Fibonacci divisibilty properties $ F_n\mid F_{kn},\,$ $\, \gcd(F_n,F_m) = F_{\gcd(n,m)}$ Can any one give a generalization of the following properties in a single proof? I have checked the results, which I have given below by trial and error method. I am looking for a general proof, which will cover the all my results below:


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*Every third Fibonacci number is even.

*3 divides every 4th Fibonacci number.

*5 divides every 5th Fibonacci number.

*4 divides every 6th Fibonacci number.

*13 divides every 7th Fibonacci number.

*7 divides every 8th Fibonacci number.

*17 divides every 9th Fibonacci number.

*11 divides every 10th Fibonacci number.

*6, 9, 12 and 16 divides every 12th Fibonacci number.

*29 divides every 14th Fibonacci number.

*10 and 61 divides every 15th Fibonacci number.

*15 divides every 20th Fibonacci number.


I found that @David's answer here is very clear to understand, but he omitted some minor details. I reproduce @David's answer with full details. All credits are given to @David.


Lemmas:
$f_{s+t}=f_{s-1}f_t+f_sf_{t+1}$ (I presented a proof for this lemma here)
$s\mid t\implies f_s\mid f_t$ (One user presented a proof here)

Let $g=\gcd(m,n)$ and $d$ be any common divisor of $f_m,f_n$. From the definition of Greatest Common Divisor, it's clear that $$\gcd(f_m, f_n) = f_{\gcd(n, m)}=f_g\iff\begin{cases}
f_g\mid f_m\text{ and } f_g\mid f_n\\d\space\space\mid f_g\end{cases}$$


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*$f_g\mid f_m\text{ and } f_g\mid f_n$


From the second lemma and the fact that $g=\gcd(m,n)$, we have $$g\mid m\Rightarrow f_g\mid f_m\quad\hbox{and}\quad
  g\mid n\Rightarrow f_g\mid f_n$$.


*$d\mid f_g$


From Bézout's identity, there exists $x,y\in\mathbb Z$ such that $g=mx+ny$.
$f_m\mid f_{mx}$ and $d\mid f_m\implies d\mid f_{mx}$. Similarly, $d\mid f_{ny}$. Thus $d\mid f_{mx-1}f_{ny}+f_{mx}f_{ny+1}$, and from the second lemma $d\mid f_{mx+ny}$, or equivalently $d\mid f_g$.
