Chen-Gendron Spin-Parity Identity for k-Differentials

Precise statement

For odd $k$ with $\gcd(n,k) = \gcd(n+1,k) = 1$, is $N_k(n) \equiv \lfloor (k+1)/4 \rfloor \pmod 2$, where $N_k(n)$ counts pairs $1 \le b_i \le (k-1)/2$ with $b_1 + b_2 \ge (k+1)/2$ and $b_2 \equiv n b_1 \pmod k$? Conjectured by Chen and Gendron; its proof removes a conditional step in the genus-zero and genus-one spin-parity classification.

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Proved by the AxiomProver system with a Lean-checked core.

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Verification boundary

lean checked statement unaudited

Core argument Lean-checked, with an expert-written exposition. Tier: the Lean-checked core comes from the proving system itself; the statement correspondence is not independently audited.

Correctness: supported · statement fidelity: unaudited · peer review: none

Timeline

1. 2026-02-03

   arXiv:2602.03722 - Parity of k-differentials in genus zero and one

   VibeMathed reports this item as resolved. VibeMath preserves that report as a source assertion and has not independently authored a plain-language mathematical explanation.

Known method families

argument (source-reported)

Source-reported tools: argument.

Independent: unknown · difference confidence: 0

What remains uncertain

VibeMath has not independently audited the mathematical statement, proof, or novelty claim.

• VibeMath has not independently verified the mathematical claim.
• AI-attempt independence and training-data exposure are unknown unless explicitly documented.
• VibeMath has not independently audited the mathematical statement, proof, or novelty claim.