Recurrence Fibonacci Sequence Proof I'm having troubles proving that in a fibonacci sequence if n is divisible by four, then Fn is divisible by three
So when Fn is 6, n is 8 and so on. I was thinking maybe I could use mod 3 or mod 4 but don't really know what to do with it.

This follows from the matrix formulation, which is well worth knowing and easily proved:
$$
\begin{pmatrix}1&1\\1&0\end{pmatrix}^n=
\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}
$$
We will prove by induction that $3$ divides $F_{4k}$.
Let
$
A=\begin{pmatrix}1&1\\1&0\end{pmatrix}
$.
Then $A^{4(k+1)}=A^{4k}A^4$ :
$$
\begin{pmatrix}F_{4(k+1)+1}&F_{4(k+1)}\\F_{4(k+1)}&F_{4(k+1)-1}\end{pmatrix}
=
\begin{pmatrix}F_{4k+1}&F_{4k}\\F_{4k}&F_{4k-1}\end{pmatrix}
\begin{pmatrix}F_{5}&F_4\\F_4&F_{3}\end{pmatrix}
$$
and so, by looking at position $2,1$,
$$
F_{4(k+1)}=F_{4k}F_{5}+F_{4k-1}F_4
$$
By induction, $F_4=3$ divides $F_{4k}$ and so $F_4=3$ divides $F_{4(k+1)}$.
