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Fermat's Last Theorem related question

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that [itex]x^{n}+y^{n}=z^{n}[/itex] has a nontrivial solution if and only if the equation [itex]\frac{1}{x^{n}}+\frac{1}{y^{n}}=\frac{1}{z^{n}}[/itex] has a nontrivial solution.

    2. Relevant equations
    By nontrivial solutions, it is implied that they are integer solutions.


    3. The attempt at a solution
    I was able to solve in one direction
    Given an integer solution to [itex]\frac{1}{x^{n}}+\frac{1}{y^{n}}=\frac{1}{z^{n}}[/itex]
    Then it follows that:
    [itex]\frac{x^{n}+y^{n}}{x^{n}y^{n}}=\frac{1}{z^{n}}[/itex]
    [itex]x^{n}y^{n}=z^{n}x^{n}+z^{n}y^{n}[/itex]
    [itex](xy)^{n}=(zx)^{n}+(zy)^{n}[/itex]
    Therefore, since the set of integers is closed under multiplication, then [itex]x^{n}+y^{n}=z^{n}[/itex] has a nontrivial solution.

    However, I can't seem to prove the other direction. Working the above proof backwards doesn't work unless I assume that d = gcd(x,y) and somehow prove that z = xy.
     
  2. jcsd
  3. Feb 26, 2013 #2

    haruspex

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    Applying exactly the same method in the reverse direction appears to work.
     
  4. Feb 26, 2013 #3
    I might note that ##\frac{1}{\frac{1}{x^n}}+\frac{1}{\frac{1}{y^n}}=\frac{1}{\frac{1}{z^n}}## is identical to writing ##x^n+y^n=z^n##. Then you can refer back to your earlier proof.
     
  5. Feb 26, 2013 #4

    tiny-tim

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    alternatively, since the formula you start with is symmetric, try doing something symmetric to it :wink:
     
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