[tex] \displaystyle\\
\frac{n^2+n+4}{n+1} \in Z \text{ daca }n\in Z\\\\
\frac{n^2+n+4}{n+1} = \frac{n^2+n}{n+1}+\frac{4}{n+1} =\frac{n(n+1)}{n+1}+\frac{4}{n+1} =n+\frac{4}{n+1}\\\\
n \in Z \text{ daca }n\in Z ~~~~:)) \\\\
\frac{4}{n+1} \in Z \text{ daca }(n+1)\in D_4\\\\
D_4 = \{-4;~-2;~-1;~1;~2;~4\}\\\\
n+1=-4 \Rightarrow n=-5\\
n+1=-2 \Rightarrow n=-3\\
n+1=-1 \Rightarrow n=-2\\
n+1=1 \Rightarrow n=0\\
n+1=2 \Rightarrow n=1\\
n+1=4 \Rightarrow n=3\\\\
\boxed{n \in \{-5;~-3;~-3;~0;~1;~3\} \subset Z} [/tex]
