Fibonacci numbers modulo cubes of primes

Details Fibonacci numbers modulo cubes of primes Fibonacci numbers modulo cubes of primes

2009-11-16

arxiv.org/abs/0911.3060v8

Mathematics

Number Theory

21 pages. Polished version

Scientific paper

Let $p$ be an odd prime. It is well known that $F_{p-(p/5)}=0 (mod p)$ where {F_n}_{n\ge 0} is the famous 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: $$\sum_{k=0}^{(p-1)/2}\binom{2k}{k}/(-16)^k=(p/5)(1+{F_{p-(p/5)}/2) (mod p^3).$$ 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 divisible by p; in particular, we obtain $$\sum_{k=0}^{(p-1)/2}\binom{2k}{k}/16^k=(3/p) (mod p^2).$$ In addition, we raise two conjectures for further research.

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