Find all n for which $2^n + 3^n$ is divisible by $7$? I am trying to do this with mods.  I know that:
$2^{3k+1} \equiv 1 \pmod 7$, $2^{3k+2} \equiv 4 \pmod 7$, $2^{3k} \equiv 6 \pmod 7$ and $3^{3k+1} \equiv 3 \pmod 7$, $3^{3k+2} \equiv 2 \pmod 7$,  $3^{3k} \equiv 6 \pmod 7$, so I thought that the answer would be when $n$ is a multiple of $3$ since then $2^{3k} + 3^{3k} \equiv 1 + 6 \pmod 7$, but it doesn't work for $n = 6$

We need $$2^n+3^n\equiv0\pmod7\iff3^n\equiv-2^n\iff-1\equiv(3\cdot4)^n\equiv(-2)^n$$ using $2\cdot4\equiv1\pmod7\iff2^{-1}\equiv4$
$\implies(-2)^n\equiv-1\equiv(-2)^3\iff(-2)^{n-3}\equiv1$
$\implies n-3\equiv0\pmod6$ as $(-2)^3\equiv-1,(-2)^6\equiv1$
