What is all N that make $2^N + 1$ divisible by 3? When $N = 1$, $2^N + 1 = 3$ which is divisible by $3$.
When $N = 7$, $2^N + 1 = 129$ which is also divisible by $3$.
But when $N = 2$, $2^N + 1 = 5$ which is not divisible by $3$.
A quick lookout can determine that when $N$ is an odd number, $2^N + 1$ is divisible by $3$. If is that correct, how can it be proven? Otherwise, what is the counterexample and the correct $N$?
Can this be made into a more general form, for example $2^N + C$ or something similar?

If $2^N+1$is divisible by $3$ then in other words $2^N+1$ is a multiple of $3$ .
$2^N+1=3k \implies 2^N=3k-1$ so we want $3k-1$ to be even then we want  $3k$ to be odd. 
$3k$ is odd only when $k$ is odd.
