<s>[INST] 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 
 [/INST] 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 
$a^{2k+1}+b^{2k+1}=(a+b)\cdot\left(a^{2k}+a^{2k-1}\cdot(-b)+a^{2k-2}\cdot(-b)^2+...+b^{2k}\right)$
$$...=(243+32)\cdot Q1 \cdot (1024+243)\cdot Q2=\\
275 \cdot 1267 \cdot Q1\cdot Q2=
2^5 \cdot \color{red}{11} \cdot 7 \cdot \color{red}{181} \cdot Q1\cdot Q2=\\
\color{red}{1991}\cdot 2^5 \cdot 7\cdot Q1\cdot Q2$$
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