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Prove that $\forall p \in \Bbb P,n \in \Bbb Z^+;p \ne 5,$ $F_{p^n - \left(\frac{5}{p}\right)p^{n-1}} \equiv 0 \mod p^n$ and $F_{5^n} \equiv 0 \mod 5^n$, where $\left(\dfrac{5}p\right)$ is the Legendre symbol and $F_m$ is the $m$th Fibonacci number, with $F_1=F_2=1$. This is a generalization of the identity $F_{p-\left(\frac{5}p\right)} \equiv 0 \mod p$. Is the generalization well-known? I was trying to find an upper bound on the index of the first Fibonacci number dividing any integer $m$, which would then be the product of the indices associated with the largest power of each prime dividing $m$.

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    $\begingroup$ I would guess that in the case $p\neq5$ a formula like this can be proven by studying the Galois ring $\mathbb{Z}_{p^n}[\sqrt5]$ (so either the ring or its quadratic extension). The structure of the unit groups of those rings is known. Thinking... (this will take a while, also we have a department-wide junket today, so I won't be posting a solution any time soon). +1 to a fun question. $\endgroup$ Aug 19, 2013 at 6:54

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It is known that if $p$ is an odd prime factor of $F_n$, then $\frac{F_{p n}}{F_n}$ is divisible by $p$ but not by $p^2$. For $p=2$, the situation has to be handled more carefully, but is also well known (cf. e.g. N.N. Vorobiev, Fibonacci Numbers, p. 76).

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  • $\begingroup$ Thank you. I'm having difficulty finding an english version of this online. Is it worth buying? $\endgroup$ Aug 19, 2013 at 8:02
  • $\begingroup$ The result should online be available as Thm. 1 in J.H. Halton, On Divisibility Properties of Fibonacci Numbers, Fib. Quat. 4.3 (1966), 217-240. The book is nice to have, but not necessary. $\endgroup$ Aug 19, 2013 at 8:30

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