Fibonacci Number whose Index is Plus or Minus Prime p is Multiple of p

Theorem
Let $p$ be a prime number distinct from $5$.

Let $F_n$ denote the $n$th Fibonacci number.

Then either $F_{p - 1}$ or $F_{p + 1}$ (but not both) is a multiple of $p$.

Proof
First we consider the edge cases:

For $p = 2$ we have that:
 * $F_{2 + 1} = F_3 = 2$ which is a multiple of $2$.
 * $F_{2 - 1} = F_1 = 1$ which is not a multiple of $2$.

For $p = 3$ we have that:
 * $F_{3 + 1} = F_4 = 3$ which is a multiple of $3$.
 * $F_{3 - 1} = F_2 = 1$ which is not a multiple of $3$.

Let $p$ be an odd prime.

From Cassini's Identity:
 * $(1): \quad F_{p - 1} F_{p + 1} - {F_p}^2 = \left({-1}\right)^p = -1$

Then:

Hence either $F_{p - 1}$ or $F_{p + 1}$ is a multiple of $p$.

But as $F_{p + 1} = F_p + F_{p - 1}$, only one of them can be.

Finally, note that if $p = 5$:
 * $F_{5 + 1} = F_6 = 8$ which is a multiple of $5$.
 * $F_{5 - 1} = F_4 = 3$ which is not a multiple of $5$.