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Thread: Fermat Last Theorem

  1. #1
    Junior Member
    Sep 2010

    Fermat Last Theorem


    I am not sure if I am approaching the problem below in the right way. Is there any chance that someone can check it for me,please?

    Show that the equation

    x^2=3(y^2 + z^2) has \ no \ zero \ integer \ solution.

    I assumed by contradiction that  (x,y,z) is an integer solution to our problem and it has the smallest value of z of any such solution.
    We can also note that (x, \sqrt{3}y, \sqrt{3}z) is in fact a Pythagoran triple as 3y^2+3z^2=x^2.
    Suppose that
    \sqrt{3}y=2st\\ \sqrt{3}z=s^2+t^2\\x=s^2+t^2

    for some co-prime integers s \ and \ t. Let s be even and t be odd, but that also imlpies that either s or t is divisible by  \sqrt{3}. Hence contradict the fact that they are integer.

    Thank you
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  2. #2
    MHF Contributor
    Dec 2012
    Athens, OH, USA

    Re: Fermat Last Theorem

    If I were grading your answer, I wouldn't accept it. What do you mean by \sqrt3 divides an integer?

    Suppose x, y and z satisfy the equation. As you point out, we can assume gcd(x,y,z)=1. Clearly 3 divides x, so write x=3x_0, and the equation becomes 3x_0x_0=y^2+z^2. So 3 divides y^2+z^2. Since gcd(x,y,z)=1, 3 does not divide both y^2 and z^2. So 3 can divide neither. But then both y^2 and z^2 must be 1 mod 3, and so y^2+z^2 is 2 mod 3, and not 0. Contradiction.
    Thanks from Gibo
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