Let $f\in\mathbb{Z}[x]$ be monic of degree $n\ge 1$, and suppose $f$ factors as $$ f=(x-r_1)\cdots (x-r_n) $$ where $r_1,...,r_n$ are algebraic integers.
For each positive integer $k$, let $$ s_k=\sum_{i=1}^n r_i^k $$ By properties of symmetric functions of the roots, each $s_k$ is an integer.
We have the following analog of Fermat's little Theorem . . .
Claim:$\;$If $p$ is prime, then $p{\,\mid\,}(s_p-s_1)$.
Proof:
Let $p$ be a prime.$\;$Then \begin{align*} s_p-s_1&= \sum_{i=1}^n r_i^p-\sum_{i=1}^n r_i \\[4pt] &\equiv \left(\Bigl(\sum_{i=1}^n r_i\Bigr)^p-\sum_{i=1}^n r_i\right)\;(\text{mod}\;p) \\[4pt] &\equiv 0\;(\text{mod}\;p) \\[4pt] \end{align*} which completes the proof of the claim.
Next consider $(s_{p-1}\;\text{mod}\;p)$ . . .
If $p$ is prime and $r_1,...,r_n\in\mathbb{Z}$ are such that $p{\,\not\mid\,}r_i$ for all $i$, then by Fermat's little Theorem we have $s_{p-1}\equiv n\;(\text{mod}\;p)$.
I'll conjecture a sort of converse . . .
Conjecture:$\;$If the congruence $s_{p-1}\equiv n\;(\text{mod}\;p)$ holds for all but finitely many primes $p$, then $r_1,...,r_n\in\mathbb{Z}{\setminus}\{0\}$.
Question:$\;$Does the conjecture hold?