Diophantine Proof Help using Pythagorean Triples

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.)