$x^4 - y^4 = 2z^2$ 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?
 A: 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 $4x^2y^2$.
A: Using
$$x^2+y^2=(x+y)^2-2xy\tag{$*$}$$
and similar things, show that $x^2+y^2$ and $x+y$ and $x-y$ are all odd, or all even.
They can't all be odd since their product is even.  So they are all even.
This also means that $z$ is even.
Using $(*)$ again, if $p\mid\frac{x^2+y^2}2$ and $p\mid\frac{x+y}2$, then $p$ is a factor of either $x$ or $y$, and hence of both, which is impossible.  Similar reasoning shows that
$$\frac{x^2+y^2}2\ ,\quad\frac{x+y}2\ ,\quad \frac{x-y}2$$
are pairwise relatively prime, and their product is a square.
I'm sure you can fill in the bits I have omitted and take it from here.
