Negative Pell’s Equation

We want to understand how many discriminants D have solutions to negative Pell’s equation. First of all we have some local restrictions on the
the prime factors of D if there is a solution. There cannot be prime factors 3 modulo 4. If we restrict to these prime factors for D, (special discriminants) how likely is it to actually have a solution to the equation.

$\displaystyle d=\left\{\begin{array}{ll} D & \text { if } D \text { is odd } \\ D / 4 & \text { if } D \text { is even } \end{array}\right.$

$\displaystyle x^{2}-d y^{2}=-1 \text { with } x, y \in \mathbb{Z}$

$\displaystyle \mathscr{D}=\{D>0 \text { fundamental discriminant }: p | D \Rightarrow p \equiv 1 \text { or } 2 \bmod 4\}$

The total number of special discriminants less than ${X}$ is given by

$\displaystyle \mathscr{D}(X) \sim c_{1} \cdot \frac{X}{\sqrt{\log X}}$

${\alpha:=\prod_{i, \text { odd }}\left(1-2^{-j}\right)=\prod_{i=1}^{\infty}\left(1+2^{-j}\right)^{-1}=.4194224417951 \cdots}$

Stevenhagen conjectured that the number of positive squarefree d ${\leq X}$ for which the negative Pell equation s solvable is asymptotic to

$\displaystyle (1-\alpha) \frac{4}{3} \cdot \frac{9}{8 \pi} \prod_{p \equiv 1 \bmod 4}\left(1-p^{-2}\right)^{\frac{1}{2}} \cdot \frac{X}{\sqrt{\log X}}$

Fouvry, Kluners detect existence of solutions to negative Pell’s equation as equality of the narrow and ordinary class group of ${\mathbb{Q}(\sqrt{D})=\mathbb{Q}(\sqrt{d})}$ ${\mathrm{C}_{D} \cong \mathrm{Cl}_{D}}$

${\mathcal{N}\left(\varepsilon_{D}\right)=-1 \Leftrightarrow {rk}_{2^{k}}\left(\mathrm{C}_{D}\right)={rk}_{2^{k}}\left(\mathrm{Cl}_{D}\right) \forall k \geq 2}$

Now they compute joint distribution of the 4-ranks of class groups. ${\delta(a, b):=\lim _{X \rightarrow \infty} \frac{\#\left\{D \in \mathscr{D}: D<X, \mathrm{r}_{4}\left(\mathrm{C}_{D}\right)=a \text { and } \mathrm{rk}_{4}\left(\mathrm{Cl}_{D}\right)=b\right\}}{\mathscr{D}(X)}}$

To access these, they use the following formulae for the ranks in terms of characters and factorization of discriminants.

$\displaystyle 2^{\mathrm{rk}_{4}\left(\mathrm{C}_{D}\right)}=\frac{1}{2 \cdot 2^{\omega(D)}} \sum_{D=D_{0} D_{1} D_{2} D_{3}}\left(\frac{-1}{D_{3}}\right)\left(\frac{D_{2}}{D_{0}}\right)\left(\frac{D_{1}}{D_{3}}\right)\left(\frac{D_{0}}{D_{3}}\right)\left(\frac{D_{3}}{D_{0}}\right)$

$\displaystyle 2^{\mathrm{rk}_{4}\left(\mathrm{Cl}_{D}\right)}=\frac{2^{\mathrm{rk}_{4}\left(\mathrm{C}_{D}\right)}}{2}+\frac{1}{4 \cdot 2^{\omega(D)}} \sum_{D=a b c d}\left(\frac{\mathrm{a} \overline{\mathrm{b}}}{\mathrm{c} \overline{\mathrm{d}}}\right)_{4}^{2}$

On the analytic side, the use mely Siegel-Walfisz theorems and double oscillllation of the character involved.

The main term is as predicted by Cohen-Lenstra Heuristics: The prime to 2 part of the class group behaves like the prime to 2 part of a random finite abelian group– (a group with large automorphism is chosen with less probability)

$\displaystyle \lim _{X \rightarrow \infty} \frac{\sum_{-X<D<0} f\left(\mathrm{Cl}_{D}^{\prime}\right)}{\sum_{-X<D<0} 1}=\lim _{X \rightarrow \infty} \frac{\sum_{G, \# G \leq X} f\left(G^{\prime}\right) / \# A u t(G)}{\sum_{G, \# G \leq X} 1 / \# A u t(G)}$