I had seen an elementary proof for Fermat's last theorem at Quora.
I had checked all the steps (around one page only), where I couldn't catch any error, but I was confused about the last step only that includes the main idea that depends on a right angle triangle (in integers), are impossible to be with all sides being as powerful integers, the author claims this is too elementary to prove, but I don't see why this must be true?, there must be a counter example in large numbers (say more than six digits)! or this might have a simple proof as stated!
So, can we find such a counter example or prove it simply?
The Quora "proof" does not actually claim to be a proof of Fermat's Last Theorem - that there is no solution of $$x^n+y^n=z^n$$ for $n>2$.
It claims to be a proof that there is no solution of $$x^{p-1}+y^{p-1}=z^{p-1}$$ for $p-1>2$. Thus it would say nothing about (for example) $x^3+y^3=z^3$ or $x^5+y^5=z^5$.
So the result only deals with certain even values of $n$, and not with any of the odd ones.
The start is promising. Eliminating awkward cases by simple modular arguments has a long and respectable history.
However, although the arguments in part (a) are ingenious, they are not necessary. If it is true in general that $$(x^m)^2+(y^m)^2=(z^m)^2$$ has no solutions, then this does not depend on the parity of $x$, $y$, and $z$. In fact it doesn't even require $n=p-1$, only $n=2m$, and it is a good simple proof of the $n=2m$ case.
But I would appreciate a reference to the sides of a Pythagorean triangle not all being perfect $m$th powers.