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?

$((-1)+3))^N+1= $
$\sum_{k=0}^{N}\binom{N}{k}(-1)^{N-k}3^k+1=$
$(-1)^N +1+\sum_{k=1}^{N}(-1)^{N-k}3^k.$
The expression is divisible by $3$ $\iff$
$(-1)^N+1=0.$
Hence?
