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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.

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    Such a result would imply Fermat's Last Theorem, so that there are two possibilities: it is true and has a terribly hard proof, or it is false. The second thing happens, for example $$7^3+9^3=3^3 \pmod{11}$$ – 2017-01-08
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    It is quite easy to construct counterexamples, indeed. Pick $a,b,c,n$ ANY four positive integers $\ge 3$, and let $p$ be any prime divisor of $a^n+b^n-c^n$. Then you have $$a^n+b^n=c^n \pmod{p}$$ by construction. – 2017-01-08
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    @Crostul as 3 is relatively prime to 11 this is exactly the example I mentioned. I don't see how it implies Fermat's last theorem. – 2017-01-08
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    @yoni It implies the **failure** of Fermat's Last Theorem - it says there is a **counterexample**. In fact, you've already shown, in your question, that FLT does not hold in finite fields, but I don't think you've realized that. – 2017-01-08
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    @DustanLevenstein right. corrected – 2017-01-08
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    @yoni Well, you've still said it's an automorphism, which is, as Noah said, a benign falsehood for the purpose of this problem. – 2017-01-08
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    @yoni Do you understand that the fact that (for appropriate $p, n$) $a\mapsto a^n$ is a bijection, already implies that Fermat's Last Theorem **fails** in general $\mathbb{F}_p$, since it implies that - for all $x, y\in\mathbb{F}_p$ - we have $x^n+y^n$ is an $n$th power? – 2017-01-08
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    See http://math.stackexchange.com/questions/1515236/last-fermats-theorem-modulo-m and http://math.stackexchange.com/questions/697685/fermats-last-theorem-fails-in-mathbbz-p-mathbb-z-for-p-sufficiently-larg – 2017-01-08
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    @yoni: The usual analog of FLT to consider is not whether it holds for finite fields, but for the polynomial rings $\mathbb{F}_p[x]$ or algebraic extensions thereof. – 2017-01-08
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    @Hurkyl could you elaborate or share a link? – 2017-01-08

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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.

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    Thanks, that's the sort of thing I was looking for – 2017-01-08
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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.