$x^4 - y^4 = 2z^2$ intermediate step in proof I am ultimately trying to prove, for an Exercise in Burton's Elementary Number Theory, that $x^4 - y^4 = 2z^2$ has no solution in the positive integers.
I can establish that if there is a solution, the solution with the smallest value of x has gcd(x,y)=1 
I see that $x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x + y)(x - y)$
I also have the fact that $uv = w^2$ and $gcd(u,v)=1$ implies $u$ and $v$ are each squares.
If I can establish (along the lines of the hint from the textbook) that 
 $(x^2 + y^2) = 2a^2$ , $(x + y) = 2b^2$ , and $(x - y)=2c^2$ for some integers a,b,c then I can derive a contradiction via a previous theorem.
However, I am stuck on how to establish the above equalities.  Can anyone offer me some direction?

If you want to establish your equalities,
Note that $\gcd(x,y)=1$. This implies that 
$$\gcd(x^2+y^2,x+y)=2,1$$$$\gcd(x^2+y^2,x-y)=2,1$$$$  \gcd(x-y,x+y)=2,1$$
This implies that either $x+y=2a^2, x-y=b^2,x^2+y^2=c^2$
Or that $x+y=a^2, x-y=2b^2, x^2+y^2=c^2$
Or that $x+y=a^2, x-y=b^2, x^2+y^2=2c^2$
Or that $x+y=2a^2, x-y=2b^2, x^2+y^2=2c^2$. I suggest you proceed from here. 
However, the shorter way would be to squaring both sides and adding $4x^2y^2$.
