Fibonacci $\equiv -1 \mod p^2$

The set of such primes $p$ is probably infinite but very sparse, and there are no such $p < 2 \cdot 10^{14}$.

We show that $p$ must be a "Fibonacci-Wieferich prime", i.e. a prime for which $F_k \equiv 0 \bmod p^2$ for some $k \not\equiv 0 \bmod p$; as with ordinary Wieferich primes (primes such as $1093$ and $3511$ for which $2^p \equiv 2 \bmod p^2$), the number of such $p \leq x$ is expected to grow as $\log \log x$ as $x \rightarrow \infty$. Conversely, any Fibonacci-Wieferich prime $p$ will admit a congruence $F_{np} \equiv -1 \bmod p^2$.

Suppose $p>5$. Recall that $F_m = (\varphi^m - \overline\varphi^m) / \sqrt{5}$, where $\varphi, \overline\varphi = (1 \pm \sqrt5) / 2$ with $\varphi \overline\varphi = -1$. Hence if $F_m \equiv -1 \bmod p^2$ then $\varphi^m \bmod p^2$ is a root of $X^2 + \sqrt5 \, X - (-1)^m = 0$. Thus if $m$ is odd then $$ \varphi^m = \frac{-1 \pm \sqrt{5}}{2} = -\varphi \ \ \text{or} \ \ \varphi^{-1}, $$ while if $m$ is even then $$ \varphi^m = \frac{-3 \pm \sqrt{5}}{2} = -\varphi^2 \ \ \text{or} \ \ -\!\varphi^{-2}. $$ [This even case is where we must assume $p \neq 3$, because the discriminant of $X^2 + \sqrt5 \, X - 1$ is $9 \equiv 0 \bmod 3$, so $\phi^m$ can be congruent to one of its roots only modulo 3 but still satisfy the quadratic equation modulo 9.] Thus if $m$ is odd then $\varphi^{m+1}$ or $-\varphi^{m-1}$ is $1 \bmod p^2$, while if $m$ is even then $-\varphi^{m+2}$ or $-\varphi^{m-2}$ is $1 \bmod p^2$. In each of these four cases, then, if $m=np$ then $\varphi^k \equiv 1 \bmod p^2$ for some $k$ that is not a multiple of $p$ (namely $k = m+1$, $2m-2$, or $2m\pm 4$). This makes $p$ a Fibonacci-Wieferich prime. The paper

Richard J. McIntosh and Eric L Roettger: A search for Fibonacci-Wieferich and Wolstenholme primes, Math. of Computation 76 #260 (2007), 2087-2094.

explains why we expect the $\log \log x$ behavior, and reports on an exhaustive search over $p < 2 \cdot 10^{14}$ that came up empty.

solutions $m$ of $F_m \equiv -1 \bmod p^2$ should include multiples of $p$.

Conversely, if $p$ is a Fibonacci-Wieferich prime then there exists some even $k \not\equiv 0 \bmod p$ such that $\varphi^k \equiv 1 \bmod p^2$. (If the smallest $k$ was odd then double it.) By "Chinese Remainder" $k$ has a multiple $k' \equiv 1 \bmod p$, and this $k'$ is again even with $\varphi^{k'} \equiv 1 \bmod p^2$. Therefore $F_{np} \equiv -1 \bmod p^2$ with $np = k'-1$ odd, QED.