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Fermat's Theorm extension

  1. Jan 29, 2008 #1
    Is it true that [tex]2z^n(z^n + x^n + y^n)[/tex] can never be a perfect square if n is a prime greater than 2 and x,y,z are prime to each other?
     
  2. jcsd
  3. Jan 29, 2008 #2

    CRGreathouse

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    If there is a perfect square of the form [itex]2x^p(x^p+y^p+z^p)[/itex] for p an odd prime, then [itex]2x(x^p+y^p+z^p)[/itex] is also a perfect square, so consider that instead.

    Since x, y, and z are coprime, at most one can be odd. If all three were odd then [itex]2x(x^p+y^p+z^p)\equiv2\pmod4[/itex] and so it is not a perfect square. Thus exactly one of x, y, and z is even -- and without loss of generality, we can assume that z is odd.
     
  4. Jan 30, 2008 #3
    ramsey2879: Is it true that [tex]2z^n(z^n + x^n + y^n)[/tex]
    can never be a perfect square if n is a prime greater than 2 and x,y,z are prime to each other?


    If x^p + y^p = z^p, then [tex]2z^p(z^p + x^p + y^p)[/tex]
    =[tex]2z^p(2z^p)[/tex]

    However, does this imply the converse? Not at all, consider this case of cubes:
    x=2, y=3, z=5:

    [tex](2)(5^3)(2^3+3^3+5^3)=250x 160 = 25(100)(16)=(200)^2[/tex]

    Or look at: [tex](2)(19^3)(5^3+14^3+19^3) =(2)(19^4)(151+19^2)=(2^{10})(19^4)[/tex]
     
    Last edited: Jan 30, 2008
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