Fermat's last theorem over finite fields I am taking a course in number theory that deals with quadratic and (higher order) reciprocity and have been wondering about the following question: Fermat's last theorem states that there are no integer solutions to $x^n+y^n=z^n$ for $n>2$. Does a similar result exists over finite fields?
My first thought was looking$\mod p$ and then for $n$ such that $\gcd(n,p-1)=1$ the map $$\phi :F_p \rightarrow F_p$$ $$\phi(a)=a^n$$ is an automorphism and therefore a solution exists.
Also, for every finite field we could reduce the equation to $$1+y^n=z^n$$ by multiplying by $(x^{-1})^{n}$.
I can't seem to get much further and would be glad to hear about existing results on the subject.
 A: Fermat's last theorem is in general false in finite fields.
For example, in $\mathbb{F}_5=\mathbb{Z}/5\mathbb{Z}$, every element is a cube:


*

*$0^3=0$

*$1^3=1$

*$2^3=8=3$

*$3^3=27=2$

*$4^3=64=4$
So any two elements $x, y\in \mathbb{Z}/5\mathbb{Z}$ satisfy $x^3+y^3=z^3$ for some $z$.

More is true: Fermat's Little Theorem states that $a^p\equiv a$ mod $p$, for all $a$ and all primes $p$. So for all $x, y\in\mathbb{F}_p$, we have that $x^p+y^p=(x+y)^p$. Now choose a prime $p$ which is $>2$ . . .
And, in fact, if $\mathbb{F}$ is a finite field of characteristic $p$, the Frobenius map $x\mapsto x^p$ is an automorphism of $\mathbb{F}$ - so, again, we'll have $x^p+y^p=(x+y)^p$ for all $x, y\in\mathbb{F}$. So indeed Fermat's Last Theorem fails in every finite field of characteristic $\not=2$. See here for an explanation of why this is the case.
A: Schur proved that for every $n$, if $p$ is a large enough prime, then there is a nontrivial solution to $x^n+y^n\equiv z^n\bmod p$. See this link. 
