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Fermat's Last Theorem

  1. Feb 24, 2004 #1
    I am trying to get my father's ingenious proof of Fermat's last theorem the recognition that it deserves. Please visit the link below.



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  3. Feb 24, 2004 #2

    matt grime

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    I'm not sure who you talked to about the mathematics, but we'll assume you're serious and not a troll.

    1. Let r=2ab be the decomposition into subfactors (a=p_1p_2... and b=p_3p_4... which is by the way appallingly badly chose notation)

    then according to you the triples are of the form if I can decipher you slightly ambiguous notation (it is not clear what the dots should represent in your expressions)


    y=2ab +a^2


    This observation is drawn from the trivial assertion that the squares are the sum of the odd numbers. And is a moderately interesting variation on the usual formula for generating primitve triples, however, I can't seem to find 5,12,13 as one such triple. wlog x=5, then a=2 b=1 is the only solution, and implies y=8.

    2. Your history lesson probably ought to include the notice that 5 years after he claimed to have a general proof, he produced a sepific proof for n=4 I think, which would be surprising if Fermat actually had a general result. Something which practically no serious mathematician believes. Besides, we'd all look idiotic if there were a proof from the book. Surely one can say that given the huge number of plausible but subtley flawed proofs (presumption of unique factoization) it is likely Fermat was also wrong.

    I haven't read the generalization to higher powers, give me a minute and I'll point out the first error.
  4. Feb 24, 2004 #3

    matt grime

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    Hmmm, not as easy as usual. Will think about it.

    OK, now I've got it. The detritus that must fit into the r^3 space. You never actually calculate what that is,and I'll not believe the proof until you do. I mean, you do some fiddling with it, and say it must do this that and the other, but never write down what it is. It's z^3-x^3 - x^2(z-y)-something else I can't be buggered to work out, and you claim that isn't possible because it reduces to a quadratic that can't be true. But hey, you don't do that at all, as far as I can tell.

    Also even if this is true for n=3, it doesn't imply a reductio ad absurdum because you can't from it deduce that n=4 reduces to n=3
    Last edited: Feb 24, 2004
  5. Feb 25, 2004 #4
    Hello digiflux, thanks for putting this interesting mathematics on the internet.

    I agree that Fermat could have been onto something...

    Here is a quote from the paper at www.fermatproof.com :

    So far, so good. The generalized Pythagorean equation is this:

    [a^2 - b^2]^2 + [2*a*b]^2 = [a^2 + b^2]^2

    All odd numbers, 2x+1, can be represented as a^2 - b^2

    [x+1]^2 - x^2

    x^2 + 2x +1 - x^2 = 2x+1

    The equation [a^2-b^2]^2 + [2*a*b]^2 = [a^2 + b^2]^2 generates all Pythagorean triples, but it is not the only one that does. There are other expressions and relations that can generate all Pythagorean triples, for example:

    An infinite stack of predictable algebraic expressions to generate
    positive integer solutions for the Pythagorean Theorem.

    a b c

    4n(n+1)-3 , 4(2n+1) , 4n(n+1)+5

    4n(n+2)-5 , 4(3n+3) , 4n(n+2)+13

    4n(n+3)-7 , 4(4n+6) , 4n(n+3)+25

    4n(n+4)-9 , 4(5n+10) , 4n(n+4)+41

    etc. etc. etc.

    a^2 + b^2 = c^2

    Very interesting at www.fermatproof.com

    2[5*1] + 1^2 = 11

    2[5*1] + 2[5]^2 = 60

    2[5*1] + 2[5*1]^2 + 1 = 61

    A + B = C

    2A + 2B + 1 = 2C + 1

    11^2 + 60^2 = 61^2

    3 = 2^2 - 1^2

    3^2 = 5^2 - 4^2

    3^3 = 14^2 - 13^2

    3^4 = 41^2 - 40^2 = 9^2

    3^p = A^2 - B^2

    [2x+1]^p = A^2 - B^2
    Last edited: Feb 25, 2004
  6. Feb 28, 2004 #5
    Thanks for looking at my father's proof. I have absolute confidence in his abilities. He was a genius who worked at Vought developing rocket missile systems. I on the other hand have a degree in Fine Arts...lol. Oh well. Please forward my father's link to others who are interested in a Fermat proof. Oh and sign the guestbook.

    Thanks Again...
  7. Feb 29, 2004 #6

    matt grime

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    But your confidence appears to be misplaced. As I said it contains two obviously false assertions about generating pythagorean triples and claiming a reductio ad absurdum proof. The proof in case n=3 is half reasoned and not convincing and appears to make some unjustified conclusions about possible cases. That isn't to say the n=3 argument is wrong, just that it needs more examing. If you are genuine and not just a troll you should look at it and think about it - it uses nothing more than high school maths.
    Last edited: Feb 29, 2004
  8. Feb 29, 2004 #7
    [zz)] [zz)] [zz)]

    Actually the proof does generate all Pythagorean triples.

