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This is a problem that has haunted me for more than a month. Not all the time - but from time to time, and always on windy or rainy days, it suddenly reappears in my mind:

Assume that the sequence $\{F_{n}\}$ of Fibonacci numbers is defined by the recurrence relation:

$$F_n = F_{n-1} + F_{n-2} ,F_{1}=F_{2}=1.$$

Prove that there exist $i\in\mathbb{N}\setminus\{0\}$, such that $\dfrac{F_{i}}{2015}\in \mathbb{Z}$, and find the minimun of such $i$s.

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    Related: http://math.stackexchange.com/questions/872071/fibonacci-number-that-ends-with-2014-zeros/872077#872077 – Jack D'Aurizio May 04 '15 at 09:52
  • Oh,Nice ,you mean this? http://mathworld.wolfram.com/PisanoPeriod.html,and use Chinese remainder theorem?But How to find minium of this $i$ also use Chinese remainder theorem? –  May 04 '15 at 10:00

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Here we have a complete solution. It is well-known (and easy to prove) that the Fibonacci sequence is periodic modulo any prime. To require $2015\mid F_n$ is the same as requiring $$ 5\mid F_n\quad\wedge\quad 13\mid F_n\quad\wedge\quad 31\mid F_n. \tag{1}$$ Now, by direct inspection of the sequence modulo $5,13,31$, we have: $$ 5\mid F_n \Longrightarrow 5\mid n, $$ $$ 13\mid F_n \Longrightarrow 7\mid n, $$ $$ 31\mid F_n \Longrightarrow 30\mid n,\tag{2} $$ hence the minimum positive $n$ such that $2015\mid F_n$ is $n=7\cdot 30 = \color{red}{210}.$ $$\begin{eqnarray*} F_{210}&=&34507973060837282187130139035400899082304280\\&=&2015\cdot 17125544943343564360858629794243622373352.\tag{3} \end{eqnarray*}$$

Jack D'Aurizio
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