(I edited)
Let $\big(f_{n}\big)$ be the Fibonacci sequence which defined by $f_{n}=f_{n-1}+f_{n-2}$ with $f_{1}=1$ and $~f_{2}=1$.
Now I have deduced that for all prime $p\ge 7$, either $p~|~f_{p-1}$ or $p~|~f_{p+1}$.
Here's my attempt :
As prime $p\ge 7$, we see on account of the Binet formula that \begin{align*} f_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}=\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}}{\sqrt 5}=\frac{1}{2^{n}\sqrt{5}}\bigg(\left(1+\sqrt{5}\right)^{n}-\left(1-\sqrt{5}\right)^{n}\bigg)\tag{*}\\ \end{align*} and by the Binomail theorem that $$\left(1{\pm\sqrt{5}}\right)^{n}=\sum_{k=0}^{n}{{n}\choose{k}}\left(\pm\sqrt{5}\right)^{k}=1\pm{{n}\choose{1}}{\sqrt{5}}+{{n}\choose{2}}\left(\sqrt{5}\right)^{2}\pm{{n}\choose{3}}\left(\sqrt{5}\right)^{3}+\cdots+(-1)^{n}\left(\sqrt{5}\right)^{n}$$ From $(*)$, we know that if $n$ is odd then \begin{align*} 2^{n-1}f_{n}&=\frac{1}{2\sqrt{5}}\left(\left(1+\sqrt{5}\right)^{n}-\left(1-\sqrt{5}\right)^{n}\right)\\ &=\frac{1}{2\sqrt{5}}\left(\sum_{k=0}^{n}{{n}\choose{k}}\left(\sqrt{5}\right)^{k}-\sum_{k=0}^{n}{{n}\choose{k}}(-1)^{k}\left(\sqrt{5}\right)^{k}\right)\\ &=\frac{1}{2\sqrt{5}}\sum_{k=0}^{n}{{n}\choose{k}}\left(\left(\sqrt{5}\right)^{k}-(-1)^{k}\left(\sqrt{5}\right)^{k}\right)\\ &=\frac{1}{2\sqrt{5}}\sum_{m=0}^{\frac{n-1}{2}}{{n}\choose{2m+1}}2\left(\sqrt{5}\right)^{2m+1}\\ &=\sum_{m=0}^{\frac{n-1}{2}}{{n}\choose{2m+1}}\left(\sqrt{5}\right)^{2m}\\ &=\sum_{m=0}^{\frac{n-1}{2}}{{n}\choose{2m+1}}5^{m} \end{align*} Now let $n=p$ be an odd prime and since $p~|~{{p}\choose{i}}$ for all $1\le i<p$.
Then from above, one has that
$$2^{p-1}f_{p}\equiv 5^{\frac{p-1}{2}}~(mod~p)$$
But $gcd(p,2^{p-1})=1$ since $p$ is odd number, then we have
$$f_{p}\equiv 5^{\frac{p-1}{2}}~(mod~p)$$
and hence that
$$f^{2}_{p}\equiv 5^{p-1}\equiv 1~(mod~p)$$
, where the last modulo holds by Fermat's little theorem .
Note that $f_{p}^{2}-f_{p-1}f_{p+1}=1$.
Here's a detail: \begin{align*} f_{p}^{2}-f_{p-1}f_{p+1}&=\left(\frac{\alpha^{p}-\beta^{p}}{\alpha-\beta}\right)^{2}-\frac{\alpha^{p-1}-\beta^{p-1}}{\alpha-\beta}\frac{\alpha^{p+1}-\beta^{p+1}}{\alpha-\beta}\\ &=\frac{\alpha^{2p}-2\alpha^{p}\beta^{p}+\beta^{2p}}{\left(\alpha-\beta\right)^{2}}-\frac{\alpha^{p-1+p+1}-\alpha^{p-1}\beta^{p+1}-\beta^{p-1}\alpha^{p+1}+\beta^{p-1+p+1}}{\left(\alpha-\beta\right)^{2}}\\ &=\frac{\alpha^{2p}-2\alpha^{p}\beta^{p}+\beta^{2p}}{\left(\alpha-\beta\right)^{2}}-\frac{\alpha^{2p}-\left( \alpha^{p-1}\beta^{p+1}+\alpha^{p+1}\beta^{p-1}\right)\beta^{2p}}{\left(\alpha-\beta+\right)^{2}}\\ &=\frac{1}{\left(\alpha-\beta\right)^{2}}\bigg(-2\alpha^{p}\beta^{p}+\alpha^{p-1}\beta^{p+1}+\alpha^{p+1}\beta^{p-1}\bigg)\\ &=\frac{1}{\left(\alpha-\beta\right)^{2}}\bigg(\left(\alpha\beta\right)^{p-1}\beta^{2}-2(\alpha\beta)^{p}+(\alpha\beta)^{p-1}\alpha^{2}\bigg)\\ &=\frac{1}{\left(\alpha-\beta\right)^{2}}\bigg((-1)^{p-1}\beta^{2}-2(-1)^{p}+(-1)^{p-1}\alpha^{2}\bigg)\\ &=\frac{1}{\left(\alpha-\beta\right)^{2}}\bigg(\alpha^{2}+\beta^{2}+2\bigg)\\ &=\frac{1}{\left(\sqrt{5}\right)^{2}}(3+2)\\ &=1\\ \end{align*} Therefore, as $f_{p}^{2}\equiv 1$ (mod p) then we get
$$0\equiv f^{2}_{p}-1\equiv f_{p-1}f_{p+1}~(mod~p)$$
Whence,
$$\color{red}{either~~p~|~f_{p-1}~~or~~p~|~f_{p+1}}$$
since $$gcd(f_{p-1},f_{p+1})=f_{gcd(p-1,p+1)}=f_{2}=1$$ Note that as $p$ is odd then $p=2q+1$ for some $q\in{\bf N}$.
