Fibonacci numbers modulo $p$ If $p$ is prime, then $F_{p-\left(\frac{p}{5}\right)}\equiv 0\bmod p$, where $F_j$ is the $j$th Fibonacci number, and $\left(\frac{p}{5}\right)$ is the Jacobi symbol.
Who first proved this?  Is there a proof simple enough for an undergraduate number theory course?  (We will get to quadratic reciprocity by the end of the term.)

Here's a proof that only uses a little Galois theory of finite fields (and QR). I don't know if it's any of the proofs referenced by Gerry. Recall that
$$F_n = \frac{\phi^n - \varphi^n}{\phi - \varphi}$$
where $\phi, \varphi$ are the two roots of $x^2 = x + 1$. Crucially, this formula remains valid over $\mathbb{F}_{p^2}$ where $p$ is any prime such that $x^2 = x + 1$ has distinct roots, thus any prime not equal to $5$. We distinguish two cases:


*

*$x^2 = x + 1$ is irreducible. This is true for $p = 2$ and for $p > 2, p \neq 5$ it's true if and only if the discriminant $\sqrt{5}$ isn't a square $\bmod p$, hence if and only if $\left( \frac{5}{p} \right) = -1$, hence by QR if and only if $\left( \frac{p}{5} \right) = -1$. In this case $x^2 = x + 1$ splits over $\mathbb{F}_{p^2}$ and the Frobenius map $x \mapsto x^p$ generates its Galois group, hence $\phi^p \equiv \varphi \bmod p$. It follows that $\phi^{p+1} \equiv \phi \varphi \equiv -1 \bmod p$ and the same is true for $\varphi$, hence that $F_{p+1} \equiv 0 \bmod p$.

*$x^2 = x + 1$ is reducible. This is false for $p = 2$ and for $p > 2, p \neq 5$ it's true if and only if $\left( \frac{p}{5} \right) = 1$. In this case $x^2 = x + 1$ splits over $\mathbb{F}_p$, hence $\phi^{p-1} \equiv 1 \bmod p$ and the same is true for $\varphi$, hence $F_{p-1} \equiv 0 \bmod p$. 
The case $p = 5$ can be handled separately. Maybe this is slightly ugly, though. 
