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X^p + p*M + [Y-X] = Y^p

  1. Aug 29, 2004 #1
    A general equation for Y > X :

    X^p + p*M + [Y-X] = Y^p

    3^3 + 3*12 + 1 = 4^3


    3^3 + 3*32 + 2 = 5^3


    3^3 + 3*62 + 3 = 6^3


    3^3 + 3*104 + 4 = 7^3


    3^3 + 3*160 + 5 = 8^3


    3^3 + 3*232 + 6 = 9^3



    [...]
    4^3 + 3*20 + 1 = 5^3


    4^3 + 3*50 + 2 = 6^3


    4^3 + 3*92 + 3 = 7^3


    4^3 + 3*148 + 4 = 8^3


    4^3 + 3*220 + 5 = 9^3



    [...]

    X^p + p*M + N = Y^p


    It also works for p = 2,5,7, etc...?


    3^2 + 2*3 + 1 = 4^2


    3^2 + 2*7 + 2 = 5^2


    3^2 + 2*12 + 3 = 6^2


    3^2 + 2*18 + 4 = 7^2


    [...]


    4^5 + 5*420 + 1 = 5^5

    4^5 + 5*1350 + 2 = 6^5

    4^5 + 5*3156 + 3 = 7^5

    [...]


    X^p + p*M + [Y - X] = Y^p
     
  2. jcsd
  3. Aug 29, 2004 #2

    Zurtex

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    Correct me if I am wrong, but to prove that you need to show that for y > x that p is a factor of [itex]y^p - y - x^p + x[/itex]. I don't know how to do that, but perhaps you do.
     
  4. Aug 29, 2004 #3

    matt grime

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    Don't you? You just pick examples where it's true and don't bother about the obvious fact it's false.
     
  5. Aug 29, 2004 #4

    Zurtex

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    Really? I've not got something on me where I can easily write a program to find a counter example but running a few million values through Excel I can't find one.
     
  6. Aug 29, 2004 #5

    matt grime

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    actually, i'm doing a slight disservice there, since fermat's little theorem gives you that it is always true. i got carried away with the fact it was something russell wrote and was probably either trivial or false. had i looked more closely i'd have realized it was the first of those. if he ever actually explained using words what he meant he might be clearer
     
    Last edited: Aug 29, 2004
  7. Aug 29, 2004 #6

    X^p + p*M + [Y-X] = Y^p

    Y^p - X^p = p*M + [Y-X]

    p*M + [Y-X]

    is not a "pth" power for p > 2
     
  8. Aug 29, 2004 #7

    Zurtex

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    Eh? Matt said Fermat's little theorem not Fermat's last theorem.
     
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