For which $n ≥ 0$ is $2^n + 2 · 3^n$ divisible by $8$? Stuck on this problem for some time:
For which $n ≥ 0$ is $2^n + 2 · 3^n$ divisible by $8$?
I've reached the conclusion that $n = 1$ is the only solution to the question at hand, but I cant quite prove that/why this is the case. 
The properties of the evaluation of the expression as $n = 1$, $n = n + 1$ gives that the first term increases by a multiple of $2$, while the second by a multiple of 3, how can this help proving the solution?
Showing that $2^n + 2 · 3^n \equiv 0$ (mod 8) only when $n = 1$, why is it not enough to show the cases $0...7$? How would I go about proving the solution by modular arithmetic?
Going by induction, what would be the best way to prove that indeed 1 is the only solution - that the expression is not divisible by $8$ for any $n \geq 1$, in that case - how would this look? Or is there a better, different angle to attack it? 

If $n\geq 3$, $2^n+2\cdot 3^n$ is either $2$ or $6\pmod{8}$, since $3^2=9\equiv 1\pmod{8}$ and $8\mid 2^n$.
So we just have to check by hand the cases $n\in\{0,1,2\}$.
