The $n$th Fibonacci number $F_n$ is defined as follows,$$F_1=F_2=1\mbox{ and } F_{n+2}=F_{n+1}+F_{n}\mbox{ for } n\geq 1$$ I want to know how many of the first $1000$ Fibonacci numbers are divisible by $9$ ? Is it possible to classify all Fibonacci numbers which are divisible by $9$ ?
I have searched in Google and found in an article that says every $12$th Fibonacci number is divisible by $9$ but no proof is given. Are these the only such numbers ? Any idea for the soloution would be helpful.
Looking at the Fibonacci numbers $\bmod\ 9$ yields $$ \color{#C00000}{0,1,}1,2,3,5,8,4,3,7,1,8,\color{#0000FF}{0,8,}\dots\tag{1} $$ Since the blue terms are the negative of the red terms, we get $$ F_{n+12}\equiv-F_n\pmod{9}\tag{2} $$ Since the only Fibonacci number in the first $12$ that is $0\bmod9$ is $F_0$, we get $$ F_n\equiv0\pmod{9}\Longleftrightarrow n\equiv0\pmod{12}\tag{3} $$ Therefore, $84$ of the first $1000$ Fibonacci numbers are divisible by $9$ if you start with $F_0$, but if you start with $F_1$, only $83$ of the first $1000$ are.
We know from here ,here or here, $F_m\mid F_n \iff m\mid n$ or $m=2$
now $F_{12}=144$ so, $144\mid F_{12k}$ where $k$ is natural number.
As $F_{12k}\equiv 0\pmod 9$ and if $F_{12k-1}\equiv a\pmod 9, (a,9)=1$ as $(F_m,F_{m+1})=1$
So, $F_{12k+1}\equiv (a+0)\pmod 9\equiv a\cdot F_1$ $F_{12k+2}\equiv (a+a)\pmod 9\equiv a\cdot F_2 ,$ $F_{12k+3}\equiv (2a+a)\pmod 9\equiv a\cdot F_3$ and so on upto $F_{12k+11}\equiv a\cdot F_{11}$
again, $F_{12k-2}=F_{12k}-F_{12k-1}\equiv 0-a\pmod 9\equiv -a\cdot F_1,$ $F_{12k-3}\equiv -a-a\pmod 9\equiv -a\cdot F_2$ and so on upto $F_{12k-11}\equiv -a\cdot F_{10}$
As observed by Hendrik Jan, the period is $24$,
so $9\mid F_n\iff 12\mid n$
Hence, the numbers required Fibonacci number is $\lfloor \frac{1000}{12}\rfloor=83$
First, let's prove if $n$ divisible by $m$, then $F_n$ is also divisible by $F_m$. To prove it we need formula $$\Large{F_{n+k} = F_{k+1} \cdot F_n + F_k \cdot F_{n-1}} \; (1) $$
Therefore, $F_{2n} = F_{n+1} F_n + F_n F_{n-1}$ is divisible by $F_n$, and $F_{3n} = F_{2n + n} = F_{2n+1} F_n + F_{2n} F_{n-1}$ is divisible by $F_n$, and so on.
Using induction method, let's assume $F_{k \cdot n}$ is divisible by $F_n$, therefore, $$\Large{F_{(k+1)n} = F_{kn +n} = F_{n+1} F_{kn} + F_n F_{kn-1}} \; (2) $$ is also divisible by $F_n$.
Now, we need to find all Fibonacci numbers which are divisible by 9. The first Fibonacci number which is divisible by 9 is $F_{12}=144$.
Therefore, $F_{12n}$ is divisible by 9.
Now, let's prove, that there is no other Fibonacci number which is divisible by 9.
$$ \begin{array}{|c|c|} \hline F_{1} \equiv 1 \; (mod \; 9) & F_{13} \equiv 8 \; (mod \; 9) \\ F_{2} \equiv 1 \; (mod \; 9) & F_{14} \equiv 8 \; (mod \; 9) \\ F_{3} \equiv 2 \; (mod \; 9) & F_{15} \equiv 7 \; (mod \; 9) \\ F_{4} \equiv 3 \; (mod \; 9) & F_{16} \equiv 6 \; (mod \; 9) \\ F_{5} \equiv 5 \; (mod \; 9) & F_{17} \equiv 4 \; (mod \; 9) \\ F_{6} \equiv 8 \; (mod \; 9) & F_{18} \equiv 1 \; (mod \; 9) \\ F_{7} \equiv 4 \; (mod \; 9) & F_{19} \equiv 5 \; (mod \; 9) \\ F_{8} \equiv 3 \; (mod \; 9) & F_{20} \equiv 6 \; (mod \; 9) \\ F_{9} \equiv 7 \; (mod \; 9) & F_{21} \equiv 2 \; (mod \; 9) \\ F_{10} \equiv 1 \; (mod \; 9) & F_{22} \equiv 8 \; (mod \; 9) \\ F_{11} \equiv 8 \; (mod \; 9) & F_{23} \equiv 1 \; (mod \; 9) \\ F_{12} \equiv 0 \; (mod \; 9) & F_{24} \equiv 0 \; (mod \; 9) \\ \hline \end{array}$$
It will continue in the same sequence. Which means only $F_{12n}$ is divisible by 9.
Prove (1) using induction, also.
