<s>[INST] 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 $n = 6$
 [/INST] Powers of $2$ modulo $7$ are $1,2,4,1,2,4,\cdots$ repeating every $3$ and powers of $3$ modulo $7$ are $1,3,2,6,4,5,1,\cdots$ repeating every $6$ 
Adding these sequence gives $n \equiv 3 \pmod{6}$ so $7$ divides $2^n+3^n$ if and only if $n=3+6m$ (for $ m \in \mathbb{N}_0$).
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