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Math Help - diophantine equation problem

  1. #1
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    diophantine equation problem

    Solve: x^2 - 7y^2 = 3z^2

    I set it congruent (mod 3) to reduce it to x^2 - y^2 \equiv 0 (mod 3)

    This does not fit the sum of two squares theorem and I can not seem to find any difference of two squares theorem. Also, I can not make -y^2 \equiv a^2 (mod 3) for any a, so I can not make it sum of two squares that way either.
    The answer is supposed to be x=y=z=0, which is obvious. But setting it congruent (mod 3) seems to give me additional answers. How do I go about solving this problem? Thanks
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    Hello,
    Quote Originally Posted by Proggy View Post
    Solve: x^2 - 7y^2 = 3z^2

    I set it congruent (mod 3) to reduce it to x^2 - y^2 \equiv 0 (mod 3)

    This does not fit the sum of two squares theorem and I can not seem to find any difference of two squares theorem. Also, I can not make -y^2 \equiv a^2 (mod 3) for any a, so I can not make it sum of two squares that way either.
    The answer is supposed to be x=y=z=0, which is obvious. But setting it congruent (mod 3) seems to give me additional answers. How do I go about solving this problem? Thanks
    Set it congruent (mod 4) (in my opinion, it's not good making disappear a variable, but my knowledge in diophantine equations is not wide enough to be certain)

    -7 \equiv 1 \bmod 4 and 3 \equiv -1 \bmod 4.

    The equation becomes : x^2+y^2=-z^2 \implies x^2+y^2+z^2=0

    This means that x \equiv y \equiv z \equiv 0 \bmod 4

    x=y=z=0 is obviously a solution.

    Now, if they're all \neq 0, let's say x=4x', y=4y' and z=4z', the equation is :
    16x'^2-7*16y'^2=16*3z'^2 \implies x'^2-7y'^2=3z'^2 and we're back to the same equation...
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