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X^3 + y^3 = z^3 ?

  1. Sep 26, 2012 #1
    we know the equation above does not have integer solutions ( Fermat last theorem having been proven few years back ). I am wondering if the following is a new way to prove that.

    let y = x +α and z = x + β
    now by simple substitution we get the following cubic equation:

    x^3 + 3(α-β)x^2 + 3(α^2 - β^2)x + α^3 - β^3 = 0

    So all that is left to prove is that the above cubic equation can never have integer solutions.
    Here we can try to use the work done by Konstantine Zelator here

    http://arxiv.org/ftp/arxiv/papers/1110/1110.6110.pdf

    where he gives conditions for a cubic polynomial to have integer roots.

    The problem is that I am having a bit of a problem applying his theorem to the case above. Any help will be greatly appreciated.

    Another question that comes to mind is: Can this method be generalized to the case with exponent greater than 3?
     
  2. jcsd
  3. Sep 27, 2012 #2
    So why am I not getting any feedback?

    1: first about the idea itself, that we can in fact transform the original Fermat equation into a cubic equation
    2: that it is just a matter of doing the algebra to prove it.

    if 1: is wrong, there is no point in considering point 2:

    And something I should have added in the original post
    α < β
     
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