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Math Help - 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
    Joined
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    Firstly, w^2=x^2+y^2+2xy, so z^2=2xy.

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

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

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

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

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