7. If $n$ is a positive integer such that $n^{6}+206$ is divisible by $n^{2}+2$, find the sum of all possible values of $n$.
(1 mark)
若 $n$ 是正整數, 且 $n^{6}+206$ 可被 $n^{2}+2$ 整除, 求 $n$ 所有可能值之和。
(1 分)
7. 32
7. Note that $n^{6}+8=\left(n^{2}\right)^{3}+2^{3}$ is divisible by $n^{2}+2$. Hence $n^{2}+2$ must divide $206-8=198$. Among the factors of 198 , we check that only five of them are of the form $n^{2}+2$ for positive integer $n$, namely, $3=1^{2}+2,6=2^{2}+2,11=3^{2}+2,18=4^{2}+2,66=8^{2}+2$ and $198=14^{2}+2$. It follows that the possible values of $n$ are $1,2,3,4,8,14$ and the answer is $1+2+3+4+8+14=32$.