Cyclotomic polynomial in $\mathbb{Z}/(p)$

Question

Let $p$ be prime, $n \in \mathbb{N}$ and $p \nmid n$.

$\Phi_n$ is the $n$-th cyclotomic polynomial.

How can I find the maximum $n \in \mathbb{N}$ (with $p \nmid n)$ so that $\Phi_n$ splits into linear factors over $\mathbb{Z}/(p)$.

Answer 1

A root of $\Phi_n$ is an element $a$ such that $a^n=1$ but $a^d\ne 1$ for all $d\mid n$, $d<n$. This suggests $n=p-1$.

Answer 2

A heavy artillery answer: for $p\nmid n$, $\Phi_n$ splits completely over $\mathbf Z/p\mathbf Z$ if and only if $(p)$ splits completely in $\mathbf Q(\zeta_n)$, if and only if the Frobenius at $p$ is trivial, if and only if $p\equiv 1 \pmod{n}$. The maximal such $n$ is therefore $p-1$.