Prime Number divides Fibonacci Number

Theorem
For $n \in \Z$, let $F_n$ denote the $n$th Fibonacci number.

Let $p$ be a prime number.

Then:
 * $p \equiv \pm 1 \pmod 5 \implies p \mathrel \backslash F_{p - 1}$
 * $p \equiv \pm 2 \pmod 5 \implies p \mathrel \backslash F_{p + 1}$

where $\backslash$ denotes divisibility.

Thus in all cases, except where $p = 5$ itself:
 * $p \mathrel \backslash F_{p \pm 1}$

Proof
It is worth noting the one case where $p = 5$:
 * $5 \mathrel \backslash F_5 = 5$