Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fermat's marginal proof

  1. Sep 23, 2012 #1
    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?
     
    Last edited: Sep 23, 2012
  2. jcsd
  3. Sep 23, 2012 #2
    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.
     
  4. Sep 23, 2012 #3

    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.
     
  5. Sep 23, 2012 #4
    Thanks, would it still be a reductio if it could hold for only some natural numbers?
     
  6. Sep 23, 2012 #5
    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?
     
  7. Sep 23, 2012 #6
    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?
     
  8. Sep 23, 2012 #7
    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.
     
    Last edited: Sep 23, 2012
  9. Sep 24, 2012 #8
    I get it now, thanks again :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fermat's marginal proof
Loading...