AoPS - Problem

I think, it should be $2a_{n+2} =3a_n +\sqrt{5 (a_n^2+a_{n+1}^2)}$ $a_1=2,a_2=11,a_3=\frac{31}{2}$ $20(a_{n+2}^2+a_{n+1}^2)=(4a_{n+2}-2a_{n+1})^2+(2a_{n+2}+4a_{n+1})^2=\left(6a_n-2a_{n+1}+2\sqrt{5(a_n^2+a_{n+1}^2)}\right)^2+\left(3a_n+4a_{n+1}+\sqrt{5(a_n^2+a_{n+1}^2)}\right)^2 = (6a_n-2a_{n+1})^2+(3a_n+4a_{n+1})^2+25(a_n^2+a_{n+1})^2+2 (2*(6a_n-2a_{n+1})+3a_n+4a_{n+1})\sqrt{5(a_n^2+a_{n+1}^2)}=70a_n^2+45a_{n+1}^2+30a_n \sqrt{5(a_n^2+a_{n+1}^2)}=\left(5a_n+3\sqrt{5(a_n^2+a_{n+1}^2)}\right)^2=(6a_{n+2}-4a_n)^2$ So $\sqrt{5(a_{n+2}^2+a_{n+1}^2)}=3a_{n+2}-2a_n$ And so $2a_{n+3}=3a_{n+2}+3a_{n+1}-2a_n$ is always rational