# Math Help - Diophantine Proof Help using Pythagorean Triples

1. ## 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.

2. 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.)