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

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

    But note that $\displaystyle z^2+4y^4=x^4\ \implies\ z,\ 2y^2,\ x^2$ form a Pythagorean triple. This would imply that $\displaystyle x$ 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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