Chen-Gendron Spin-Parity Identity for k-Differentials

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.

Result

Proved

Status

Resolved

AI contribution

AI-discovered

Method

Argument

Field

Flat surfaces & moduli

Posed by

Dawei Chen & Quentin Gendron

Year posed

2022

Years open

4y

Solved

2026-02-03

Model

AxiomProver

Vendor

—

Collaborators

—

Verification

Lean-verified

Publication

Preprint

Significance

10 / 100*

Disclosed cost

—

Wikipedia

No dedicated article