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If one takes in general $(\star)\, \,x^2-dy^2=c$ where $d$, $c$ in $\mathbb{N}$.

Taking $d=w^2p^2+p$ with $w\in \mathbb{Q}\ge 1$ and $p\in \mathbb{Z}$ which is verified (explained later), for the matrix $$A=\begin{pmatrix}2w^2p+1&2w(w^2p^2+p)\\2w&2w^2p+1\end{pmatrix}$$ if $X_0$ is a solution to $(\star)$ then $AX_0$ is another one.

Now $w$ could be taken in a cool way basicaly say $d=a^2b^2+cb$ with $c\in\mathbb{Z}$, $|c|<|a|$ and $c$ coprime with $b$ and $a$, letting $w=\frac{a}{c}$ and $p=cb$ the matrix $A$ is in $\mathbb{Q}$ but can have a power $A^n$ with integer entries. So to say that i didn't find any reference for this idea which is surprising. This is related and known of course as a Pell equation when $w\in \mathbb{N}$.

A question is if there is a related topic discussion to this approach since Pell equations are known, and as a conjecture to give certain family of $A$ with $A^n$ of integer entries. (It appears there are many). Thanks

Edit, i'll illustrate this in an example just for clarity: $$x^2-2021y^2=d^2$$ one solution is $(d,0)$, i took $2021$ by chance as it is within what i can get, (i don't know if it should work for $2020$) since $2021=\frac{45^2}{4^2}4^2-4$. An easy argument says if the numerator of $w$ here $45$ is $5 \pmod{8}$ Then $A^3\in \mathbb{M}_2(\mathbb{Z})$ so

$$A=\begin{pmatrix}-1011.5&45472.5\\22.5&-1011.5\end{pmatrix}$$ and $$A^3=\begin{pmatrix}-4.139590049\times 10^9&1.8609747948\times 10^{11}\\9.2081880\times 10^7&-4.139590049\times 10^9\end{pmatrix}.$$

Edit. It seems such $A$ has an all integer power $A^n$ if and only if $c$ is a power of two and mainly $|c|= 1, 2, 4$,

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    $\begingroup$ I am not sure if there is a question in this post. $\endgroup$ – Stanley Yao Xiao Jul 13 at 11:17
  • $\begingroup$ A question is implicitly if there is a related topic discussion to this approach since Pell equations $(\star)$ are known, besides a natural question rises as a conjecture to give certain $A$ with $A^n$ of integer entries for some $n$ etc. $\endgroup$ – Toni Mhax Jul 13 at 11:29
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    $\begingroup$ Why not edit the question to explicitly ask what you want to know? $\endgroup$ – David Roberts Jul 13 at 11:38
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If I understand correctly, your question is the following: suppose that for a given positive integer $d$ the equation

$$\displaystyle x^2 - dy^2 = c \text{ } (\ast)$$

has a solution in integers $x,y$ for some integer $c$. Then does there exist an infinite family of solutions generated by $A^k (x,y)^T$ for some $A \in \text{GL}_2(\mathbb{Z})$ having infinite order?

The answer is yes, and was answered completely by Siegel. Indeed, the equation $(\ast)$ has finitely many solutions modulo the action induced by the unit group of the ring of integers of $\mathbb{Q}(\sqrt{d})$, which always has rank one. See the following paper of Siegel: The average measure of quadratic forms with given determinant and signature.

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  • $\begingroup$ ok but here i gave the $A$ along. It is related to this. I know there are others given in terms of fundamental solutions theory. Thanks. $\endgroup$ – Toni Mhax Jul 13 at 13:09
  • $\begingroup$ @ToniMhax Siegel also gave an explicit $A$ in terms of a fundamental solution of the equation. It's in the paper. $\endgroup$ – Stanley Yao Xiao Jul 13 at 13:11
  • $\begingroup$ but one should compute a fundamental non trivial solution i guess. Here $A$ is in $\mathbb{M}_2(\mathbb{Q})$ that we elevate to a certain power iteration... $\endgroup$ – Toni Mhax Jul 13 at 13:21
  • $\begingroup$ @ToniMhax determining whether the equation $x^2 - dy^2 = c$ has a solution in the first place is extremely difficult. Indeed, even in the case of $c = -1$ things are very challenging. We don't know how to count the number of $d \leq X$ for which $x^2 - dy^2 = -1$ has an integer solution for example, and it was only 10 years ago that an upper and lower bound of the correct order of magnitude was established by Fouvry and Kluners. $\endgroup$ – Stanley Yao Xiao Jul 13 at 14:11
  • $\begingroup$ Yes of course i'll illustrate the simple algo in an exemple just for trial and record. $\endgroup$ – Toni Mhax Jul 13 at 14:16

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