[Math] Fermat’s last theorem over finite fields

diophantine equationsfinite-fieldsnumber theory

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.

Best Answer

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.

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