Determine whether all characteristic factors of the $n$th Fibonacci number, which are primes $p_1, p_2,..., p_k$ such that $p_i \mid F_n$ and $p_i \not\mid F_m \hspace{3 mm} \forall i \in [1,k], m \in [1,n)$, we have $p_i \equiv \pm 1 \bmod n \hspace{3 mm} \forall i \in [1,k]$.
Second, I notice that for each Fibonacci prime $F_p$, if $q$ is a prime factor of either of $F_{p \pm 1}$, then $F_p \equiv \pm 1 \bmod q$.
----------The following has been shown to be false----------
Lastly, $F_n \equiv \pm 1 \mod n$ iff n is prime or twice a prime, with the exceptions $F_5 \equiv 0 \mod 5$ and $F_{10} \equiv 5 \mod 10$. These are patterns I found by examining a list of Fibonacci number factorizations, and I'm not sure where to start to prove any of them.
Edit: @Dietrich Burde has provided the correct form of the third congruence. The first two are still open.
Hoggat and Bicknell proved in $1974$ that $F_p\equiv \pm 1(p)$ if and only if $p\neq 5$ is a prime. More precisely, $F_p\equiv \left( \frac{p}{5}\right) \mod p$ with the Legendre symbol. So this is known. We also have for an odd prime $F_{2p}\equiv (\frac{p}{5})\equiv (\frac{5}{p}) \mod p$, since for an odd prime $(5/p)=(p/5)$. This is not true in general mod $2p$. There are a lot of other congruences discussed in the literature, but one has to be a bit careful sometimes with the "proofs".