[tex]\it a)\ \ Fie\ d\in\mathbb{N}^{*},\ pentru\ care:\\ \\ d|n\ \ \ \ (1)\\ \\ d|n+1\ \ \ \ (2) \\ \\ (1),\ (2) \Rightarrow d|n+1-n \Rightarrow d|1 \Rightarrow d=1 \Rightarrow \dfrac{n}{n+1}=ireductibil\breve{a},\ \forall n\in\mathbb{N}^{*}[/tex]
[tex]\it b)\ d|(3n+119) \Rightarrow d|(3n+119)\cdot 2 \Rightarrow d|6n+238\ \ \ \ (1) \\ \\ d|(2n+79) \Rightarrow d|(2n+79)\cdot 3\Rightarrow d|6n+237\ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow d|6n+238-n-237 \Rightarrow d|1 \Rightarrow d =1 \Rightarrow \\ \\ \Rightarrow \dfrac{3n+119}{2n+79}=\ ireductibil\breve{a},\ \forall n \in\mathbb{N}^{*}[/tex]
