$2n+8\vdots n+1$
$\rightarrow 2n+2+6\vdots n+1$
$\rightarrow 2(n+1)+6\vdots n+1$
Ta nhận thấy: $2(n+1)\vdots n+1$ mà $2(n+1)+6\vdots n+1$
$\rightarrow 6\vdots n+1$
$\rightarrow n+1\in Ư(6)=\{\pm 1;\pm 2; \pm 3; \pm 6\}$
$\begin{array}{|c|c|c|}\hline n+1&1&2&3&6&-1&-2&-3&-6\\\hline n&0&1&2&5&-2&-3&-4&-7\\\hline\end{array}$
Vậy $n=\{0;1;2;5;-2;-3;-4;-7\}$