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So I've been studying the FLT and I am able to prove by descent that there are no solutions with $x,y,z$ natural numbers to $x^4+y^4=z^2$

I am trying to now prove by descent that there are no solutions with $x,y,z$ natural numbers to the equation $x^4+2y^4=z^2$.

I get to the point where i've labelled $X=x^2, Y=y^2$ and $Z=z$ to give $X^2+2Y^2=Z^2$.

So now in the case that $y$ is even meaning that $x$ is odd, then we have there $\exists a,b$ of opposite parities with $gcd(a,2b)=1$ such that

$$ X=|a^2-2b^2| $$ $$ Y=2ab $$ $$ Z=a^2+2b^2 $$

This gives the possibility of two Pythagorean triples of $x^2 +2b^2=a^2$ and $x^2+a^2=2b^2$.

I'm unsure if what I've done so far is correct and then what the next step would be.

Thanks for your help

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