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Short Proof of FLT

  1. Jun 24, 2011 #1
    I've been enormously impressed with the core members of this forum, especially hurkyl. There was a guy who claimed to have a proof of FLT, and hurkyl and others gave him a lot of time over two weeks and were much more fair to him than he deserved. Since I'm hoping to get a fair hearing for my own claim, I decided to post here.

    I have a proof, using only high school algebra and elementary number theory, that I believe to be valid for all odd values of n, n being the common exponent. It's clear, simple, and perhaps even a bit elegant. It took me far too long to find, but hey, it's a tough problem.

    I'd like to share it with everyone here and invite intelligent comments. However, I'd prefer to post the proof on my own blog, and then discuss specific details here if necessary. Does that sound like a good way to do it?
     
  2. jcsd
  3. Jun 24, 2011 #2
    if you actually have a proof like that, just post it wherever you want.
     
  4. Jun 24, 2011 #3
    Hey there, I have a proof that I believe works for [itex]n=3[/itex] and is equally elegant, though I failed to extend it to all odd numbers of [itex]n[/itex] (especially since [itex]n=1[/itex] of FLT is obviously false, i.e. [itex]x^1+y^1=a^1[/itex] has solutions). Here it goes:

    Proof of Fermat's Last Theorem for [itex]n=3[/itex]

    Suppose [itex]x[/itex], [itex]y[/itex], and [itex]a[/itex] are some natural number solutions to the equation [itex]x^3+y^3=a^3[/itex].
    Let [itex]b=x+y[/itex] for some natural number [itex]b[/itex] ([itex]x[/itex] and [itex]y[/itex] are by assumption both natural numbers, which are closed under addition). Then [itex]y=-x+b[/itex].

    [itex]x^3+y^3=a^3[/itex]
    [itex]x^3+(-x+b)^3=a^3[/itex]
    [itex]x^3-x^3+3x^2b-3xb^2+b^3=a^3[/itex]
    [itex]3bx^2-3b^2x+(b^3-a^3)=0[/itex]

    By the quadratic formula (solving in terms of [itex]b[/itex] and [itex]a[/itex]):

    [itex]x=\frac{3b^2\pm\sqrt{9b^4-12b^4+12ba^3}}{6b}[/itex]=[itex]\frac{3b^2\pm\sqrt{-3b^4+12ba^3}}{6b}[/itex]

    We now seek to express b and a in the expression above once again in terms of [itex]x[/itex] and [itex]y[/itex].

    [itex]-3b^4=-3(x+y)^4=-3(x^4+4x^3y+6x^2y^2+4xy^3+y^4)[/itex]
    [itex]=-3x^4-12x^3y-18x^2y^2-12xy^3-3y^4[/itex]

    [itex]12ba^3=12x^4+12x^3y+12xy^3+12y^4[/itex]

    [itex]-3b^4+12ba^3=9x^4-18x^2y^2+9y^4=3^2(x^2-y^2)^2[/itex]

    We rewrite the first expression for x above as follows:

    [itex]x=\frac{3(x^2+y^2
    )\pm\sqrt{3^2(x^2-y^2)^2}}{6(x+y)}[/itex]=[itex]\frac{3(x^2+y^2)\pm3(x^2-y^2)}{6(x+y)}[/itex]=[itex]\frac{(x^2+y^2)\pm(x^2-y^2)}{2(x+y)}[/itex]

    The two solutions then are [itex]x=\frac{x^2}{(x+y)}[/itex] and [itex]x=\frac{y^2}{(x+y)}[/itex].

    The first reduces to [itex]x^2+xy=x^2[/itex] or simply [itex]xy=0[/itex]. However, [itex]x[/itex] and [itex]y[/itex] must be natural numbers, neither being 0, so there are no solutions from this first equation which fit our assumption.

    The second reduces to [itex]x^2+xy-y^2=0[/itex]. By applying the quadratic formula again, solving for [itex]x[/itex] in terms of [itex]y[/itex], we have [itex]x=\frac{-y\pm\sqrt{5x^2}}{2}[/itex].

    This reduces to [itex]x=(\frac{(-1+\sqrt{5})}{2})y[/itex] and [itex](\frac{(-1-\sqrt{5})}{2})y[/itex].

    Clearly, both [itex]x[/itex] and [itex]y[/itex] cannot be natural numbers if one is an irrational expression, for an irrational expression times a natural number expression produces an irrational product.

    Since we assumed [itex]x[/itex], [itex]y[/itex], and [itex]a[/itex] to be natural number solutions to the equation [itex]x^3+y^3=a^3[/itex], but found that they could not be such, we are forced to conclude that the equation [itex]x^n+y^n=a^n[/itex] has no natural number solutions for [itex]n=3[/itex].
     
