Prove that $23$ does not divide $2^n + 3^m$, for any $m, n \in \mathbb{N}$ Prove that $23$ does not divide $2^n + 3^m$, for any $m, n \in \mathbb{N}$.
I've tried to prove that the equation $2^n + 3^m = 0$ has no solutions in $\mathbb{Z}_{23}$, but didn't succed.
Thank you!

The residues $\mod23$ of the sequences $(2^n)_n$ and $(3^n)_n$ are, respectively,
$$
D=\{2^n\mod23:n\in\mathbb{N}\}=\{1,2,4,8,16,9,18,13,3,6,12\}
$$
$$
T=\{3^m\mod23:m\in\mathbb{N}\}=\{1,3,9,4,12,13,16,2,6,18,8\}.
$$
You need to show that $\{23\}\not\in D+T$, which can be easily accomplished by observing that whenever $a$ is in $D$, $23-a$ is not in $T$.
