I need help with this exercise, I don't know how to aproach it.
Find $\gcd(a_n,a_{n+1})$ for each $n \in \mathbb{N}$
$(a_n)_{n\in\mathbb{N}} :\left\{\begin{matrix} a_1 = 2 \\ a_2 = 4 \\ a_{n+2} = 4a_{n+1}+3a_n \end{matrix}\right.$
I could only discard 3 as an option proving that $\forall n\in\mathbb{N} \;\;3 \not|\;\; a_n \;\; $ and guess that 2 and 4 are potencial options but don't know how to proceed on proving that or if it is actually correct.
If $d= \gcd (a_n,a_{n+1})$ then $d\mid a_{n+1}$ and $d\mid a_n$ so $$d\mid a_{n+1}-4a_n = 3a_{n-1}$$
And since you proved that $3\not{\mid} a_n$ we have $d\mid a_{n-1}$. So
$$\gcd (a_n,a_{n+1}) = \gcd(a_{n-1},a_n)=...= \gcd(a_1,a_2)=2$$
Proof that $3\not{\mid} a_n$ for all $n$:
Suppose that is not true and let $a_{n+1}$ be first in sequence divisible by $3$ Then $4a_n = a_{n+1}-3a_{n-1}$ so $3\mid 4a_n$ so $3\mid a_n$ a contradicition.