Prove $a_{n+2}=3a_n+2\sqrt{2a_n^2+2a_{n+1}^2}$ is an integer Given the sequence
$$a_1=a_2=1;\ a_ {n+2} = 3a_n + 2\sqrt{2a_n^2 + 2a_{n+1}^2}$$
prove that $a_n$ is an integer for all $n\in\mathbb N$.
Attempt
It is enough to show that $2a_n^2 + 2a_{n+1}^2$ is a perfect square. That means it's an even perfect square and so divisible by 4; thus $a_n^2+a_{n+1}^2=2k^2$ for some integer $k$. I think can solve this diophantine equation but I can't relate the solution to the original problem. Can anyone help?

We have that:
$$a_{n+2}-3a_n=2\sqrt{2a_n^2+2{a_{n+1}}^2}\Rightarrow (a_{n+2}-3a_n)^2=(2\sqrt{2a_n^2+2{a_{n+1}}^2})^2\Rightarrow\\ {a_{n+2}}^2-6a_na_{n+2}+9a_n^2=8a_n^2+8{a_{n+1}}^2\Rightarrow {a_{n+2}}^2+{a_{n+1}}^2=-a_n^2+6a_na_{n+2}+9{a_{n+1}}^2\Rightarrow \\{a_{n+2}}^2+{a_{n+1}}^2=-a_n^2+6a_n(3a_n+2\sqrt{2a_n^2+2{a_{n+1}}^2})+9{a_{n+1}}^2\Rightarrow\\ {a_{n+2}}^2+{a_{n+1}}^2=17a_n^2+9{a_{n+1}}^2+12a_n\sqrt{2a_n^2+2{a_{n+1}}^2}\Rightarrow\\ 2({a_{n+2}}^2+{a_{n+1}}^2)={(4a_n)}^2+\left(3\sqrt{2a_n^2+2{a_{n+1}}^2}\right)^2+2\cdot (4a_n)\cdot \left(3\sqrt{2a_n^2+2{a_{n+1}}^2}\right)\Rightarrow\\ 2({a_{n+2}}^2+{a_{n+1}}^2)=\left(4a_n+3\sqrt{2a_n^2+2{a_{n+1}}^2}\right)^2\ (1)$$
Now we will prove the claim with induction.
It is a matter of a few calculations to verify that $a_3$ and $a_4$ are integers. 
Now we suppose that $a_n$ is an integer for every $4\leq n\leq k$. Then we immediately get that $\sqrt{2{a_{k-2}}^2+2{a_{k-1}}^2}$ is an integer as the difference of the integers $a_k$ and $3a_{k-2}$. From $(1)$ it follows that $\sqrt{2{a_{k-1}}^2+2{a_k}^2}$ is an integer too. 
Hence $a_{k+1}=3a_{k-1}+\sqrt{2{a_{k-1}}^2+2{a_k}^2}\in \mathbb Z$ as a sum of integers and the inductive proof is complete. Thus we are done. 