    Look at it again.

    http://hometown.aol.com/parkerdr/math/pythagor.htm [Broken]


    Last edited by a moderator: May 1, 2017
  9. Feb 29, 2004 #8

    matt grime

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    please read the first reply i gave on this where i asked if the generation was of the form I rewrote without the ambibuigty in the decompositions into prime factors (P_1P_2.... is used as is p_1p_2p_3p_4... and p_3p_4... which is ambiguous as it is not clear what the ...means) where I prove that 5,12,13 is not of the form specified. or post a correction to my interpetation (which is just an interpretation; i'm prepared to accept i misread it).

    There is no dimensionless argument so it is not correct to argue reductio ad absurdum, do you dispute that? Given your own bizarre postings on FLT I'll require some direct refutation like posting the correct generating function (there are I believe many, and i could well accept that this is very close to one and it is just my pedantic reading of mathematics by a non-mathematician) - the spirit of its argument may be correct but the object he writes down at the end doesn't look correct.
  10. Feb 29, 2004 #9

    x = r + odd

    y = r + even

    z = r + even + odd

    Let a = 2

    Let b = 1

    r = 4 = 2*(2*1)

    x = 2ab + b^2

    y = 2ab + 2a^2

    z = 2ab + 2a^2 + b^2

    x = 2*(2*1) + (1)^2 = 5

    y = 2*(2*1) + 2*(2)^2 = 12

    z = 2*(2*1) + 2*(2)^2 + (1)^2 = 13

    I still am not sure about the reducto ad absurdum though [?]

    Very intersting still...
  11. Mar 1, 2004 #10

    matt grime

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    yes, i figured out shortly after i wrote it that it should be 2ab+b^2 and 2ab+2a^2, which works.

    It isn't relevant to the rest of the proof that it generates all triples anyway.

    The proof for n=3 ought to be rewritable in a purely algebraic form omitting the diagrams, where it ought to become clear if it is correct (my gut feeling is that it isn't).

    There is no obvious way from geometrical arguments to induct.
  12. Mar 4, 2004 #11
    o = odd, e = even

    oag +/- obg is not ocg

    eag +/- ebg = ecg can be divided with 2g.

    eag +/- ebg is not ocg.

    oag +/- obg = ecg (this must be proven wrong for g>2).

    ecg is devideable with 2nx.

    (oa/2x)g +/- (ob/2x)g = ocg

    Exchange oa with 2n - 1, and ob with 2s - 1.

    ((2n-1)a/2x)g +/- ((2s-1)b/2x)g = ocg

    Try to prove it with "binomials".

    It actually works.

    Try with g = 3 first.

    Since no 2gx higher than 4 can work, it's actually a proof.

    42 = 52 - 32 still works.

    12 = (5/4)2 - (3/4)2

    12 = ((2s - 1)/4)2 - ((2n - 1)/4)2

    ( n = 2, s = 3 then x = 2 and g = 2 )


    I sweare this can be used to solve Fermat's last theorem.
    Last edited: Mar 9, 2004
  13. Mar 4, 2004 #12
    Please, reply for me?
  14. Mar 4, 2004 #13

    matt grime

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    it's unclear what you are attempting to say, but it appears, that you've divided odd numbers by two so you're immediately out of the realm of FLT, and there ie nothing special about n=2 so you've proved that there are no pythagorean triples despite writing one down
  15. Mar 4, 2004 #14

    oan +/- obn = ecn. This must be proven wrong for n>2.

    all the other cases are impossible.

    ecn is divideable with 2n.

    e = even, o = odd
    Last edited: Mar 4, 2004
  16. Mar 4, 2004 #15

    matt grime

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    yes, that is rather obviously what you must prove. but you've not done so.
  17. Mar 4, 2004 #16
    Here you are deviding an odd number by a power of two. As a result, you are no longer dealing with whole numbers.
  18. Mar 4, 2004 #17
    well two odd numbers cannot become a third odd number.

    And two even numbers cannot become an odd.

    ofcourse two even numbers can become a third, then you can divide both sides with 2nx
  19. Mar 4, 2004 #18
    No, but now we are dealing with binomials instead!

    Ain't that so?
  20. Mar 4, 2004 #19
    I guess I am just dumb. Binomial coefficients are of the form

    [tex]\frac{a!}{b!\; (a-b)!} \equiv \left(\begin{array}[c] a a \\ b \end{array}\right)[/tex]

    What has that to do with Fermat or with your division by 2^x ?
  21. Mar 4, 2004 #20

    matt grime

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    what are you blathering about, sariaht?

    if p^n+q^n=r^n then we may assuming that after dividing out by the largest power of 2 as necessary that at most one of these numbers is even. yes we know this. that is why we often look for p,q,r coprime - if any two of p,q and r are divisible by a number so is the third, therefore we may divide through and reduce to the case of coprime numbers, in particular at most one can be even, at most one is a multiple of 3 etc.


    your reduction in the complexity of the problems doesn't reduce to anything at all that we didn't know already.

    so why is it that expanding

    (x/2+y/2)^n tells you anything?

    bearing in mind you're now in the ring Z[1/2]
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