Can Fermat's Little Theorem be applied to non-integer rational numbers?

Question

Let $m$ and $n$ be relatively prime integers, with $n \ne 0,1$, so that $a=m/n$ is a non-integer rational fraction. Let $p$ be an odd prime.

QUESTION 1: Can Fermat's Little Theorem be applied, i.e., can one say $a^p \equiv a\!\pmod{p}$, without any further consideration/explanation?

To that end,

QUESTION 2: Are there any proofs (preferably well-known results) that do this?

Answer 1

If $p$ divides $n$, $m/n \mod p$ makes no sense. If $p$ divides $m$, $a^p \equiv a \equiv 0 \mod p$. If $p$ divides neither, $a^{p-1} \equiv m^{p-1}/n^{p-1} \equiv 1 \mod p$.

Answer 2

It can be applied in the following sense.

• You need to assume that $p\nmid n$ (division by $p$ amounts to division by zero when doing arithmetic $\pmod p$, so is a no-no).
• We can work in the ring $$R_p=\{\frac mn\in\Bbb{Q}\mid \gcd(n,p)=1\}.$$
• This ring has an ideal $$P_p=\{\frac mn\in\Bbb{Q}\mid \gcd(n,p)=1,\ \text{$m$ is divisible by $p$}\}.$$ Basically the ideal consists of the numbers $px, x\in R_p$, so we can say that the ideal $P_p$ consists of the elements of $R_p$ that are divisible by $p$.
• We have $$x^p\equiv x\pmod{P_p}$$ for all $x\in R_p$. In other words, the difference $x^p-x$ is an element of $P_p$ (divisible by $p$ if you wish).

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Alternatively (when $\gcd(n,p)=1$) the residue class of $n$ has an inverse modulo $p$. IOW there exists an integer $n'$ such that $nn'\equiv1\pmod p$. We can then define $m/n=mn'$ (think: $m/n=m\cdot n^{-1}$). Of course, the Little Fermat relation $a^p\equiv a\pmod p$ holds for $a=mn'$, and in this sense the result also makes sense.