[tex]A = \left\{n\in \mathbb{Z}\,\big|\,\frac{3n+2}{2n+3}\in \mathbb{Z}\right\}\\ \\[/tex]
Aduc fracția la o formă mai convenabilă:
[tex]\frac{3n+2}{2n+3} = \frac{1}{2}\cdot \frac{2(3n+2)}{2n+3}=\frac{1}{2}\cdot \frac{6n+4}{2n+3}=\\ \\=\frac{1}{2}\cdot \left(\frac{3(2n+3)-5}{2n+3}\right) = \frac{1}{2}\cdot \left(3-\frac{5}{2n+3}\right)[/tex]
Sunt două condiții:
[tex]\boxed{1}\,\,\,\frac{5}{2n+3} \in \mathbb{Z} \Rightarrow 2n+3\,|\,5\Rightarrow 2n+3\in D_{5} \\\\ \Rightarrow 2n+3 \in \{-5;\,-1;\, 1;\, 5\} \Rightarrow 2n \in \{-8;\,-4;\,-2;\,2\} \Rightarrow\\ \\ \Rightarrow n\in \{-4;\,-2;\,-1;\, 1\}\\ \\ \boxed{2}\,\,\,\left(3-\frac{5}{2n+3}\right) \, \vdots\,\, 2\,\,\, (A),\quad \forall n\in \{-4;\,-2;\,-1;\, 1\}[/tex]
Din ① și ②:
⇒ A = {-4; -2; -1; 1}
