IMO 1988 Q6 $a_n = ...$ My question is in bold below. Here is a summary of my working on this problem which is question 6 from the IMO 1988 paper:
For $\frac{a^2+b^2}{ab+1}=n$, where $a,b,n \in Z^+$ 
let $n=x^2$
then for all $x \in Z^+, (a,b)$ have infinite solutions such that
$\hspace{2cm}$$a_m = a_0,a_1,a_2,... = 0, x,x^3,x^5-x,x^7-2x^3,x^9-3x^5+x,...$
$\hspace{2cm}$and $b=a_{m-1}, a_{m+1}$ for all $a_m$.
e.g. when $x=2$
valid solutions include: 
$\hspace{2cm}$$(2,0)$ and $(2,8)$
$\hspace{2cm}$$(8,2)$ and $(8,30)$
$\hspace{2cm}$$(30,8)$ and $(30, 112)$ etc.
when $x=3,$
$\hspace{2cm}$$(3,0)$ and $(3,27)$
$\hspace{2cm}$$(27,3)$ and $(27,240)$
$\hspace{2cm}$$(240,27)$ and $(240,2133)$ etc.
An observation is that $a_m$ is a list of infinite solutions. For all $x$ (for all positive integers), $n$ must be a square. Another way to state this would be to say $n=x$, so then $a_m= \sqrt{x},x^\frac{3}{2},...$ and so on, $\therefore$ not integer solutions.
I am hoping I can either prove $a_m$ is a list of all solutions, therefore n must be a square integer, or to find the $m^{th}$ term for $a_m$. 
My main question is where does this sequence for $a$ come from? It is easily reproducible, and looks like some sort of binomial expansion, but an expansion of what? And could I possibly prove this is a list of all solutions?
Some more terms of $a_m$ for curiosity:
$x^{11}-4x^7+3x^3$
$x^{13}-5x^9+6x^5-x$
$x^{15}-6x^{11}+10x^7-4x^{3}$... a clear pattern emerges.
I have briefly read that this is connected to the geometry of conics. If anybody has a good reference to more reading on this it would be greatly appreciated.
Also is there any links with Vieta jumping and $a_m$?
 A: $$ \frac{n_{i}^2+x^2}{1+n_{i} x}=s \tag{1}$$
$ \ $
$$n_{i}^2+x^2=s(1+n_{i} x) \\ x^2 + (-s n_{i})x+(n_{i}^2-s)=0 \\
x_{1,2}=\frac{1}{2}\bigg(n_{i} s\pm \sqrt{n_{i}^2 s^2+4(s-n_{i}^2)}\bigg)
$$
if $(n_{i}s) \neq 0$
$$x_{1,2}=\frac{n_{i} s}{2}\bigg(1\pm \sqrt{1+4 \bigg( \frac{s-n_{i}^2}{n_{i}^2 s^2} \bigg)} \bigg) \tag{2}
$$
Note that the element under the square root looks a lot like the square of a binomial:
$$
1+4 \bigg( \frac{s-n_{i}^2}{n_{i}^2 s^2}\bigg)=\bigg(1-2 \frac{ q}{n_{i} s}  \bigg)^2=1+4 \frac{q^2}{n_{i}^2 s^2}-4\frac{q}{n_{i} s}
$$
So we can rewrite $(2)$ as:
$$ x_{1} =\frac{n_{i} s}{2} \Bigg(1 - \bigg(1-\frac{2 q}{n_{i} s}  \bigg)   \Bigg)=q \\ 
 x_{2} =\frac{n_{i} s}{2} \Bigg(1 + \bigg(1-\frac{2 q}{n_{i} s}  \bigg)   \Bigg)= n_{i} s -q
$$

Since the equation $(1)$ is symmetric for $n_{i}$ and $x$, the procedure done in $ (2) $ to get $ x $ can be used equivalently to get $ n_ {i} $ and vice versa, we can write
$$  n_{i+1}=n_{i} s -n_{i-1} \tag{3} $$

$$  \frac{0^2+n_{i+1}^2}{1+0  n_{i+1}} =n_{i+1}^2=s \Leftarrow\Rightarrow n_{i+1}=\sqrt{s}
$$
The solutions $ n_ {i} $ must be positive so the solution $ n_ {i} = 0 $ will surely be the smallest that can be found therefore we call it $ n_ {0} = 0 $ we will then have $ n_ {1 } = \sqrt {s} $. We have found two solutions, which if compared with all the others, with the same $ s $ take smaller values:
$$ \forall i , s>1 : 0=n_{0}< \sqrt{s}=n_{1}<n_{i}.
$$
Known the first two solutions we can find the third and so on:
$$n_{0}=(0 )\sqrt{s} \\
n_{1}=(1) \sqrt{s} \\s
n_{2}=s n_{1}-n_{0}=(s) \sqrt{s}\\
n_{3}=(s^{5}-1)\sqrt{s}  \\
\vdots$$
Equation $(3)$ has the following solution:
$$ 
n_{i}=\frac{\sqrt{s}\bigg( \big(s+\sqrt{s^2-4}\big)^i - \big(s-\sqrt{s^2-4}\big)^i \bigg)}{2^{i} \sqrt{s^2-4}}
$$
which for $ i> 1 $ has the following series representation:
$$
n_{i}=\sum_{k=0}^{\frac{1}{4}(2i+i-(-1)^i)} (-1)^k \binom{i-k-1}{k} s^{\frac{1}{2} (2i-1-4k)}
$$
For more, look at:
Simple proof for the Legendary Question 6. International Mathematical Olympiad (IMO) 1988
