$x^4 - y^4 = 2z^2$ has no solution How do I prove that the equation $x^4 - y^4 = 2 z^2$ has no solutions using the fact that the equations $x^4 + y^4 = z^2$ and $x^4 - y^4 = z^2$ have no solutions.
I cant think of a method of reducing the above equation to one of these forms.
 A: IDENTITY--The identity  $(X^4  - Y^4)^2  + 4(XY)^4   = (X^4 + Y^4)^2 $ solves the proposer's question.
The proposer assumes that there is a non-zero triple of integers X,Y,Z
such that $(X^4 - Y^4 ) = 2(Z^2)$.   WOLOG we may assume $(X,Y,Z) =1$.
We substitute " $2(Z^2)$ " for $(X^4  -Y^4)$ in the above IDENTITY.
This translates to $(2Z^2)^2  + 4(XY)^4  = (2B)^2$  where $X^4 + Y^4 = 2B$.
( Note that X,Y, are both odd as $X^4 = Y^4 = 2(Z^2)$ by hypothesis. Hence we may write $X^4 + Y^4) = 2B$ for some nonzero integer $B$ in the above IDENTITY.)
Dividing both sides by $4$ yields
$Z^4 + (XY)^4 = B^2$, which Fermat proved is impossible in non-zero integers $Z$,$(XY)$, and $B$  and is the connection that the proposer was seeking.
Many thanks to JW Tanner and Tinlyx for improving the formatting of this proof.
I wanted to share the above IDENTITY equation with the readership because it is an important tool to shorten and simplify Fermat-type equations of exponent $4$ which may or may not have a "$2$" as a coefficient.
The first answer given to this problem above is correct, but this solution is also very elegant, and simpler as well.
A: I think you meant nonexistence of positive solutions. 
Suppose that there exist some positive solutions(meaning that all of $x$, $y$, $z$ are positive). 
Then there is a positive solution $(x_0,y_0,z_0)$ with $x_0$ smallest. 
First, notice that parity of $x_0$ and $y_0$ cannot be different. 
So, either both even or both odd. Both even case is not possible, because otherwise
$(x_0/2,y_0/2,z_0/4)$ is a positive solution solution with smaller $x_0$. 
Similarly with odd prime $p$, suppose $p|gcd(x_0,y_0)$, then you also get smaller solution $(x_0/p, y_0/p, z_0/p^2)$. Thus we can assume $(x_0,y_0)=1$.  
Thus we can now assume that $x_0$ and $y_0$ are both odd, and coprime. 
Then by looking at the expression 
$$
\frac{x_0^2-y_0^2}{2}\cdot \frac{x_0^2+y_0^2}{2}=\frac{z_0^2}{2}
$$
The right side must be an integer, so it must be $2Z^2$ , and any prime $p$ cannot divide both $\frac{x_0^2-y_0^2}{2}$ and $\frac{x_0^2+y_0^2}{2}$(otherwise we would have some prime $p$ dividing both $x_0$ and $y_0$. Then we have 
$$
x_0^2-y_0^2 = u^2$$
, and 
$$x_0^2+y_0^2 = 2v^2$$
for some positive integers $u,v$. 
Now we solve for the Pythagorian triple in the first equation. 
$$
x_0=s^2+t^2\\
y_0=s^2-t^2\\
u=2st
$$
for some positive integers $s,t$. 
Then $x_0^2+y_0^2=2(s^4+t^4)=2v^2$. Hence we obtain $s^4+t^4=v^2$. However, this cannot have positive integer solution.  
