Let $p$ be an odd prime. It is well known that $F_{p-(\frac p5)}≡ 0\pmod{p}$, 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: $∑_{k=0}^{(p-1)/2}\frac{\binom{2k}k}{(-16)^k}≡\left(\frac{p}5\right)\left(1+\frac{F_{p-(\frac {p}5)}}2\right)\pmod{p^3}.$ We also use Lucas quotients to determine $∑_{k=0}^{(p-1)/2}\binom{2k}k/m^k$ modulo $p^2$ for any integer $m\not≡0\pmod{p}$; in particular, we obtain $∑_{k=0}^{(p-1)/2}\frac{\binom{2k}k}{16^k}≡\left(\frac3{p}\right)\pmod{p^2}.$ In addition, we pose three conjectures for further research. ; Comment: 21 pages. Final published version