[tex]\displaystyle\it\\2n+1|2n+11.\\dar,~evident,~2n+1~divide~si~pe~el~insusi,~adica:2n+1|2n+1.\\2n+1|2n+11,~iar~2n+1|2n+1 \implies 2n+1~va~divide~si~diferenta\\celor~doua~numere \implies 2n+1|2n+11-2n-1 = 10.\\2n+1|10,~cum~2n+1\in\mathcal{D}_{10} \implies 2n+1 \in \left\{1,2,5,10\right\},~\\dar~2n+1~este~impar \implies 2n+1~este~divizor~impar~al~numarului~10 \implies\\2n+1 \in \left\{1,5\right\} \implies \boxed{\it n\in\left\{0,2\right\}}~.[/tex]
