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?

You can easily prove by induction on $k$ that $2\times 4^k+1$ is divisible by $3$: It is for $k=1$, and if it is for some $k$, then
$$2\times 4^{k+1}+1=2\times 4^k\times 4+1=4(2\times 4^k+1)-3$$
Therefore $2^{2m+1}+1$ is divisible by $3$ for every natural $m$.
