- #1

The Seeker

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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?

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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