  Let $p$ be an odd prime. It is well known that $F_{p-(\frac p5)}\equiv
0\pmod{p}$, where $\{F_n\}_{n\ge0}$ is the Fibonacci sequence and $(-)$ is the
Jacobi symbol. In this paper we show that if $p\not=5$ then we may determine
$F_{p-(\frac p5)}$ mod $p^3$ in the following way:
$$\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k}{(-16)^k}\equiv\left(\frac{p}5\right)\left(1+\frac{F_{p-(\frac
{p}5)}}2\right)\pmod{p^3}.$$ We also use Lucas quotients to determine
$\sum_{k=0}^{(p-1)/2}\binom{2k}k/m^k$ modulo $p^2$ for any integer
$m\not\equiv0\pmod{p}$; in particular, we obtain
$$\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k}{16^k}\equiv\left(\frac3{p}\right)\pmod{p^2}.$$
In addition, we pose three conjectures for further research.
