Proving x⁴ + y⁴ = z² Has No Integer Solutions

Intro to Proofs · Axiom Academy

EXAMPLE Proving x^4 + y^4 = z^2 Has No Integer Solutions Fermat's method of infinite descent, worked one step at a time on a concrete quartic Prove that the equation x^4 + y^4 = z^2 has no solution in positive integers . (Because z^4 = (z^2)^2 , this immediately settles the n = 4 case of Fermat's Last Theorem, x^4 + y^4 = z^4 .) The tool is infinite descent : from any positive solution we manufacture a strictly smaller one — which is impossible. Each solution forces a strictly smaller hypotenuse z' < z of the same form. Positive integers cannot decrease forever — they hit a floor at 1 — so no solution can exist at all. Nicely done — you walked Fermat's infinite descent all the way down. The engine of the proof: Well-ordering: the positive integers have a least element, so a strictly decreasing chain of them cannot continue forever — that's the contradiction every descent argument cashes in. Minimal counterexample: assume a solution exists, take the one with the smallest z , then build a smaller one. The "smallest" can't have a smaller — so none exists. Pythagorean structure, used twice: is one primitive triple; its odd-leg relation x^2 = m^2 - n^2 hides a second triple . Both parametrizations are needed. Coprime factors of a square are squares: y^2 = 4pq(p^2+q^2) with pairwise coprime forces — delivering a^4+b^4=c^2 . The descent is genuine: , so c < z strictly.

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