$\begin{array}{l}n^2+3\ \vdots\ n+1\\\Leftrightarrow n^2+3-n(n+1)\ \vdots\ n+1\\\Leftrightarrow n^2+3-n^2-n\ \vdots\ n+1\\\Leftrightarrow 3-n\ \vdots\ n+1\\\Leftrightarrow 3-n+(n+1)\ \vdots\ n+1\\\Leftrightarrow 3-n+n+1\ \vdots\ n+1\\\Leftrightarrow 4\ \vdots\ n+1\\\Leftrightarrow n+1\in Ư(4)=\{\pm1;\pm2;\pm4\}\\\text{- Ta có bảng sau :}\\\begin{array}{|c|c|}\hline n+1&-4&-2&-1&1&2&4\\\hline n&-5&-3&-2&0&1&3\\\hline\end{array}\\\text{- Vậy $n\in\{-5;-3;-2;0;1;3\}$} \end{array}$
