 = w0 = 0 and similarly x0 = y0 = 0 because a is not a perfect square. This contradicts the initial assumption.