Conjecture 2: $H_a(2)=1$

Open

HilbertSixteenth.quadratic_algebraic_conjecture

by Lucas · Sep 30, 2026 · Mathlib 0df444a (Lean v4.33.1)

dynamical-systemshilbert-16th-problemlimit-cycles

Every quadratic polynomial vector field (degree at most $2$ ) has at most one algebraic limit cycle, and some quadratic vector field has exactly one:

H_a(2)=1.

This is the simplest case of Conjecture 1 and is open.

Preamble

import Definitions.Def_HilbertSixteenth_PolyFields

Formal statement

namespace HilbertSixteenth
theorem quadratic_algebraic_conjecture : algebraicHilbertNumber 2 = 1 := by sorry
end HilbertSixteenth

Source

J. Llibre, *Sobre el problema 16 de Hilbert*, La Gaceta de la RSME 18 (2015), no. 3, pp. 543–554, §7, Conjecture 2.

Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic) — same agent as the drafter; NON-BLIND

Disclosure — NON-BLIND read-back. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statements of this proposal, at the explicit request of the proposal owner. It is not independent testimony: its author knew the intended meaning of the code while writing it, so it may unconsciously read that intent into the code. Reviewers should compare the Lean code against the source directly and should not treat this text as an independent audit.

The algebraic Hilbert number at $d=2$ equals $1$ : the supremum, over all polynomial vector fields $V=(P,Q)$ with $\max(\operatorname{totdeg}P,\operatorname{totdeg}Q)\le2$ , of the number (in $\mathbb N\cup\{\infty\}$ ) of algebraic limit cycles of $V$ (limit cycles contained in the real zero set of a non-zero polynomial) is exactly $1$ . Equivalently: every such $V$ has at most one algebraic limit cycle, and at least one such $V$ has exactly one.

Human review

Endorsed by Shuze Chen · Sep 30, 2026
Confirmed by the moderator at approval.
Endorsed by Lucas · Sep 30, 2026
Confirmed by the mission captain (proposal self-audit).

View graph

Conjecture 2: Ha(2)=1H_a(2)=1Ha​(2)=1

Conjecture 2: $H_a(2)=1$