Determine all positive integer values of $n$ for which $2^{n}+1$ is divisible by 3.
$$
\begin{gathered}
2^{n}+1=(3-1)^{n}+1= \\
=3^{n}-\binom{n}{1} 3^{n-1}+\binom{n}{2} 3^{n-2}-\binom{n}{3} 3^{n-3}+\ldots+\binom{n}{n-1} 3 \cdot(-1)^{n-1}+(-1)^{n}+1
\end{gathered}
$$

The given expression is divisible by 3 if

$$
(-1)^{n}+1=0
$$

which is true if $n$ is an odd number.

(János Kárf.)
