[tex]\it n\in\mathbb{Z},\ \ \ \dfrac{3n^2+11n+17}{n+2} \in\mathbb{Z}\ \ \ \ (1)\\ \\ \\ 3n^2+11n+17=3n^2+11n+10+7=3n^2+6n+5n+10+7=\\ \\ \\ =3n(n+2)+5(n+2)+7=(n+2)(3n+5)+7\ \ \ \ \ (2)[/tex]
[tex]\it \dfrac{3n^2+11n+17}{n+2}\ \stackrel{(2)}{=}\ \dfrac{(n+2)(3n+5)+7}{n+2}=\dfrac{(n+2)(3n+2)}{n+2}+\dfrac{7}{n+2}=\\ \\ \\ = 3n+2+\dfrac{7}{n+2}\ \stackrel{(1)}{\Longrightarrow} \dfrac{7}{n+2}\in\mathbb{Z} \Rightarrow n+2|7 \Rightarrow\\ \\ \\ \Rightarrow n+2\in D_7 \Rightarrow n+2\in\{\pm1,\ \pm7\} \Rightarrow n+2\in \{-7,\ -1,\ 1,\ 7\}|_{-2} \Rightarrow\\ \\ \\ \Rightarrow n\in \{-9,\ -3,\ -1,\ 5\}[/tex]
