Conjecture regarding Pisano period of prime numbers I observed an interesting pattern in the pisano periods of primes.
It seems to me that for any prime p such that p%10 = 3 or 7, its pisano period is a factor of p+1. And for any prime p such that p%10 = 1 or 9, the pisano period is a factor of p-1. 
I tested this for primes upto 6000 but I am not entirely convinced.
I need help with either proving this mathematically or finding a counter-example.
 A: This proof likely requires more number theory than I'm guessing you're familiar with, but in any case, here we go.
Let $p$ be an odd prime other than $5$. If we adjoin a root $\varphi$ of the polynomial $x^2 - x - 1$ to $\mathbb{F}_p$, we get either $\mathbb{F}_p$ (if this polynomial is reducible $\bmod p$) or the finite field $\mathbb{F}_{p^2}$ (if it's irreducible). Either way, write $\phi = - \varphi^{-1}$ for the other root. The proof of Binet's formula continues to go through $\bmod p$, and we get that
$$F_n = \frac{\phi^n - \varphi^n}{\phi - \varphi}$$
where this equality is to be understood as being in either $\mathbb{F}_p$ or $\mathbb{F}_{p^2}$ as above. (Exercise: what goes wrong with this argument when $p = 5$?)
Now, $x^2 - x - 1$ has discriminant $5$; it follows that it's irreducible $\bmod p$ if and only if $5$ is not a quadratic residue $\bmod p$, which by quadratic reciprocity occurs if and only if $p \equiv 2, 3 \bmod 5$. We now divide into two cases:
Case 1. $p \equiv 1, 4 \bmod 5$. Here $x^2 - x - 1$ is reducible, so $\phi, \varphi \in \mathbb{F}_p$. Then $\phi^{p-1} = \varphi^{p-1} = 1$ by Fermat's little theorem, so
$$F_{p-1} = \frac{\phi^{p-1} - \varphi^{p-1}}{\phi - \varphi} \equiv 0 \bmod p$$
$$F_p = \frac{\phi^p - \varphi^p}{\phi - \varphi} \equiv 1 \bmod p$$
from which it follows by induction (or repeated application of Fermat's little theorem) that $F_{(p-1) + k} \equiv F_k \bmod p$; that is, 

the Pisano period of a prime $p \equiv 1, 9 \bmod 10$ divides $p - 1$. 

Case 2. $p \equiv 2, 3 \bmod 5$. Here $x^2 - x - 1$ is irreducible, so $\phi, \varphi \in \mathbb{F}_{p^2}$. The Frobenius map in this case must swap them: that is, $\phi^p = \varphi$ and $\varphi^p = \phi$. This gives
$$F_p = \frac{\phi^p - \varphi^p}{\phi - \varphi} \equiv -1 \bmod p$$
$$F_{p+1} = \frac{\phi^{p+1} - \varphi^{p+1}}{\phi - \varphi} = 0 \bmod p$$
from which it follows by induction that

the Pisano period of a prime $p \equiv 3, 7 \bmod 10$ divides $2(p + 1)$. 

It's not true that the Pisano period divides $p + 1$; according to this table, the Pisano period of $13$ is $28 = 2(13 + 1)$. 
