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Diophantine equation x^4 - y^4 = 2 z^2

  1. Jun 23, 2010 #1
    I need to prove that the equation x^4 - y^4 = 2 z^2 has no positive integer solutions.

    I have tried to present this equation in some known from (like x^2+y^2=z^2 with known solutions, or [tex]x^4 \pm y^4=z^2[/tex] that has no integer solutions) without success.

    Any hints ?
     
  2. jcsd
  3. Jun 23, 2010 #2
    I think I have a solution. I'll post a hint and take another look tomorrow. Rewrite

    [tex]x^4 - y^4 = 2z^2[/itex]

    as

    [tex][(x + y)^2 + (x - y)^2](x + y)(x - y) = (2z)^2[/tex].

    Then let

    [tex] X = x + y, Y = x - y, Z = 2z[/tex]

    so we have

    [tex]XY(X^2 + Y^2) = Z^2[/tex].

    Show that this equation has no non-trivial solutions (several additional steps are still required).

    Petek
     
  4. Jun 24, 2010 #3

    disregardthat

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    I dont know if this works, but looking at different residues modulo 8 seems promising.
     
  5. Jun 24, 2010 #4

    CRGreathouse

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    Yes, it works.
     
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