Let {$F_n$} be Fibonacci Series defined by $F_{n+2}=F_{n+1}+F_n$ with $F_1=F_2=1$.I have proved that $3^m|F_n \Leftrightarrow 4\cdot 3^{m-1}|n$ for another problem.Then I began to think about the necessary and sufficient condition of $p^m|F_n$,where $p$ is a prime number and $m$ is a positive integer.By analyzing the first sixty terms of {$F_n$},I suppose that
where $\tau (p)$ is the least positive integer s.t. $p|F_{\tau(p)}$.
I think we could prove this with induction through several steps:
①Prove that $(F_m,F_n)=F_{(m,n)}$,where $(a,b)$ is the greastest common divisor of $a$ and $b$.
②Strengthen the proposition by proving $p^m||F_{\tau (p) \cdot p^{m-1}}$ $\ (m \geqslant 3 \ $when $p=2)$.
③Prove the proposition when $p=2$.
④Prove $p||F_{\tau (p)}$.
⑤Prove that $p^{m+1}|F_n \Leftrightarrow \tau (p) \cdot p^m|n $ ,and that $p^{m+1}||F_{\tau (p) \cdot p^m}$ ,when $p^{m}|F_n \Leftrightarrow \tau (p) \cdot p^{m-1}|n $ ,and $p^{m}||F_{\tau (p) \cdot p^{m-1}}$$\ \ $ for all odd primes $p$.
For ①,it's easy to prove $(F_n,F_{n+1})=1$ with the fact that $(F_1,F_2)=1$.And we have $$\begin{align}F_m&=F_{m-1}+F_{m-2}=F_2F_{m-1}+F_1F_{m-2}=F_2(F_{m-2}+F_{m-3})+F_1F_{m-2}\\&=(F_2+F_1)F_{m-2}+F_2F_{m-3}=F_3F_{m-2}+F_2F_{m-2}\\&=\cdots\\&=F_nF_{m-n+1}+F_{n-1}F_{m-n}\end{align}$$ when $m>n$.Thus we have $(F_m,F_n)=(F_{m-n},F_n)$ when $m>n$.So by Euclid algotithm,we have $(F_m,F_n)=F_{(m,n)}$.
For ③,it's easy to prove that $2|F_n\Leftrightarrow3|n\ $,$4|F_n\Leftrightarrow6|n\ $,$8|F_n\Leftrightarrow6|n\ $,and $8||F_6$.When $2^m|F_n\Leftrightarrow3\cdot2^{m-2}|n$ and $2^{m}||F_{3\cdot2^{m-1}}$ $(m>2)$,we have $$\begin{align}\frac{F_{3\cdot2^{m-1}}}{F_{3\cdot2^{m-2}}}&=\frac{(\frac{1+\sqrt{5}}{2})^{3\cdot2^{m-1}}-(\frac{1-\sqrt{5}}{2})^{3\cdot2^{m-1}}}{(\frac{1+\sqrt{5}}{2})^{3\cdot2^{m-2}}-(\frac{1-\sqrt{5}}{2})^{3\cdot2^{m-2}}}=(\frac{1+\sqrt{5}}{2})^{3\cdot2^{m-2}}+(\frac{1-\sqrt{5}}{2})^{3\cdot2^{m-2}}.\end{align}$$ Let {$Z_n$} be a sequence where $Z_1=1$,$Z_2=3$ and $Z_{n+2}=Z_{n+1}+Z_{n}$.Then we have $Z_n=(\frac{1+\sqrt{5}}{2})^n+(\frac{1-\sqrt{5}}{2})^n$,and $$\begin{align}&Z_1\equiv1,Z_2\equiv1,Z_3\equiv0,Z_4\equiv1,Z_5\equiv1,Z_6\equiv0,\cdots(mod2)\\ &Z_1\equiv1,Z_2\equiv3,Z_3\equiv0,Z_4\equiv3,Z_5\equiv3,Z_6\equiv2,Z_7\equiv1,Z_8\equiv3,\cdots(mod4) \end{align}$$ which tolds us that if $6|n$,then $2||Z_n$.So we have $$\begin{align}2||Z_{3\cdot2^{m-2}}=\frac{F_{3\cdot2^{m-1}}}{F_{3\cdot2^{m-2}}}.\end{align}$$ And $2^{m}||F_{3\cdot2^{m-1}}\ \ $,thus $2^{m+1}||F_{3\cdot2^{m}}$.Obviously,the multiple of $2^{m+1}$ is the multiple of $2^{m}$.So if $2^{m+1}|F_n$,then $2^{m}|F_n$.So we have $3\cdot2^{m-2}|n$.Let $n=3k\cdot2^{m-2}$,where k is a positive integer.Then with ①,we have $(F_{3\cdot2^{m-1}}\ ,\ F_{3k\cdot2^{m-2}}\ )=F_{3\cdot2^{m-2}\ \ (2,k)}$.If $2\nmid k$,then $(F_{3\cdot2^{m-1}}\ ,\ F_{3k\cdot2^{m-2}}\ )=F_{3\cdot2^{m-2}}\ \ $,which means $2^m||F_{3k\cdot2^{m-2}}\ \ $;if $2|k$,then $(F_{3\cdot2^{m-1}}\ ,\ F_{3k\cdot2^{m-2}}\ )=F_{3\cdot2^{m-1}}\ \ $,which means $2^{m+1}|F_{3k\cdot2^{m-2}}\ \ $.So $2^{m+1}|F_n\Leftrightarrow3\cdot2^{m-1}|n$.By the induction principle,we've achieved ③.
