Some heuristics about the Pisano Period, primes and Fibonacci primes. What reasons are behind them? I started to read about the Pisano Period, $\pi(n)$, applied to the classic Fibonacci sequence and made some simple tests looking for possible properties of the sequence. I have observed the following ones, tested for the first 10000 terms:


*

*$\pi(n)=n-1 \implies n\in\Bbb P$


*$\pi(n)=(n-1)/2 \implies n\in\Bbb P$


*$\pi(n)=(n+1)\cdot 2 \implies n\in\Bbb P$


*$k \gt 5\ ,\ F_k \in \Bbb P \implies \pi(F_k)/4=k$

I do not understand the reasons for the results: points $1\sim3$ would work as a primality test, but it does not detect all the possible primes, only a subset of them, e.g. $\{2, 5, 47, 107, 113, 139,\ldots\}$ do not comply with points $1\sim3$ and are not detected. And specially the last point, if the test is correct, would mean that the Pisano period of a Fibonacci prime is exactly four times the index of the Fibonacci prime in the Fibonacci sequence when the index is greater than $5$ (being $F_5=5$) . For instance: $\pi(1597)= 68$ and $\frac{68}{4}=17$ which is exactly the index of $1597$ in the Fibonacci sequence, $F_{17}=1597$.
I would like to ask the following questions:

(a) Is there a counterexample? Initially I think the tests are correct, but I am not very sure about point 4. If somebody could confirm would be great.
(b) What are the reasons behind the observations? I guess that it is related with the relationship of the Pisano periods and the divisibility of the Fibonacci numbers by prime numbers.
(c) If the observations are correct, would we find pseudoprimes in the lists of primes detected by the rules $1 \sim 3$?

Probably the reasons behind the observations (if no counterexamples are found) are based on some simple properties of the Fibonacci numbers, but I do not see it clearly. Any hints or ideas are very welcomed. Thank you!
Update 2016/01/14: I have modified the information about point $4$ just to keep the correct information. After testing again, there are other $n$'s complying with $4$ and not being Fibonacci primes, so I have rewritten the statement: the Pisano period of a Fibonacci prime seems to be four times its Fibonacci index (position in the Fibonacci sequence) but that also holds for some other numbers.
Addendum: Below is the graph $n \rightarrow \pi(n)$ including the fist $100$ numbers showing the rules $1\sim3$. Rule $1$: $\color{red}{Red}$, Rule $2$: $\color{blue}{Blue}$, Rule $3$: $\color{green}{Green}$ (click to widen).

 A: I think first three statements are all false. Numbers with these properties are analogous to Fermat pseudoprimes, and in particular there's no reason to expect that they should in fact always be prime, although counterexamples might be quite large. 
Using Binet's formula as in Jyrki's comments, you can prove results like the following. Let $p \neq 5$ be a prime. We will need the Legendre symbol $\left( \frac{5}{p} \right)$, which is equal to $1$ if $p \equiv 1, 4 \bmod 5$ and $-1$ if $p \equiv 2, 3 \bmod 5$. For reasons that will become apparent I'll write $F_n$ as $F(n)$. 
First,
$$F(p) \equiv \left( \frac{5}{p} \right) \bmod p.$$
Next,
$$F \left( p - \left( \frac{5}{p} \right) \right) \equiv 0 \bmod p.$$
These are the two basic results, analogous to Fermat's little theorem. Together they allow you to bound the Pisano period of primes as follows: if $\left( \frac{5}{p} \right) = 1$, then the Pisano period divides $p - 1$. If $\left( \frac{5}{p} \right) = -1$, then the Pisano period divides $2(p + 1)$. 
This is a partial explanation of your first observation. For your second two observations we have the following slightly harder result. If $p \equiv 1 \bmod 4$, then
$$F \left( \frac{p - \left( \frac{5}{p} \right)}{2} \right) \equiv 0 \bmod p.$$
A: I think it is not hard to prove that. In fact, only by using the definition. For example, if $\pi(n)=n-1$ and $n=4k+1$, then $4k+1$ divides $F_{4k}$ and $4k+1$ divides $F_{4k+1}-1=F_{2k+1}L_{2k-1}$. Let $p$ be a prime dividing $4k+1$, then $\pi(p)\mid 4k$ and $\pi(p)\mid 2k+1$ (without loss of generality, since $\gcd(F_{2k+1},L_{2k-1})=1$). Thus, $pi(p)\mid 2(2k+1)-4k=2$. Thus, $\pi(p)=1$ or $2$ and the conclusion follows.
