<s>[INST] 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?
 [/INST] When $N = 1 + 2k$;
\begin{align}2^{1+2k} +1 \pmod 3 & \equiv 2(2^{2k})+1 \pmod 3\\
&\equiv 2(4^{k})+1 \pmod 3\\
&\equiv 2+1 \pmod 3\\
&\equiv \pmod 3
\end{align}
When $N = 2k$
$$2^{2k} +1 \pmod 3\equiv 2 \pmod 3$$
</s>