For ⑤,similar to ③,we have only to prove $p^{m+1}||F_{\tau (p)\cdot p^{m}}$ ,when $p^{m}|F_n \Leftrightarrow \tau (p) \cdot p^{m-1}|n $ ,and $p^{m}||F_{\tau (p) \cdot p^{m-1}}\ $.But actually I can only do this when $p$ is a certain prime such as $3,5$.I can't find a way to prove it for a general $p\ $(constructing a sequence is not suitable to an odd prime ,or my poor knowledge of sequence can't support me to step further).When $p=3$,I proved it like that:
It's easy to know $\tau (3)=4$.And it's also easy to prove that $3|F_n\Leftrightarrow4|n$,and $3||F_4$.If $3^{m}|F_n\Leftrightarrow4\cdot3^{m-1}|n$ and $3^m||F_{4\cdot3^{m-1}}\ $,then: we have $$\begin{align}\frac{F_{4\cdot3^{m}}}{F_{4\cdot3^{m-1}}}&=\frac{(\frac{1+\sqrt{5}}{2})^{4\cdot3^m}-(\frac{1-\sqrt{5}}{2})^{4\cdot3^m}}{(\frac{1+\sqrt{5}}{2})^{4\cdot3^{m-1}}-(\frac{1-\sqrt{5}}{2})^{4\cdot3^{m-1}}}=(\frac{1+\sqrt{5}}{2})^{8\cdot3^{m-1}}+(\frac{1+\sqrt{5}}{2})^{4\cdot3^{m-1}}(\frac{1-\sqrt{5}}{2})^{4\cdot3^{m-1}}+(\frac{1-\sqrt{5}}{2})^{8\cdot3^{m-1}}\\ &=\frac{(1+\sqrt{5})^{8\cdot3^{m-1}}+(-4)^{4\cdot3^{m-1}}+(1-\sqrt{5})^{8\cdot3^{m-1}}}{2^{8\cdot3^{m-1}}}=\frac{2\sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{2t}}\cdot5^t}+4^{4\cdot3^{m-1}}}{2^{8\cdot3^{m-1}}}. \end{align}$$ And because $$\begin{align}2\sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{2t}}\cdot5^t}+4^{4\cdot3^{m-1}}&\equiv 2\sum_{t=0}^{2\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{4t}}}+2\sum_{t=0}^{2\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{4t+2}}\cdot2}+1\\&\equiv 2\sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{2t}}}+2\sum_{t=0}^{2\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{4t+2}}}+1\\&\equiv 2^{8\cdot3^{m-1}}+2^{8\cdot3^{m-1}-1}-2^{4\cdot3^{m-1}}+1\\&\equiv 1+2-1+1 \equiv 0 (mod3) \end{align}$$ and it's easy to prove that $2\sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{2t}}\cdot5^t}+4^{4\cdot3^{m-1}}\equiv 3(mod9)$when $m=1$,and when $m>1$ $$\begin{align}2\sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{2t}}\cdot5^t}+4^{4\cdot3^{m-1}}&\equiv 2\Biggl( \sum_{t=0}^{2\cdot3^{m-1}}{\biggl( {{8\cdot3^{m-1}}\choose{2t}}\cdot5^t+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-2t}}\cdot5^{8\cdot3^{m-1}-2t} \biggl)}-{{8\cdot3^{m-1}}\choose{4\cdot3^{m-1}}}\cdot5^{2\cdot3^{m-1}} \Biggr)+1\end{align}$$ $$\begin{align}&\equiv 