Show that $4\mid f_n$ if and only if $6\mid n$. We define the Fibonacci sequence via $f_1 = f_2 = 1$, and if $n > 2$, then $f_n = f_{n−1} + f_{n−2}$. Show that $4\mid f_n$ if and only if $6\mid n$.
I have been trying to make a copy of the following solution for the exercise "$ 3 \mid f_n $ iff $ 4 \mid n $".

But so far I have not found anything satisfactory. In this question $f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$ they answer it by redefining the fibonacci sequence, but I wonder if it can be done as I am trying.
By the way, what they do here $f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$  I don't understand very well.

You can just go on and see that
$$
f_{n+6}=8f_{n+1}+5f_n
$$
If $f_n$ is divisible by $4$, then also $f_{n+6}$ is.
If $f_n$ is not divisible by $4$, then neither $f_{n+6}$ is.
(In simpler terms: $f_{n+6}\equiv f_n\pmod{4}$.)
Now just look at $f_0=1,f_1=1,f_2=1,f_3=2,f_4=3,f_5=5$, which is the start of the induction.
