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 
 A: If $m/5$ is odd, $11|2^5+3^5|2^m+3^m$ while $181|3^5+4^5|3^m+4^m$, so$$\operatorname{lcm}(11,\,181)=1991|(2^m+3^m)(3^m+4^m)=6^m+8^m+9^m+12^m.$$
A: From
$$m \equiv 5 \pmod{10}\iff m=10\cdot k +5=5\cdot(2k+1)\tag{1}$$
and
$$12^m + 9^m + 8^m + 6^m=\\
3^m \cdot 4^m + 3^m\cdot 3^m + 2^m \cdot 4^m + 2^m \cdot 3^m=\\
3^m(4^m+3^m)+2^m(4^m+3^m)=\\
(3^m+2^m)\cdot(4^m+3^m)\overset{(1)}{=}\\
(243^{2k+1}+32^{2k+1})\cdot(1024^{2k+1}+243^{2k+1})=...$$
From the well known 
$a^{2k+1}+b^{2k+1}=(a+b)\cdot\left(a^{2k}+a^{2k-1}\cdot(-b)+a^{2k-2}\cdot(-b)^2+...+b^{2k}\right)$
$$...=(243+32)\cdot Q1 \cdot (1024+243)\cdot Q2=\\
275 \cdot 1267 \cdot Q1\cdot Q2=
2^5 \cdot \color{red}{11} \cdot 7 \cdot \color{red}{181} \cdot Q1\cdot Q2=\\
\color{red}{1991}\cdot 2^5 \cdot 7\cdot Q1\cdot Q2$$
A: Hint $\, \  \{8^5,\ \ \,9^5\} \, \equiv\, \{-6^5,\,\  {-}12^5\}\ \ \bmod  11\ \&\ 181,\: $ so by the Congruence Power Rule
$ \overset{n\ \rm odd}\Longrightarrow\,  \{8^{5n},\, 9^{5n}\}  \equiv \{-6^{5n},{-}12^{5n}\}\ {\rm mod}\,\ 11\ \&\ 181,\, $ so also $\!\bmod 1991 = {\rm lcm}(11,181)$
A: 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.
