On the Fibonacci sequence: is there an infinite number of primes $p$ dividing $F_{p-1}$? 
Let $\{F_n\}_{n\geq 0}$ be the Fibonacci sequence.
Prove that the number of primes $p$ so that $p\mid F_{p-1}$ is infinite.


I tried to use induction, to no avail.
 A: Assume that $p$ is a prime for which $5$ is a quadratic residue.
That is equivalent, by quadratic reciprocity, to $p\equiv\pm 1\pmod{10}$.
Since the explicit formula for Fibonacci numbers gives:
$$ F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right),$$
provided that $\sqrt{5}\in\mathbb{F}_p$, from Fermat's little theorem it follows that:
$$ F_{p-1}\equiv F_0 \equiv 0\pmod{p}$$
as wanted. We may recover an infinitude of odd primes $\equiv 1\pmod{5}$ by factoring $\Phi_5(n)=\frac{n^5-1}{n-1}=n^4+n^3+n^2+n+1$ for different values of $n$, for instance.
A: Hint 1: Prove that if the equation 
$$x^2-x-1=0 \pmod{p}$$
has two solutions $x_1 \neq x_2$ then there exists constants $C_1, C_2$ such that
$$F_n \equiv C_1x_1^n+ C_2x_2^n \pmod{p}$$
Hint 2: In this case what can you say about $F_{p-1}$ and $F_0$?
Hint 3: For $p >5$ the $$x^2-x-1=0 \pmod{p}$$
has two solutions $x_1 \neq x_2$ if and only if $5$ is a quadratic residue modulo $p$. 
Quadratic reciprocity tells you that this happens exactly when $p$ has certain remainders modulo $5$, and Dirichclet Theorem telss you that there are infinitely many such primes..
