# Fermat's marginal proof

The Seeker
In trying to work out what Fermat may have conceived of as his proof, using the mathematics available at the time I have the following suggestion:

Fermat's Last Theorem can be expressed the following way:

There are no natural numbers A, B, C, N >1 for which a non-trivial solution of the following formula exists:

(1) A ^(2N+1) + B ^(2N+1) = C ^(2N+1)

To prove this we begin by assuming (1) is true and rewriting it as follows:

(2) A x A ^(2N) + B x B ^(2N) = C x C ^(2N)

We now proceed to prove that (1) cannot hold by creating a contradiction. Firstly we reasonably assume that for any valid triple of the form (1) and (2) C must be greater than A and B according to the logic of valid pythagorean triples. Secondly we introduce the general function:

(3) P x (A ^M + B ^M) = P x C ^M

And similarly assume it holds for all* natural numbers A, B, C, P, M. It therefore follows that there must be natural numbers M and P for which P=C. This would imply:

(4) C x A ^M + C x B ^M = C x C ^M = C ^(M+1)

Which, taking (1) to be true, and substituting 2N for M implies C=A=B. But for valid triples C>A and C>B and so a contradiction has been established which proves that (1) cannot hold.

* That should be at least one set of natural numbers A, B, C, P, M, where P=C.

It appears far too simple but why?

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Number Nine
The general consensus is that Fermat was mistaken. Wiles nearly had to invent entirely new fields of mathematics to construct his proof; almost no one actually believes that Fermat constructed one as simple as he claims.

SteveL27
In trying to work out what Fermat may have conceived of as his proof, using the mathematics available at the time I have the following suggestion:

Fermat's Last Theorem can be expressed the following way:

There are no natural numbers A, B, C, N >1 for which a non-trivial solution of the following formula exists:

(1) A ^(2N+1) + B ^(2N+1) = C ^(2N+1)

To prove this we begin by assuming (1) is true and rewriting it as follows:

(2) A x A ^(2N) + B x B ^(2N) = C x C ^(2N)

We now proceed to prove that (1) cannot hold by creating a contradiction. Firstly we reasonably assume that for any valid triple of the form (1) and (2) C must be greater than A and B according to the logic of valid pythagorean triples. Secondly we introduce the general function:

(3) P x (A ^M + B ^M) = P x C ^M

And similarly assume it holds for all natural numbers A, B, C, P, M.

I see a logic problem right at this step. To do your reductio proof, you are assuming there exist a, b, c, and n such that an + bn = cn. That's fine.

But that certainly does not imply that this relationship holds for all a, b, c, n, and p as above. Even if FLT were false, there would be particular a-b-c-n examples. The Fermat relationship would surely not hold for all numbers a, b, c, and n.

The Seeker
Thanks, would it still be a reductio if it could hold for only some natural numbers?

The Seeker
I suppose what I'm wondering is - if we assume there is at least one natural number solution for which P=C for (1) and (3) would that immediately establish the contradiction?

SteveL27
Thanks, would it still be a reductio if it could hold for only some natural numbers?

FLT says that there is NO 4-tuple (a,b,c,n) satisfying an + bn =cn +an (with n > 2 etc.)

The negation of that statement is that there exists SOME 4-tuple satisfying the Fermat condition.

Surely you can see that it would be unreasonable to assume that ALL 4-tuples satisfied the Fermat condition. Can you see why that wouldn't make any sense?

The Seeker
I think I understand you, I should really have said that in order to establish the reductio I assume the existence of a particular set of natural numbers A, B, C, P, M such that in (2) and (3) P=C which would assert both that C > A and B and C=A=B. I'm a novice and I'm in the dark about so much, I'm grateful for your time and comments.

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The Seeker
I get it now, thanks again :)