Find the integers $n$ such that $37 \mid 2 \cdot 6^{4 n+3}+2^{n}$.
Here there is nothing magical, we are going to look at the sequences of $6^{m}$ and $2^{n}$ modulo 37:

$$
6 \rightarrow 36=-1 \rightarrow-6 \rightarrow 1 \rightarrow 6 \rightarrow-1 \rightarrow \ldots
$$

so $2 \cdot 6^{4 n+3}$ is always congruent to $12 \bmod 37$.

$2 \rightarrow 4 \rightarrow 8 \rightarrow 16 \rightarrow 32 \rightarrow 27 \rightarrow 17 \rightarrow 34 \rightarrow 31 \rightarrow 25 \rightarrow 13 \rightarrow 26 \rightarrow 15 \rightarrow 30 \rightarrow 23 \rightarrow 9 \rightarrow 18 \rightarrow-1$

with after the same sequence but with all negatives. So the order of 2 is 36, and $2^{n} \equiv$ $-12[37]$ if and only if $n \equiv 10[36]$.