FIBONACCI NUMBERS MODULO CUBES OF PRIMES

Abstract

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