Fibonacci Prime has Prime Index except for 3

Theorem

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

Let $F_n$ be a prime number.

Then, apart from $F_4 = 3$, $n$ is a prime number.

Let $F_n$ be a prime number.

Aiming for a contradiction, suppose $n$ is a composite number greater than $4$.

Then $n = r s$ for some $1 < r, s < n$.

Except for the case where $n = 4$, at least one of $r$ and $s$ is greater than $2$.

$F_r \divides F_n$

and:

$F_s \divides F_n$

where $\divides$ denotes divisibility.

When $k > 2$ we have that $F_k > 1$.

Thus when $n$ is composite such that $n > 4$, $F_n$ has at least one proper divisor.

That is, $F_n$ is not prime.

The exception is when $n = 4$, as noted above in which case its only proper divisor is $2$.

But $F_2 = 1$, and so divisibility by $F_2$ does not preclude primality.

Indeed, $F_4 = 3$, which is prime.

$\blacksquare$