Find all integers $n$ such that 19 divides $2^{3 n+4}+3^{2 n+1}$.
The conditions of the statement imply that $9^{n} \equiv 8^{n} \bmod 19$. But the inverse of 8 modulo 19 is 12. We deduce that $13^{n} \equiv 108^{n} \equiv (9 \times 8)^{n} \equiv 1 \bmod 19$. Since 13 is a primitive root modulo 19, the integers sought are therefore the set of multiples of 18.