If $m \equiv 5\mod 10 $ prove that $1991 \mid 12^m + 9^m + 8^m + 6^m$ I tried to find the remainder of each one of $12$,$9$,$8$ and $6 \mod 5$ and then combine them but I didn’t get the answer 

Let $m=5+10n$ and let $x_n = 12^m + 9^m + 8^m + 6^m = 12^5 (12^{10})^n + 9^5 (9^{10})^n + 8^5 (8^{10})^n + 6^5 (6^{10})^n$.
Then $x_n$ satisfies a linear recurrence of the form $x_{n+4}=a_3 x_{n+3} + a_2 x_{n+2} + a_1 x_{n+1} + a_0 x_{n}$, where $t^4=a_{n+3}t^3 + a_2 t^2 + a_1 t^1 + a_0 $ is the equation having $2^{10}$, $9^{10}$, $8^{10}$, $6^{10}$ as roots. Hence, the $a_i$ are integers.
The claim that $1991$ divides $x_n$ for all $n$ follows by induction once you have verified it for $n=0,1,2,3$.
Not nice, but doable. WA can help. If you have to do it manually, it's probably easier to do it mod $11$ and $181$ separately.
