Solving the sequence $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$: proving that $2+2a_n$ is a perfect square Question:

Let $a_1=a_2=97$ and $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$ for $n>1$. Prove that
(a) $2+2a_n$ is a perfect square, and (b) $2+\sqrt{2+2a_n}$ is a perfect square.

I changed the given recursive formula by squaring, and the result was as follows:
$$(a_{n+1}^2+a_n^2+a_{n-1}^2)-2a_{n+1}a_na_{n-1}-1=0$$
$\Rightarrow$ $$(a_{n+1}+a_n+a_{n-1})^2-2(a_{n+1}a_n+a_na_{n-1}+a_{n-1}a_{n+1}+a_{n+1}a_na_{n-1})-1=0$$
$\Rightarrow$ $$(1+a_{n+1})^2+(1+a_n)^2+(1+a_{n-1})^2-2(a_{n+1}+a_n+a_{n-1}+a_{n+1}a_n+a_na_{n-1}+a_{n-1}a_{n+1}+a_{n+1}a_na_{n-1}+1)-2=0$$
$\Rightarrow$ $$(1+a_{n+1})^2+(1+a_n)^2+(1+a_{n-1})^2-2(1+a_{n+1})(1+a_n)(1+a_{n-1})-2=0$$
And consequently, I lost the way :(
I thought of proving in a inductive way, that is :
$$2+2a_1=196=14^2$$
When we set $2+2a_n=k^2$, $2+2a_{n-1}=l^2$,
$$ 2+2_{n+1}=\frac{(k^2-2)(m^2-2)}{4}+\sqrt{\left(\left(\frac{k^2-2}{2}\right)^2-1\right)\left(\left(\frac{m^2-2}{2}\right)^2-1\right)}$$
As a result, I again lost the way :/
I think I sill have not got the main points. Could you give me some clues about this problem? Thanks.
 A: $  $(Essentially taken from Sequence problem on AoPS.)
The key point is to recognize the connection between the recurrence formula
$$
a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}
$$
and the addition formula for the inverse hyperbolic cosine:
$$
\DeclareMathOperator{\arcosh}{arcosh}
\arcosh u + \arcosh v=\arcosh \left(uv + {\sqrt {(u^{2}-1)(v^{2}-1)}}\right) \, .
$$
First note that $a_n > 1$ for all $n$, so that the sequence is well-defined, and we can set $x_n = \arcosh a_n$. Then
$$
 x_{n+1} = x_n + x_{n-1} \, ,
$$
which together with $x_1 = x_2$ implies that $(x_n)$ is a multiple of the Fibonacci sequence: $x_n = F_n x_1$. (In the following we need only that $x_n$ is an integer multiple of $x_1$.)
So we have
$$
 a_n = \cosh x_n = \cosh \left( F_n x_1 \right) \, ,
$$
and using the half-angle formula for $\cosh$ we conclude that
$$ 
2 + 2a_n = \left( 2 \cosh \left( \frac{F_n x_1}{2} \right)\right)^2 
$$
and then
$$
2+\sqrt{2+2a_n} = \left( 2 \cosh \left( \frac{F_n x_1}{4} \right)\right)^2 
$$
It remains to show that
$$
 y_n = 2 \cosh \left( \frac{F_n x_1}{4} \right) = e^{F_n x_1/4} + e^{-F_n x_1/4} = \alpha^{F_n} + \left(\frac{1}{\alpha}\right)^{F_n} 
$$
with $\alpha = e^{x_1/4}$ is an integer for all $n$. Setting $n=1$ allows to determine $\alpha$: 
$$
y_1 = \sqrt{2+\sqrt{2+2 \cdot 97}} = 4 = \alpha + \frac 1\alpha
$$
has the solution
$$
 \{ \alpha, \frac 1\alpha \} = \{ 2 + \sqrt 3, 2 - \sqrt 3 \} \, .
$$
So finally we get
$$
 y_n = \left(  2 + \sqrt 3\right)^{F_n} + \left(  2 - \sqrt 3\right)^{F_n}
$$
and that is indeed an integer for all $n$ (see for example The number $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is an integer).
A: Simply redoint the argument of @Martin R: so I understand it better.
If $a = \frac{1}{2}(s +1/s)$ then $a^2-1= \left(\frac{1}{2}(s - 1/s)\right)^2$. Now for
$a =  \frac{1}{2}(s + 1/s)$, $b =  \frac{1}{2}(t + 1/t)$ ( $s$, $t> 1$)  we get
$$a b + \sqrt{(a^2-1)(b^2-1)}= \frac{1}{4} (s+1/s)(t+ 1/t) + \frac{1}{4} (s-1/s)(t- 1/t)=\\ = \frac{1}{2}( s t + 1/(s t) )$$
