Ta có:
\(n^2+7\)
\(=n^2+n-n-1+8\)
\(=n\left(n+1\right)-\left(n+1\right)+8\)
\(=\left(n+1\right)\left(n-1\right)+8\)
Vì \(\left(n+1\right)\left(n-1\right)⋮n+1\)
\(\Rightarrow\) Để \(n^2+7⋮n+1\)
\(\Rightarrow8⋮n+1\)
\(\Rightarrow n+1\inƯ\left(8\right)\)
\(\Rightarrow n+1\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Rightarrow n\in\left\{0;-2;1;-3;3;-5;7;-9\right\}\)