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.

As you have asked for, this can be done using a similar argument to the one in your post:
If you repeatedly apply the recurrence relation, then $f_{n} = 8f_{n-5}+5f_{n-6}$ so that when you use strong induction on some set of elements less than $n$ in the same way by supposing $6|n$, then $6|n-6$ and by hypothesis $4|f_{n-6}$ and by the equation above, $4|f_{n}$.
If $6\nmid n$, then $6\nmid (n-6)$ and $4\nmid f_{n-6}$.
If $4|f_{n}$, then $4|8f_{n-5}+5f_{n-6}\implies4|5f_{n-6}\implies4|f_{n-6}$, a contradiction. Hence, $4\nmid f_{n}$. In this way, both implications have been shown.
For the second implication, I was unable to use the exact same argument as above since, in this case, $f_{n}$ was not able to reduced in such a convenient way.