Then, $$gcd(p-1,p+1)=gcd(2q,2q+2)=gcd\left(2q,2(q+1)\right)=2$$ ,where the last equality is true since $gcd(q,q+1)=1$.
Question:
Is it possible to use this result (the red one) to show that if $p\equiv \pm1$ (mod$10$) then $p~|~f_{p-1}~?$
Any comment or advice I will be grateful. Thanks for patient reading .
Note that, in any field $K$ such that the quadratic $x^2-x-1$ has two distinct roots $\alpha$ and $\beta$, we must have $$f_n = \frac{\alpha^n-\beta^n}{\alpha-\beta}$$ for every integer $n\geq 0$. Let $p$ be a prime natural number. We shall work in the finite field $\mathbb{F}_{p^2}$ of order $p^2$. Note that the quadratic $x^2-x-1$ splits in the prime field $\mathbb{F}_{p}$ if and only if $p=5$ or $\left(\frac{5}{p}\right)=+1$ (or equivalently, $p \equiv \pm1\pmod{10}$). If the quadratic has identical roots in $\mathbb{F}_{p^2}$, then $p=5$ is the only viable choice; otherwise, $\alpha \neq \beta$.
If $p=5$, we see that $\alpha=\beta=3$ in $\mathbb{F}_{p^2}$. Thus, the recursion has a solution $f_n = 2n\cdot3^{n}$. This shows that $5\mid f_n$ if and only if $5 \mid n$.
If $p$ is prime such that $p \equiv \pm1\pmod{10}$, then $\alpha$ and $\beta$ are elements of $\mathbb{F}_p$. That is, $$f_{p-1} = \frac{\alpha^{p-1}-\beta^{p-1}}{\alpha-\beta} = \frac{1-1}{\alpha-\beta}=0\,.$$ Likewise, $$f_p=\frac{\alpha^p-\beta^p}{\alpha-\beta}=\frac{\alpha-\beta}{\alpha-\beta}=1$$ and $$f_{p+1} = \frac{\alpha^{p+1}-\beta^{p+1}}{\alpha-\beta} = \frac{\alpha^2-\beta^2}{\alpha-\beta}=\alpha+\beta=1\,.$$
If $p$ is prime such that $p \equiv \pm 3\pmod{10}$, then $\alpha$ and $\beta$ are two distinct elements of $\mathbb{F}_{p^2}$. Note that $$\alpha^p+\beta^p=(\alpha+\beta)^p=1^p=1$$ and $$\alpha^p\beta^p = (\alpha\beta)^p=(-1)^p=-1\,.$$ Therefore, $\alpha^p$ and $\beta^p$ are also roots of $x^2-x-1$. However, since $\alpha$ and $\beta$ are not elements of $\mathbb{F}_p$, we must have $\alpha^p = \beta$ and $\beta^p=\alpha$. (If $\alpha^p=\alpha$ or $\beta^p=\beta$, then they are roots of $x^p-x$, which means that they belong to $\mathbb{F}_p$.) That is, $$f_{p-1} = \frac{\alpha^{p-1}-\beta^{p-1}}{\alpha-\beta} = \frac{\beta\alpha^{-1} - \alpha\beta^{-1}}{\alpha-\beta} = \left(-\frac{1}{\alpha\beta}\right)\left(\frac{\alpha^2-\beta^2}{\alpha-\beta}\right) = 1(\alpha+\beta)=1$$ and $$f_p=\frac{\alpha^p-\beta^p}{\alpha-\beta}=\frac{\beta-\alpha}{\alpha-\beta}=-1\,.$$ Likewise, $$f_{p+1}=\frac{\alpha^{p+1}-\beta^{p+1}}{\alpha-\beta} = \frac{\beta\alpha^{+1}-\alpha\beta^{+1}}{\alpha-\beta} = \frac{(-1)-(-1)}{\alpha-\beta}=0\,.$$
P.S. Indeed, $f_{k\Big(p-\left(\frac{5}{p}\right)\Big)}=0$ in $\mathbb{F}_p$ for every $k=0,1,2,\ldots$ and for any prime natural number $p$.