[tex]\it n\in \mathbb{N },\ \ \dfrac{36}{4n+1} \in \mathbb{N} \Rightarrow 4n+1 \in D_{36} \ \ \ \ (1)
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4n+1 = impar \ \ \ \ (2)
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(1), (2) \Rightarrow 4n+1 \in \{1,\ 3,\ 9\}|_{-1} \Rightarrow 4n \in \{0,\ 2,\ 8\} \Rightarrow n \in \{0, \ 2\}[/tex]
