Dice Question Streamline Icon: https://streamlinehq.com

Stevenhagen’s conjecture on the negative Pell equation

Establish that among positive squarefree integers d whose odd prime factors are all congruent to 1 modulo 4, the limiting proportion of d for which the negative Pell equation x^2−dy^2=−1 is solvable in integers exists and equals 1−∏_{j=1}^∞(1−2^{−2j+1}).

Information Square Streamline Icon: https://streamlinehq.com

Background

The appendix reformulates Stevenhagen’s conjecture in Selmer-theoretic terms and recalls known partial results (e.g., lower bounds of Fouvry–Klüners and the recent proof by Koymans–Pagano). Although not original to this paper, the conjecture is explicitly stated and connected to the paper’s Selmer-based framework.

This classical problem parallels the paper’s equidistribution heuristics for Selmer elements in Frobenian families, highlighting a number-field analogue of the authors’ genus-1 and Mordell–Weil predictions.

References

Conjecture (Stevenhagen) Let \cF_{\textup{Pell} denote the set of square-free integers d' for which negative pell with parameter d' is soluble. Then the limit

\lim_{X\rightarrow \infty}\frac{#{d'\in \cF_{\textup{Pell}: ~~d'<X}{#{d'\in \cF:~~d'<X}

exists and is equal to the irrational number

c_{\textup{Pell}=1-\prod_{j=1}{\infty}(1-2{-2j+1})=0.58057....

negative pell:

x2dy2=1,x^2-dy^2=-1,

Galois module structures and the Hasse principle in twist families via the distribution of Selmer groups (2508.14026 - Bartel et al., 19 Aug 2025) in Conjecture (Stevenhagen), Appendix (Stevenhagen’s heuristic via Selmer groups)