2\Biggl( \sum_{t=0}^{3^{m-2}}{\biggl( {{8\cdot3^{m-1}}\choose{12t}}+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t}} \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+2}}\cdot5+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-2}}\cdot2 \biggl)}\\ &\ +\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+4}}\cdot7+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-4}}\cdot4 \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+6}}\cdot8+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-6}}\cdot8 \biggl)}\\ &\ +\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+8}}\cdot4+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-8}}\cdot7 \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+10}}\cdot2+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-10}}\cdot5 \biggl)}-{{8\cdot3^{m-1}}\choose{4\cdot3^{m-1}}} \Biggr)+1\end{align}$$ $$\begin{align}&\equiv 2\Biggl( \sum_{t=0}^{3^{m-2}}{\biggl( {{8\cdot3^{m-1}}\choose{12t}}+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t}} \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+2}}\cdot8+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-2}}\cdot8 \biggl)}\\ &\ +\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+4}}+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-4}} \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+6}}\cdot8+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-6}}\cdot8 \biggl)}\\ &\ +\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+8}}+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-8}} \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+10}}\cdot8+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-10}}\cdot8 \biggl)}-{{8\cdot3^{m-1}}\choose{4\cdot3^{m-1}}} \Biggr)+1\end{align}$$ $$\begin{align} &\equiv 2\Biggl( \sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}}\choose{2t}}}+7\sum_{t=0}^{2\cdot3^{m-1}}{{{8\cdot3^{m-1}}\choose{4t+2}}} \Biggl)+1\\ &\equiv 2(2^{8\cdot3{m-1}-1}+7(2^{8\cdot3{m-1}-2}-2^{4\cdot3{m-1}-1}))+1\\ &\equiv 2(5+7\times(7-5))+1 \equiv 3(mod9) \end{align}$$ (I'm terribly sorry for my making it so ugly,but it seems it has been beyond its limit),we have $3||\frac{F_{4\cdot3^m}}{F_{4\cdot3^{m-1}}}$,thus $3^{m+1}||F_{4\cdot3^m}$.Then $(F_{4\cdot3^m},F_{4k\cdot3^{m-1}})=F_{4\cdot3^{m-1}\ \ (3,k)}$.Similar to ③,We have $3^{m+1}|F_n \Leftrightarrow 4\cdot3^m|n$.Then by induction principle,$3^m|F_n\Leftrightarrow4\cdot3^{m-1}|n$.
But I can't work out in this way for a general $p$.
And I have no idea about ④,which is the basic of the induction.
How could I prove or disprove my suppose?Is there any theory about this?Is there a formula about $\tau (p)$?
Appreciated for any suggestion.
And I'm sorry for my possibly strange expressions and wrong grammar for I'm still an English learner.