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Math Help - x^2+y^2 = 7z^2

  1. #1
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    x^2+y^2 = 7z^2

    Find all solutions to the Diophantine equation x^2+y^2 = 7z^2.

    Please teach me how to solve it . Thank you.
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  2. #2
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    Quote Originally Posted by beta12 View Post
    Find all solutions to the Diophantine equation x^2+y^2 = 7z^2.

    Please teach me how to solve it . Thank you.
    There are no solutions to this diophantine equation.

    7 is a number of the form 4k+3

    z^2 can have only two form, 4k,4k+1.

    Therefore,
    7z^2 has the form of 4k+3.

    The important thing here is that the sum of two squares,
    x^2+y^2 can never take the form 4k+3.

    Thus, there are no solutions to this equation.
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  3. #3
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    Hi Perfecthacker,

    I got it . Thank you very much.
    ============================
    Also, how should I prove the below question?

    Prove Fermat's last theorem for exponent three; i.e. prove that if

    x^3 + y^3 = z^3

    where x, y, and z are rational integers, then x, y, or z is 0.

    Hint: show that x^3 + y^3 = (epsilon)z^3 , where x, y, and z are quadratic integers in Q[-3], and epsilon is a unit in Q[-3] , then x, y, or z is 0.
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