Fibonacci Sequence problem. Prove that there are infinitely many prime numbers such that $p$ divides $F_{p-1}$

I tried to use quadratic reciprocity, but I don’t understand how to use the explicit formulae to end the problem. I surmised that the primes that work are congruent to $\pm1\bmod5$, but I could not generalise. where $F_{p-1}$ is the p-1th term of the Fibonacci Sequence

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Given the explicit formula
$$ F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right], $$
as soon as $p>5$ is a prime for which $5$ is a quadratic residue (i.e. a prime $p\equiv \pm 1\pmod{5}$ by quadratic reciprocity) we have
$$ F_{p-1}\equiv 0\pmod{p} $$
as a consequence of Fermat’s little theorem.
$\left(\frac{5}{p}\right)=+1$ allows us to consider $\sqrt{5}$ as an element of $\mathbb{F}_p^*$.


In general, $\sqrt{5}$ is an element of $\mathbb{F}_{p^2}$, hence $p\mid F_{p^2-1}$ for any prime $p\neq 5$.
Additionally, if $p\neq 5$ does not divide $F_{p-1}$, it divides $F_{p+1}$.