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Math Help - [SOLVED] Diophantine equations

  1. #1
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    [SOLVED] Diophantine equations

    Prove that there are no positive solutions to
    x^4 - 4y^4 = z^2.

    (Hint: Use the previous equation to first compute z^4 + (2xy)^4)

    You assume that there are positive solutions, but I still don't understand the hint. Any help would be great. Thanks!
    Last edited by mr fantastic; May 22nd 2009 at 03:07 PM. Reason: Question restored
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by curiousmuch View Post
    Prove that there are no positive solutions to
    x^4 - 4y^4 = z^2.

    (Hint: Use the previous equation to first compute z^4 + (2xy)^4)

    You assume that there are positive solutions, but I still don't understand the hint. Any help would be great. Thanks!
    What’s your previous equation?

    Suppose there are positive solutions x,y,z to the equation. We may assume WLOG that \gcd(x,y)=1. For if x=kx_0,\,y=ky_0 for some prime k, then z^2=k^4(x_0^2-4y_0^2) so if we write z=k^2z_0 we end up where we started.

    If k is an odd prime dividing both x^2-2y^2 and x^2+2y^2, then k would divide both their sum 2x^2 and their difference 4y^2. It would follow that k divides both x and y, contrary to \gcd(x,y)=1. Hence x^2-2y^2 and x^2+2y^2 are either 1 or powers of 2. It is impossible for both of them to be 1 as it would imply that y=0. Neither is it possible for one of them to be 1 and the other to be a power of 2, since the parity of x would then be indeterminate. Hence both x^2-2y^2 and x^2+2y^2 are powers of 2. This implies that x is even.

    But note that z^2+4y^4=x^4\ \implies\ z,\ 2y^2,\ x^2 form a Pythagorean triple. This would imply that x must be odd instead.

    This contradiction means that there are no positive solutions to the original equation.
    Last edited by TheAbstractionist; May 20th 2009 at 12:38 AM.
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  3. #3
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    Thanks for the help! The previous equation simply refers to x^4 - 4y^4 = z^2.
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  4. #4
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    Also, on a completely different topic: what is the sum of the series (-10)^n
    where n goes from 0 to infinity when looked at 2-adically and 5-adically?

    2-adically, I got that the sum of the series is 2
    5-adically, I got that the sum of the series is 5/4.

    Are these correct?
    Last edited by mr fantastic; May 22nd 2009 at 03:06 PM. Reason: Original post restored
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