Ta có : 3n2+5\(⋮\)n-1
\(\Rightarrow\)3n2-3+8\(⋮\)n-1
\(\Rightarrow\)n(3n-1)+8\(⋮\)n-1
Vì n(3n-1)\(⋮\)n-1 nên 8\(⋮\)n-1
\(\Rightarrow n-1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Rightarrow n\in\left\{0;-2;-1;3;-3;5;-7;9\right\}\)
Vậy n\(\in\){0;-2;-1;3;-3;5;-7;9}