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Statement of the problem


Firstly, consider the infamous Pell's equation: $x^{2}-dy^{2}=1$. Here $x$ and $y$ are integers and $d$ is a nonsquare integer. It is known ([3]) that all solutions of this equation can be obtained from the solution with least positive $x$ called the fundamental solution. By the way, it is convenient to write solutions in the form $a+\sqrt{d}b$. This way if we denote the fundamental solution by $x_{1}+\sqrt{d}y_{1}$ then the $k$-th solution will be $x_{k}+\sqrt{d}y_{k} = (x_{1}+\sqrt{d}y_{1})^{k},k\geq 1$. And I need to note here that this equation is always solvable because we can obtain the fundamental solution from the continued fraction expansion of $\sqrt{d}$.

Now, let's look at the generalised Pell's equation: $x^{2}-dy^{2}=N$. Here $x$,$y$ and $N$ are integers and $d$ is a nonsquare integer. It turns out ([1],[2]) that the solutions (if it's solvable, cause it may not have a solution) all have quite a simple structure: $ \eta_{m}\epsilon_{0}^{k}, 1\leq m\leq h,k \geq 0. $ Where $\eta_{m}$ is the fundamental (with least positive $x$ component) solution of $m$-th equivalence class and $\epsilon_{0}$ is the fundamental solution of standard Pell's equation $x^{2}-dy^{2}=1$.

I'm talking about equivalence classes so here is the definition of equivalence: two solutions $f_{1} = a_{1}+\sqrt{d}b_{1},f_{2} = a_{2}+\sqrt{d}b_{2}$ of the generalised Pell's equation are equivalent if their ratio is a solution of the standard Pell's equation. Note that it doesn't matter if we take $\frac{f_{1}}{f_{2}}$ or $\frac{f_{2}}{f_{1}}$. Because solving the Pell's equation is essentially finding numbers of required norm in the ring $\mathbb{Z}+\sqrt{d}\mathbb{Z}$.

Finally, let's denote by $f(N,d)$ that number $h$ of equivalence classes of solutions to the generalised Pell's equation. This is what I want to research. Experimental results are as follows (changing $N$,$d$ is fixed): it behaves quite randomly, but the average seems to tend to a constant. This motivates me to consider the function: $$ F(M,d) = \sum_{N = 1}^{M} f(N,d) $$ My aim here is to get a good approximate formula for this function and then take a limit $\lim_{M \rightarrow +\infty} \frac{F(M,d)}{M}$ to see if it really tends to a constant.

My current approach


If we could somehow get the Dirichlet series $g(s) = \sum_{N=1}^{+\infty} \frac{f(N,d)}{N^{s}}$ then we'd just use Perron's inversion formula to recover the sum. The hardest thing for me is that $f(N,d)$ doesn't seem to have any simple properties that I could use to get some functional equations, etc. It feels like pure randomness and I'm stuck looking from this angle. For instance, trying $\sum_{x,y >0} \frac{1}{(x^{2}-d y^{2})^{s}(x+\sqrt{d}y)^{s-1}}$ yields something that, roughly speaking, behaves like a simplified version of that Dirichlet series near $s = 1$ but summing it is problematic for me. Because we have a double sum and this means the error terms should be quite small to be able to sum it properly.

My questions


  1. How difficult is this problem?
  2. Are there any tools known (maybe from analytic number theory) to work with such problems?

Some literature


  1. Matthews K.R.: The diophantine equation $x^2-Dy^2=N, D>0.$ Expositiones Mathematicae, 18, 323-331 (2000)
  2. Mollin R.A.: Simple Continued Fraction Solutions for Diophantine Equations. Expositiones Mathematicae, 19, 55-73 (2001)
  3. Clark, Pete. Number Theory: A Contemporary Introduction. University of Georgia. 93-102
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    $\begingroup$ If I am understanding correctly, are you asking for the number of representations of $N$ by $f$ up to $\text{GL}_2(\mathbb{Z})$-equivalence? If so, then this is basically just the divisor function on $N$ provided that all of the prime factors of $N$ either split in the quadratic field or have even multiplicity. $\endgroup$ Commented Jan 23, 2023 at 1:54
  • $\begingroup$ @StanleyYaoXiao Yes, you're right. Do you know, by chance, any good papers on the way this number of representations is counted? I want to have a deeper understanding. $\endgroup$ Commented Jan 23, 2023 at 9:50
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    $\begingroup$ What you call $F(M,d)$ is more or less completely understood; see for example these notes by Andrew Granville: dms.umontreal.ca/~andrew/Courses/Chapter4.pdf $\endgroup$ Commented Jan 23, 2023 at 15:53
  • $\begingroup$ @StanleyYaoXiao Oh, thank you. So if I take an arbitrary quadratic form of the same discriminant as $x^{2}-dy^{2}$ and find the number of representations of $N$ by both up to $GL_{2}(\mathbb{Z})$-equivalence then is it going to be the same number? $\endgroup$ Commented Jan 24, 2023 at 11:57

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