Question

Ajutor ex. 8 dau coronita

Answer 1

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