Solution triples for this Pythagorean Theorem equation may be obtained using the well-known formulas for Pythagorean triples. Note: Derivation of the Pythagorean triples formula from properties of rational numbers was published by the author in 1977.
X2 = 2PQ,
Z2 = P2 + Q2
Now with this application of the Pythagorean triple formulae, we have obtained another Pythagorean triple: Y2 + Q2 = P2. By comparison with the equation (X2)2 + (Y2)2 = (Z2)2 and the formulae above, we see that P < Z2, Q< X2, and Y < Y2.
Repeated applications of the formulae will produce smaller and smaller triples, leading to contradiction by infinite descent: Since any integer solution will lead to a smaller integer solution, the smallest integer of the smallest triple must eventually equal the smallest non-zero integer, 1, yielding a triple (1,C,D), where C and D are positive integers, such that (12)2 + (B2)2 = (C2)2→ (C2)2 - (B2)2 = 1. Given that the smallest integer of the successive non-zero triples in the infinite descent will eventually reach unity, and B > A > 1, we can demonstrate the universality of the contradiction as follows: Let C = 3, and D = 2, the two smallest positive integers larger than 1. Then (32)2 - (22)2 = 1 → 81 – 16 = 1, which is a clear contradiction, and any pairs of larger integers C1 > C and D1 > D, will lead to larger discrepancies. So we have to conclude that there can be no integer solution triples for the equation (X2)2 + (Y2)2 = (Z2)2.