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Math Help - Diophantine equation

  1. #1
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    Diophantine equation

    How to solve the following Diophantine equation?
    a^2+b^2+c^2=a^2b^2

    Thank you in advance!
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  2. #2
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    Quote Originally Posted by doug View Post
    How to solve the following Diophantine equation?
    a^2+b^2+c^2=a^2b^2
    That equation has no integer solutions. Remember that x^2 is congruent to 0 (mod 4) if x is even, and congruent t0 1 (mod 4) if x is odd.

    So a^2b^2\equiv0\pmod4 unless a and b are both odd. But if a and b are both odd the a^2+b^2+c^2 will be congruent to 2 or 3 (mod 4), so that case cannot arise.

    Therefore a^2b^2\equiv0\pmod4 and hence a^2+b^2+c^2\equiv0\pmod4. But that implies that a, b and c are all even, say a=2a', b=2b', c=2c'. Then the equation becomes a'^{\,2}+b'^{\,2}+c'^2=4a'^{\,2}b'^{\,2}. But this time, the right side is obviously congruent to 0 (mod 4), so a', b' and c' must all be even.

    That argument can be continued indefinitely, so by Fermat's method of infinite descent it follows that there are no solutions.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    That equation has no integer solutions. Remember that x^2 is congruent to 0 (mod 4) if x is even, and congruent t0 1 (mod 4) if x is odd.

    So a^2b^2\equiv0\pmod4 unless a and b are both odd. But if a and b are both odd the a^2+b^2+c^2 will be congruent to 2 or 3 (mod 4), so that case cannot arise.

    Therefore a^2b^2\equiv0\pmod4 and hence a^2+b^2+c^2\equiv0\pmod4. But that implies that a, b and c are all even, say a=2a', b=2b', c=2c'. Then the equation becomes a'^{\,2}+b'^{\,2}+c'^2=4a'^{\,2}b'^{\,2}. But this time, the right side is obviously congruent to 0 (mod 4), so a', b' and c' must all be even.

    That argument can be continued indefinitely, so by Fermat's method of infinite descent it follows that there are no solutions.

    But the trivial one (0,0,0).

    Tonio
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