Question regarding Fermat's last theorem for n = 4

Question

I am reading through a proof of Fermat's last theorem for $n=4$ and I see this statement...

"Since $x^4+y^4=z^4$ has a nontrivial solution, then $x^4-y^4=z^2$ also has a solution."

I have tried to justify this statement, but keep getting stuck. Why is this statement true? Can you help show me?

Answer 1

If you have a solution $x^4+y^4 = z^4$ you can rearrange it to $\underbrace{z^4}_{X^4}-\underbrace{y^4}_{Y^4} = x^4 = \underbrace{(x^2)^2}_{Z^2}$, and with the substitution $x^2 \mapsto Z$, $y \mapsto Y$ and $z \mapsto X$ you get the desired $X^4 - Y^4 = Z^2$.

Answer 2

Let $(a,b,c)$ be such that $a^4+b^4=c^4$ then if you take $(x,y,z)=(c,b,a^2)$ you have $z^2+y^4=x^4\iff x^4-y^4=z^2$.