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

Stevenhagen’s conjecture on the proportion of solvable negative Pell equations

Determine the limiting proportion, among square‑free integers d with all odd prime divisors p≡1 (mod 4), of parameters d for which the negative Pell equation x^2−dy^2=−1 is solvable in integers. Specifically, prove that the limit exists and equals c_Pell=1−∏_{j=1}^{∞}(1−2^{−(2j−1)}).

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

Background

The appendix recalls Stevenhagen’s conjecture on the solvability of the negative Pell equation in the natural subfamily where the local norm obstruction is removed (all odd prime divisors of d are ≡1 mod 4). The conjecture predicts a precise irrational constant c_Pell≈0.58057… as the limiting proportion.

The authors reinterpret Stevenhagen’s heuristic in Selmer‑group language: the condition x2−dy2=−1 reduces to asking whether −1 arises from the global norm inside the Selmer group Sel_2(O×(χ)/Q), and the conjectural proportion is tied to the distribution of the dimension of Sel_2+(O×(χ)/Q).

References

In light of this observation, it is natural to ask not for what proportion of all d′ (negative Pell) is soluble, but for what proportion of d′∈F. The expected answer to this question is given by a beautiful conjecture of Stevenhagen. Conjecture (Stevenhagen) Let F_{Pell} denote the set of square-free integers d′ for which (negative Pell) is soluble. Then the limit lim_{X→∞} #{d′∈F_{Pell}: d′<X}/#{d′∈F: d′<X} exists and is equal to the irrational number c_{Pell}=1−∏_{j=1}{∞}(1−2{−(2j−1)})=0.58057….

— 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 Appendix, Conjecture (Stevenhagen)