Case 1. Suppose one of a or b is even and thus gcd(a + b, a − b) = 1. Then a + b | z. Furthermore, since b | z, and gcd(b, a + b) = 1 then b(a + b) | z. Let z = kb(a + b). Then x = (a/b)z = ka(a + b). A similar argument shows then that y = kb(a − b) and w = ka(a − b). (Yes, the k value is the same!) Hence n = xy = wz = k2ab(a2 − b2).