Prove that $p$ divides $F_{p-1}+F_{p+1}-1$ 
Given the Fibonacci sequence $(F_n)$, defined by $F_0=0,F_1=1, F_{n+2}=F_{n+1}+F_n$, and $p$ an odd prime number, how to prove that $p$ divides $F_{p-1}+F_{p+1}-1$?

Is induction a good idea here?
 A: Induction on primes is not going to be useful, because there isn't a simple relationship between one prime and the next.  The Binet formula might be more useful.  Note that there are two cases, depending on whether or not 5 is a square mod $p$ (if it isn't, you might want to work an extension field of the integers mod $p$).
A: Using 
$$F_{n}=F_{n-2}+F_{n-1}$$
 $$F_{2n}=F_{n}\left(2F_{n+1}-F_{n}\right)$$
 $$F_{2n}=F_{n}L_{n}$$
 where $L_{n}$
  is the n-th Lucas number, we have$$F_{p-1}+F_{p+1}=F_{p-1}+F_{p}-F_{p}+F_{p+1}=2F_{p+1}-F_{p}
 =\frac{F_{2p}}{F_{p}}=L_{p}$$
 and we have finished because, if $p$
  is prime, holds $$L_{p}\equiv1\,\textrm{mod}\, p.$$
A: There is an easier expression!
Let $G_n = F_{n+2} + F_{n} - 1$ be the sequence of numbers we are interested in. Note that 
$$
G_0 = F_2 + F_0 - 1 = 1 + 0 - 1 = 0\\
G_1 = F_3 + F_1 - 1 = 2 + 1 - 1 = 2
$$
and
$$
G_{n+2} = F_{n+2} + F_{n} - 1 \\
= F_{n+1} + F_{n} + F_{n-1} + F_{n-2} - 1 \\
= G_{n+1} + G_{n} + 1
$$
[Edit: My derivation hereafter was in error. Removed]
A: We can notice that
$$ F_{n-1}+F_{n+1} = L_n \tag{1}$$
holds by induction since equality holds for $n=1,2$ and both the LHS and the RHS satisfy the recurrence relation $f_{n+1}=f_{n}+f_{n-1}$. Given that, the problem boils down to proving that
$$ L_p \equiv 1\pmod{p}.\tag{2}$$
Due to Binet's formula, this is trivial if $5$ is a quadratic residue $\!\!\pmod{p}$. In the other case,
$$ \mathbb{F}_p[x]/(x^2-x-1)\simeq\mathbb{F}_{p^2}, \tag{3}$$
and the roots of $x^2-x-1$ are of the form $\xi,\xi^p$ by Frobenius' automorphism. On the other hand, $\xi+\xi^p=1$ by Viète's formulas, while $L_p=\xi+\xi^p$ by Binet's formula, hence $(2)$ holds also in that case.
