Fermat's Last Theorem where $n$ is a power of $2$

Question

I have seen the proof that Fermat gives for $$x^4 +y^4 \neq z^2$$ which we know also works for $z^4$.

BUT I am wondering if the same basic argument can be used for the power of $2^n$.

Thinks 8,16,32

Can we write a proof saying $$x^8+y^8 \neq z^8$$ More generally written as: $$x^{2^n} +y^{2^n} \neq z^{2^n}$$

Anyone have a proper way to go about this?? Maybe some helpful links?

Answer 1

We don't need to write a separate argument. If $x^8+y^8=z^8$ then $(x^2)^4+(y^2)^4=(z^2)^4$. But if $a^4+b^4=c^4$, then one of $a$ or $b$ is $0$.

More generally, the result about $x^4+y^4=z^2$ shows immediately that there are no non-trivial solutions for $x^{4a}+y^{4b}=z^{2c}$.