<s>[INST] 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.
 [/INST] 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
$$a b + \sqrt{(a^2-1)(b^2-1)}= \frac{1}{4} (s+1/s)(t+ 1/t) + \frac{1}{4} (s-1/s)(t- 1/t)=\\ = \frac{1}{2}( s t + 1/(s t) )$$
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