TITLE: Divisibility of Fibonacci numbers
QUESTION [6 upvotes]: This question is inspired by a Project Euler problem I was working on.  Noticing something that did not make sense led me to the conclusion that for all primes $p$ ending in $1$ or $9$, the $(p-1)$st Fibonacci number is divisible by $p$.  I haven't proven it, but it's the only conclusion that makes sense.  I'd give more information on how I arrived there but I'm afraid it might spoil the problem.
I have tested this for the first few values using an online Fibonacci calculator
$$F_{10}=55=5\times11$$
$$F_{18}=2584=136\times19$$
$$F_{28}=317811=10959\times29$$
$$F_{30}=832040=26840\times31$$
Why is this true, assuming it's true at all?

REPLY [2 votes]: By quadratic reciprocity, $x^2=5$ has a solution in $\mathbb F_p$ iff $y^2=p$ has a solution in $\mathbb F_5$, i.e. iff $p=5n\pm 1$. In that case we can substitute the solution of $x^2=5$ to $F_n=\frac{1}{\sqrt{5}}(a^n-b^n)$, where $a=(1+\sqrt{5})/2$, $b=(1-\sqrt{5})/2$. Since $c^{p-1}=1$ for every $c\in\mathbb F_p$, $c\neq 0$, we get that $F_{p-1}=0$ in $\mathbb F_p$.