Ta có: \(3n+5⋮n+1\)
\(\Rightarrow\left(3n+3\right)+2⋮n+1\)
\(\Rightarrow3\left(n+1\right)+2⋮n+1\)
\(\Rightarrow2⋮n+1\)
\(\Rightarrow n+1\in U\left(2\right)=\left\{-1;1;-2;2\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}n+1=-1\Rightarrow n=-2\+1=1\Rightarrow n=0\+1=-2\Rightarrow n=-3\+1=2\Rightarrow n=1\end{matrix}\right.\)
Vậy \(n\in\left\{-2;0;-3;1\right\}\)
