Each prime $p$ not $2$ or $5$ divides $F_{p-1}$ or $F_{p+1}$, where $(F_n)$ is the Fibonacci sequence with $F_1=F_2=1$ 
Let $\{F_n\}$ - Fibonacci sequence: $F_1=F_2=1, F_{n+1}=F_n+F_{n-1}, n\ge2$ and $p -$ prime number, $p\not =2, p \not=5$. Prove that $p|F_{p-1}$ or $p|F_{p+1}$

My work so far.
I used formula $$F_p^2-1=F_{p-1}\cdot F_{p+1}.$$
But can not solve the problem.
 A: There are just two cases to study.
Case i): $5$ is a quadratic residue $\!\!\pmod{p}$. In such a case $x^2-x-1$ completely splits over $\mathbb{F}_p$ and $p$ divides $F_{p-1}$, since
$$ F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right). $$
Case ii): $5$ is not a quadratic residue $\!\!\pmod{p}$. In such a case $x^2-x-1$ is irreducible over $\mathbb{F}_p$ and $\mathbb{F}_{p^2}\simeq \mathbb{F}_p[x]/(x^2-x-1)$. The roots of $x^2-x-1$ can be seen as elements of $\mathbb{F}_{p^2}$ and by Frobenius' automorphism they are of the form $\alpha,\alpha^p$. By the closed formula
$$ F_{p+1} = \frac{1}{2\alpha-1}\left(\alpha^{p+1}-\alpha^{p^2+p}\right)$$
but since $\alpha^{p^2}=\alpha$, $p$ is a divisor of $F_{p+1}$.
A: Let $C_n^m=\binom{n}{m}$.
$p$− prime number, $p≠2,p≠5$.
$$F_{p-1}\cdot F_{p+1}=F_p^2-1$$
$$F_p=\frac{1}{2^{p-1}} \left ( C_{p}^1\cdot 5^0+C_{p}^3\cdot 5^1+C_{p}^{5} \cdot 5^{2}+...+C_p^p \cdot 5^{\left (\frac{p-1}{2}\right )}\right )$$
$p$ - prime number $ \Rightarrow C_p^k\equiv 0\bmod p$ for $k=1,2 . . .(p-1)$
$$2^{p-1}\cdot F_p=5^{\left (\frac{p-1}{2}\right )} \mspace{15mu} \bmod \mspace{15mu} p$$
$$2^{2(p-1)}\cdot F_p^2=F_p^2=5^{p-1}=1 \mspace{15mu} \bmod \mspace{15mu} p$$
A: An elementary proof inspired by the paper linked by Kelenner:
Lemme 1: Let $a\in \mathbb{F}_p, n\geq3 $, if the polynomial $x^2-x-1|x^{n+2}+a\pmod{p}$ then $F_{n+2}\equiv0\pmod{p}$
Proof: Calculating mod $p$: we have that $(x^2-x-1)\cdot(F_0x^n+F_1x^{n-1}+...+F_{n-1}x+F_n)=$
$=F_0x^{2n+2}+(F_2-F_1)x^{2n+1}+(F_3-F_2-F_1)x^{2n}+...+$
$+(F_{i+3}-F_{i+2}-F_{i+1})x^{2n-i}+...+(-F_{n}-F_{n+1})x-F_{n+1}=$
$=x^{2n+2}-F_{n+2}x-F_{n+1}$
So if $x^2-x-1|x^{n+2}+a$ then in particular $x^2-x-1$ divide the difference between the polynomials:
$x^2-x-1|(x^{n+2}+a-x^{2n+2}+F_{n+2}x+F_{n+1})$ so
$x^2-x-1|(F_{n+2}x+F_{n+1}+a)$
Since the polynomial on the right is of first degree it must be zero. In particular $F_{n+2}=0$ $\square$
Lemme 2: Let $p$ a prime, $p\neq5$, then either $x^2-x-1|x^{p-1}-1\pmod{p}$ either $x^2-x-1|x^{p+1}+1\pmod{p}$
*Proof:*Calculating mod $p$: Let be $\phi,\phi'$ the two roots of $x^2-x-1$ respect $\mathbb{F}_p$. If $p\neq 5$ then $\phi\neq\phi'$ and $\phi\cdot\phi'=-1$ (true in the integer case too). We have that $\phi^p,\phi'^p$ are also roots of $x^2-x-1$  since $\phi^{2p}-\phi^p-1=(\phi^2-\phi-1)^p=0$.
Then evaluating at the $p$-th power otherwise exchanges the two roots otherwise keeps them fixed. In the first case we have $\phi^p=\phi'=\phi$ so $\phi^{p+1}=\phi^{p}\cdot\phi=-1$. Then the roots of $x^2-x-1$ are roots of $x^{p+1}+1$ too so the first polyinomial divide the second one. Otherwise, in tthe second case we have $\phi^{p-1}=1$ so the roots of $x^2-x-1$ are roots of $x^{p-1}-1$ too and we conclude $\square$ 
