# Math Help - Another fibonacci proposition

1. ## Another fibonacci proposition

Hey all I would really love some help with this fibonacci proposition:

For all k,m in the Natural Numbers, fib(mk) is divisible by fib(m)

Thanks a bunch!

2. Originally Posted by CaptainBlack
Now:

$F_{m(k+1)}=F_{mk}+F_{m(k-1)}$

CB
I don't think this is true. For example, $F_{2(k+1)}=F_{2k+2}=F_{2k+1}+F_{2k}$, not $F_{2k}+F_{2k-2}$.

3. Originally Posted by Bruno J.
I don't think this is true. For example, $F_{2(k+1)}=F_{2k+2}=F_{2k+1}+F_{2k}$, not $F_{2k}+F_{2k-2}$.
Opps, will probably have to go back to the other method that I discarded as too complicated (if I can recall it now..).

CB

4. I think the original problem can be solved using the identity

$F_{m+n} = F_{m-1}F_n + F_mF_{n+1}$

which holds for all positive integers m,n. This identity can be proved by fixing m, say, and inducting on n. Then, prove that

$F_m$ divides $F_{mk}$

with another inductive argument, using the above identity.