Proving that $n|m\implies f_n|f_m$ Question: Let $m,n\in\mathbb{N}$, prove that if $n|m$, $F_n|F_m$.
I've tried to use induction, but I don't really know where to start since there's $2$ numbers:  $n$ and $m\ \dots$ I did induction before with just $1$ number, like proving $1+2+\cdots+n=\frac{n(n+1)}{2}$, that's only dealing with one number $n$, but with this I have no clue how to do it.
Can anyone give me a little hint on how to start, I don't want you to do whole problem for me but can I have some hint so I know where to start?
Thanks for the help!

Hint:  Try to prove that $f_{n+m}=f_{n-1}\ f_m+f_n\ f_{m+1}$.  
Then, let $m=n\times k$ for some $k\in \mathbb{N}$.  
Now, prove that $f_m$ is divisible $f_n$ by induction on $k$.
