Integer solutions to $r\cdot a^2+t=s\cdot b^2$ forms sequences with $a_{n+1}\cdot b_n-a_n\cdot b_{n+1}$ is invariant

Question

Conjecture:
Given natural numbers $r,s,t$. Suppose there are an infinite number of solutions $a_n,b_n \in \mathbb N$ to $ra_n^2+t=sb_n^2$, where $a_n, b_n$ correspond to increasing sequences of solutions, then $a_n\cdot b_{n+1} - a_{n+1}\cdot b_n$ is invariant equal for each $n$ greater than some $N>0$.

I've done a lot of tests without exceptions. The number $N$ is typically $\le 1$. I would like to see ideas how to proceed.

Some $(r,s,t)$ values plus sets of $(a,b)$ values:

(1,2,1)
{(1,1),(7,5),(41,29),(239,169),(1393,985),(8119,5741)}

(2,1,1)
{(0,1),(2,3),(12,17),(70,99),(408,577),(2378,3363)}

(1,2,2)
{(0,1),(4,3),(24,17),(140,99),(816,577),(4756,3363)}

(3,2,2) 
{(0,1),(4,5),(40,49),(396,485),(3920,4801)}

(2,3,1)
{(1,1),(11,9),(109,89),(1079,881),(10681,8721)}

There seems to be a matrix

$ \begin{pmatrix} \alpha & ms \\ mr & \alpha \\ \end{pmatrix} $

that transform $(a_{n},b_{n})$ to $(a_{n+1},b_{n+1})$, but I haven't proved that and I'm insecure about $\alpha$ and $m$. It could be that $\alpha=\sqrt{m^2rs+1}$ and that the determinant is unital.

The conjecture is false with counter-example

(1,2,7)
{(1,2),(5,4),(11,8),(31,22),(65,46),(181,128),(379,268)}

What's left is
Given $r,s,t\in\mathbb R_+$ and $S=\{(a,b)\in\mathbb R_+^2|rs^2+t=sb^2\}$, then it exist a sequence $(a_n,b_n)\in S$ such that:
$$a_{n+1}b_n-a_nb_{n+1}=2t \tag 1$$ $$ \begin{pmatrix} a_{n+1} \\ b_{n+1} \\ \end{pmatrix} = \begin{pmatrix} \sqrt{4rs+1}& 2s \\ 2r & \sqrt{4rs+1} \\ \end{pmatrix} \begin{pmatrix} a_n\\ b_n \\ \end{pmatrix} \tag 2$$ which I consider as virtually proved by the accepted answer.

Answer 1

Let $a,b$ solve the equations and let $\begin{pmatrix} \alpha & s \\ r & \alpha \\ \end{pmatrix}\begin{pmatrix}a \\b\\\end{pmatrix}=\begin{pmatrix}A \\B\\\end{pmatrix}$ with $\alpha$ as in your post and $m=1$.

Then straightforward algebra shows that $A,B$ also solve the equation.

Then $Ab-aB=(\alpha a+sb)b-a(ra+\alpha b)=sb^2-ra^2=t$, as you require.

Note. The above proof does not assume anything about the integer nature of the parameters. One can multiply each of $r,s,t$ by any number $m$ say and obtain the corresponding result. In that case the invariant value will of course be $mt$.

I assume that is what you you are interested in and that you will wish to choose $m$ so that $(m,\alpha)$ is an integer solution of the Pell equation $$(rs)x^2+1=y^2.$$

Furthermore, so as not to 'miss' solutions, $(m,\alpha)$ should be the fundamental solution.