Question
Given a sequence $a_0 = a_1 = 97\sqrt{2}$ and $$\forall n \space \space a_{n+1}= \frac{1}{\sqrt{2}}[a_n a_{n-1} + \sqrt{(a_n^2-2)(a_{n-1}^2-2)}]$$ Proof that $$2+\sqrt{2+a_n\sqrt{2}}$$ is a perfect square
My approach
Use mathematical induction to prove that (base case are trivial, this is the inductive step)
$$2+\sqrt{2+a_na_{n-1}+\sqrt{(a_n^2-2)(a_{n-1}^2-2)}}$$ However, this does not hold true for all $a_n$, there seems to be an invariant from the initial value that makes it perfect square. Any hints would be appreciated.
This is not an answer, but I want to share some of my findings and I cannot do it in a comment. Let $$x_n=2+\sqrt{2+a_n\sqrt{2}}$$ for $n=0,1,...$. Now it is easy to prove that indeed $x_0,x_1,x_2,x_3$ are perfect squares. But there is a precise relationship between $x_n$ for $n>2$ and it is $$\sqrt{x_n}=\sqrt{x_{n-1}x_{n-2}}-\sqrt{x_{n-3}}$$ Now the base case for $k=3$ is trivial to verify. If by induction we suppose that then for $k=4,5,,...n$ we have $$\sqrt{x_k}=\sqrt{x_{k-1}x_{k-2}}-\sqrt{x_{k-3}}$$ and also $$x_k \text{ is a perfect square}$$ then it remains to show that $$\sqrt{x_{n+1}}=\sqrt{x_{n}x_{n-1}}-\sqrt{x_{n-2}}$$ which implies than that $x_{n+1}$ is a perfect square and by induction we are done. However the algebra is a bit messy, but I believe it could be a good line of attack.