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Thread: Diophantine Proof Help using Pythagorean Triples

  1. #1
    Member diddledabble's Avatar
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    Question Diophantine Proof Help using Pythagorean Triples

    Show that all solutions to x^2+y^2+z^2=w^2 with the restrictions

    w=x+y,
    x>0, y>0, z>0, w>0,
    (x,y,z,w)=1 and
    2|x (i.e. x is even)

    are given by

    x=2a^2,
    y=b^2,
    z=2ab,
    w=2a^2+b^2

    where a>0, b>0, (a,b)=1 and b is odd.
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  2. #2
    Member
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    Firstly, $\displaystyle w^2=x^2+y^2+2xy$, so $\displaystyle z^2=2xy$.

    Since $\displaystyle x$ is even let $\displaystyle x=2k$. Then $\displaystyle z^2=4ky$, so $\displaystyle z$ must be even, say $\displaystyle z=2m$.

    Now any common factor of $\displaystyle x$ and $\displaystyle y$ must be a factor of $\displaystyle w=x+y$ and of $\displaystyle z$, since $\displaystyle z^2=2xy$. But $\displaystyle (x,y,w,z)=1$, so $\displaystyle (x,y)=1$ must hold.

    Therefore $\displaystyle (k,y)=1$ also. But $\displaystyle 4ky=z^2=4m^2$, so $\displaystyle ky=m^2$. Therefore since $\displaystyle y>0$ and $\displaystyle (k,y)=1$ we have $\displaystyle k=a^2$ and $\displaystyle y=b^2$ with $\displaystyle (a,b)=1$.

    Therefore $\displaystyle x=2k=2a^2$ and $\displaystyle w=x+y=2a^2+b^2$.

    Finally $\displaystyle z^2=4ky=4a^2b^2$, so as $\displaystyle z>0$ take $\displaystyle z=2ab$ with $\displaystyle a>0$ and $\displaystyle b>0$ (letting $\displaystyle a<0$ and $\displaystyle b<0$ gives the same solutions.)
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