<s>[INST] $x^4 - y^4 = 2z^2$ 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.
 [/INST] 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.
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