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!

You can prove the desired result without induction using Binet's formula,
$$F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$
where $\varphi$ and $\psi$ are the roots of $x^2 + x + 1$, and the factoring identity
$$a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1}).$$
