n+3 \(⋮\) n-1

Ta có : n-1\(⋮\) n-1

Theo bài ra : n+3 \(⋮\)n-1

n+3-(n-1) \(⋮\)n-1

n+3-n+1 \(⋮\) n-1

4 \(⋮\)n-1

\(\Rightarrow\) n-1 \(\in\) Ư₍₄₎ = { 1; 2; 4 }

n-1=1 n-1=2 n-1 =4

n=2 n=3 n=5

Vậy n \(\in\) {2;3;5}