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Importance: Medium ✭✭

Author(s):

Subject:	Number Theory
	» Computational Number Theory

Keywords:	Fibonacci
	prime

Recomm. for undergrads: no

Posted	by:	adudzik
	on:	June 14th, 2008

Conjecture For any prime $p$ , there exists a Fibonacci number divisible by $p$ exactly once.

Equivalently:

Conjecture For any prime $p>5$ , $p^2$ does not divide $F_{p-\left(\frac p5\right)}$ where $\left(\frac mn\right)$ is the Legendre symbol.

Let $p$ be an odd prime, and let $\nu_p(n)$ denote the $p$ -adic valuation of $n$ . Let $F_{k(p)}$ be the smallest Fibonacci number that is divisible by $p$ (which must exist by a simple counting argument). A well-known result says that $\nu_p(F_n)=0$ unless $k(p)$ divides $n$ , and $\nu_p(F_{k(p)m}) = \nu_p(F_{k(p)}) + \nu_p(m)$ . This conjecture asserts that $\nu_p(F_{k(p)})=1$ for all $p$ . This has been verified up to at least $p<10^{14}$ . [EJ]