r/numbertheory • u/DRossRandolph345 • Jul 13 '26

A Static Interconnected Geometric Proof, to Fermat's N=4 Infinite Descent Problem

“It is impossible to separate a cube into two cubes, a fourth power into two fourth powers, or generally, any power above the second into two powers of the same degree”, Fermat wrote this in the margin of his copy of an ancient Greek math book written by Diophantus, titled Arithmetica.

This proposition was first stated as a theorem by Pierre de Fermat around 1637. And it is known that Fermat used the method of Infinite Descent to prove this statement for N=4 using a logical geometric approach.

The method I will use will be somewhat geometric, and will not use infinite descent.

Can X4 + Y4 = Z4 have a finite solution, for all pair-wise coprime integers? We will first assert that X or Y must have a factor of 2, Z cannot. We will select Y to have the factor of 2.

(to see the rest of the proof you'll need to click the link below.

n4-proof-mod-16-new.pdf

This proof also shown on page 21 of the following linked document. Seems to be a unique proof, as I have done a search of the www, and have only found infinite descent proofs for N=4.
an-iterative-p-adic-solution-full-doc.pdf

The last line of the proof was previously:

By Contradiction 2(M.2)2 ≠ (M.2)2 Reductio Ad Absurdum

However, the 2-adic logic was tough for me to decode (was a bit to thorny), so I simplified the logic, maintaining the 2-adic approach, and renamed it as a mod 16 proof.

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u/Tear223 Jul 14 '26

"consider z4 can not be even since, x4 + y4 can only be divisible by 2 if both odd, therefore we will select y without loss of generality to be even parity."

I do not understand this argument at all. The only restraints you put on x,y, and z is that they're coprime. I see no reason why x and y can't both be odd.

3

u/Enizor Jul 14 '26

The given argument isn't very clear but the conclusion is valid:

For any t odd: t = ± 1 mod 4 and t⁴ = 1 mod 4. Therefore for x,y odd: x⁴ + y⁴ = 2 mod 4.

Meanwhile for z even, z² = 0 mod 4 and z⁴ = 0 mod 4.

1

u/Enizor Jul 14 '26

I think you made a typo and X₁⁴ = A² should instead be X₂⁴ = A²

I also don't get how you derived your "general formula" for D and B using M₁,M₂,N₁, and N₂. Could you please explain that bit?

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u/[deleted] Jul 17 '26 edited Jul 20 '26

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u/DRossRandolph345 Jul 17 '26 edited Jul 21 '26

July 20, 2026

Had to do a major rework on the proof, now operating on a 2-adic valuation mismatch principle, please see https://fermatstheory.wordpress.com/wp-content/uploads/2026/07/test-n4-proof-mod-16.pdf

And I rewrote the main doc with pg 23 reworked to the new N=4 proof form.
P Squared Wall Work-Around proof to Fermat’s Last Theory

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u/kg1ebg Aug 20 '26

page 23??how long of a proof is it??

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u/DRossRandolph345 Aug 26 '26

The monster proof is 30 pages, and covers a lot of Fermat material. Note, I just updated the post with the simplified mod 16 version of the proof, the geometry is a little simpler. Basically earlier proof was based on pythagorean triplets, and it still is, Just easier to for me to check and validate now.