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Fermat's theorem

  1. Aug 10, 2008 #1
    I'm talking about neither his "last theorem" nor his "little theorem", but another one. He suggested that

    x^2+2=y^3 can only have one solution (if we're dealing in natural numbers), which was (5,3). Euler reproved the theorem since, like so many others of his, the proof was lost. I can't find the proof now, but I remember it had to do with complex numbers and relative primes.

    I did some work with the theorem myself in my free time last summer, and I think I found a proof of it using infinite descent. I can't be certain my proof is correct since I haven't had the chance to have anyone review it, and I'm also not sure if anyone else has proven the theorem in a similar way. Does anyone know if it has? I'm especially proud because I think this is probably the way in which Fermat would have proven the theorem.
  2. jcsd
  3. Mar 28, 2016 #2
    Here are some interesting things I have discovered relative to FLT. Probably no proof, but strong evidence that it is true.

    Parents of Pythagonal and other triple integers

    This study is concerned with establishing the parent of a set of three different relatively prime integers A, B, C where A^n -B^n -C^n = D and D is a minimum, with a parent for every value of n greater than zero.

    With n = 1 the parent is 3, 2, 1 with D = 0
    n = 2 5, 4, 3 0
    n = 3 7, 6, 5 2
    n = 4 9, 8, 7 64

    with n = 3 there is a triple 9, 8, 6 D = 1, but I am excluding triples where A, B and C are not all relatively prime.

    The parent for n = 2 leads to all possible values with D = 0, and descendents can have A-B = 1, or B-C =1 or A-C = 2 but no more than one of these features of the parent.

    Similarly with n = 3 the parent leads to all possible values where D = 2

    With n = 4 all other values lead to D having a magnitude greater than 64.

    From these facts expressions for A, B, and C of the parent of n can be deduced as A = 2*n+1,
    B= 2*n, C = 2*n-1.

    Trials show that this is true for n = 6, 7, 9, and 10 but with n = 5 triple 17,16,13 gives D = -12 instead of 11,10,9 with D = 2002

    And with n = 8 triple 22,21,19 gives D = 69,451,134 instead of 17,16,15 with D = 1,117,899,520

    Question - are there more cases with n > 10 where D can be less than with its parent?

    Applying binomial expansions to the parent set with n = 5
    Note that the total of the first two terms cancel and odd terms after that are all zero. The total is always positive for all powers except 2 where only the first three terms exist.
    The total expansion with n > 2 is always of the order p - 3

    Applying binomial expansions to A = 3*n+1, B = 3*n+1, C = 3n with n = 5

    Here the first two terms cancel, also the third and fourth, so the total expansion is linear

    There does not seem to be an expansion for the particular n = 8 case above that makes its order less than n. It seems true that with n greater than 2 , the magnitude of D is always greater than one.
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