    Last edited: Jun 24, 2011
  5. Jun 24, 2011 #4

    HallsofIvy

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    Fermat himself produced a proof for n= 3. And, I believe, the proof for n odd is also old.
     
  6. Jun 24, 2011 #5

    micromass

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    Hmm, Fermat proved the case n=4. It was Euler who proved n=3, but his proof contained a major gap.
    It can easily be shown that FLT only needs to be proven for odd primes (since if n=de is composite and if [itex]x^n+y^n=z^n[/itex], then [itex](x^d)^e+(y^d)^e=(z^d)^e[/itex]. So the proof for odd primes is the hardest part and was only completed by Wiles.
     
  7. Jun 24, 2011 #6

    micromass

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    This is incorrect. You have said that [itex]3b^2=3(x^2+y^2)[/itex] which is of course false.
     
  8. Jun 24, 2011 #7

    No, Fermat produced a proof for n=4. Euler was the first to prove it for n=3.

    There's no old proof for all odd n, only some for specific or limited values of n.

    Wiles' proof is newer, and claims to be valid for all n>2. (I've read that he had to prove 3 and 4 as special cases, with his general proof being sufficient for 5 and above.)
     
  9. Jun 24, 2011 #8
    Thanks for pointing that out! I don't have time at the moment, but I'll see if I can still obtain an answer after fixing it.

    There have definitely been many proofs of the first few [itex]n[/itex]s, but it would be amazing we could all agree if a simpler proof than Wiles's could be found.
     
  10. Jun 24, 2011 #9
    I checked it out, and with that error remedied, my proof doesn't amount to anything. (You just get 0=0). Sorry I missed that mistake and thanks for correcting it for me :/
     
  11. Jun 24, 2011 #10
    See, Middle C? These guys are great. That's why I came here.
     
  12. Jun 24, 2011 #11
    By the way, I also have a complete elementary proof for Case 1, in which none of x,y, or z are divisible by n. I'd forgotten about that one.

    I'll be posting a link soon.
     
  13. Jun 25, 2011 #12
    I forgot about that proof for good reason. It's a proof of case 1 for n=3, which is nothing extraordinary. I did use congruences, though, which I thought was interesting.

    As I said, I'll be posting a link to my other proof soon.
     
  14. Jun 30, 2011 #13
    You've got that right (yourself included). :smile:
     
  15. Jul 8, 2011 #14
    I have always believed that a "simple" proof of FLT should exist (be able to be developed) - a page or two, say, that anyone can understand.
    But, if someone here has developed such a proof, would it not be a momentous article worthy of publication and fanfare!? It seems that such a proof should first be published in a pretigious journal. Anyone's thoughts?
     
    Last edited: Jul 8, 2011
  16. Jul 8, 2011 #15
    Fermat's claim dates from 1637. In my opinion it's incredibly unlikely that the greatest mathematicians in the world missed an elementary proof for 374 years. It's just not a credible possibility.

    It will be interesting to see how Wiles's proof gets boiled down and streamlined over the next few years and decades. But a proof based on elementary algebraic manipulations is just not possible. Someone would have found it.
     
  17. Jul 8, 2011 #16

    micromass

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    Read "Rational points on elliptic curves" by Silverman and Tate. That ought to convince you that Fermat's last theorem is far from a trivial statement. Thus the proof should be expected to be far from trivial too!!
     
  18. Jul 8, 2011 #17
    So, to my question, you're implying that it would be a momentous finding worthy of prestigious publication.
     
  19. Jul 8, 2011 #18

    micromass

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    If you would find an elementary proof of FLT, then you will become a world-famous mathematician instantly, yes. But I'm quite certain that such an elementary proof does not exist.
     
  20. Jul 8, 2011 #19
    I don't know if this is the right place to post this or if this should be in a new topic, but it relates to someone claiming to have rediscovered Fermat's original proof or, at least, making a short proof with only the mathematics known in Fermat's time.

    A user of Youtube by the name of michaelcweir claims to have proven Fermat's Last Theorem using Diophantine equations. His video is http://www.youtube.com/watch?v=gwbHORdjcbQ", for everything.

    Could anyone verify that this is, indeed, a valid proof of Fermat's Last Theorem, or, if not, tell me what's wrong with it?
     
    Last edited by a moderator: Apr 26, 2017
  21. Jul 8, 2011 #20
    A quick look at the proposed proof: I see a statement in his proof, "a whole number raised to a non-whole number is an non-number number." Hmm. I think he means a non-whole number. Let me see. 4 is a whole number. 1/2 is a non-whole number. If I raise 4 to 1/2 I usually get 2, a whole number. So this part of the proposed proof is rather doubtful. I didn't study the whole proof, but this point makes me not want to bother. Maybe there is something in the context that makes his statement true in that interpretation ????.
     
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