[tex]1)...\;\;\frac{5}{a}\in\mathbb{N}\;daca\; a={D_{5}}_{(diviz.\,naturali)}=\{1\;;\;5\}\;;\\
2)...\;\;\frac{3a+1}{a}\in\mathbb{Z}\;\;daca:\;\frac{3\not{a}}{\not{a}}+\frac{1}{a}=3+\underbrace{\frac{1}{a}}_{\in\mathbb{Z}}\\
\Rightarrow\;\;a=D_{1}=\{\pm1\}[/tex]
