p is a prime and Let $F_{n}$ denotes the Fibonacci sequence.
I want to show the following properties of $F_p$:
$F_p\equiv\left (\frac{p}{5} \right ) (mod ~p)\tag{0}$
And for $F_{p+1}$:
$F_{p+1}\equiv1 ~(mod ~p)~~if~~p \equiv \pm ~1 (mod~5)\tag{1}$
$F_{p+1}\equiv0 ~(mod ~p)~~if~~p \equiv \pm ~2 (mod~5)\tag{2}$
For $F_{p-1}$:
$F_{p-1}\equiv0 ~(mod ~p)~~if~~p \equiv \pm ~1 (mod~5)\tag{3}$
$F_{p-1}\equiv1 ~(mod ~p)~~if~~p \equiv \pm ~2 (mod~5)\tag{4}$
So $F_{p-1}$ and $F_{p+1}$ look like "converse"
I have tried to start with $F_{0} = 0$ , $F_{1} = 1$, $F_{2} = 1$ and use the property of Fibonacci sequence such that: $F_{n} = F_{n-1}+F_{n-2}$ but I failed.
Any hints will be appreciated.
You may use the Binet Formula:
$$ F_n= \frac{ \Big(\frac{1+\sqrt{5}}{2}\Big)^n - \Big(\frac{1-\sqrt{5}}{2}\Big)^n} {\sqrt{ 5}},$$ from which, after expansion of the binomials, you have $$ F_n= \frac{1}{2^{n-1}}\sum_{k\ge0} {n \choose 2k+1}5^k,$$ and then you easily obtain the desired results by letting $n =p, p-1 $ and $p+1$ in the above sum and then reducing modulo $p$.