Verified answer

[tex]N = n \ast \dfrac{1}{n} = \bigg(n \cdot \dfrac{1}{n} + 1\bigg)\bigg(n + \dfrac{1}{n}\bigg) = 2 \cdot \dfrac{n^{2} + 1}{n} = \dfrac{2n^{2} + 2}{n} = 2n + \dfrac{2}{n}\\[/tex]

[tex]N \in \Bbb{Z} \iff 2n + \dfrac{2}{n} \in \Bbb{Z}[/tex]

[tex]n \in \Bbb{N^{\ast}} \iff 2n \in \Bbb{Z} \implies \dfrac{2}{n} \in \Bbb{Z} \implies n \in \Mathcal{D}_{2} \implies \boldsymbol {n \in \{1;2\}}\\[/tex]

⇒ numerele naturale n sunt 1 și 2