Ta có
\(\frac{3+n}{n-1}=\frac{n-1+4}{n-1}=1+\frac{4}{n-1}\)
Điều kiện :
n - 1 \(\inƯ_{\left(4\right)}\)
\(n\in N\Rightarrow n-1\ge-1\)
\(\Rightarrow n-1\in\left\{-1;1;2;4\right\}\)
\(\Rightarrow n\in\left\{0;2;3;5\right\}\)
Vậy \(n\in\left\{0;2;3;5\right\}\)