Spin parity for kk-differentials

About 4 years old · traced to

Let k≥3k\geq 3 be an odd integer. For any integer nn, let Nk(n)N_k(n) be the number of pairs of positive integers (b1,b2)(b_1,b_2) such that b1,b2≤(k−1)/2b_1,b_2\leq (k-1)/2, b1+b2≥(k+1)/2b_1+b_2\geq (k+1)/2, and b2≡nb1(modk)b_2\equiv nb_1\pmod{k}. If nn satisfies gcd⁡(n,k)=gcd⁡(n+1,k)=1\gcd(n,k)=\gcd(n+1,k)=1, then

Nk(n)≡⌊k+14⌋(mod2).N_k(n)\equiv \left\lfloor\frac{k+1}{4}\right\rfloor\pmod{2}.
References

Progress summary

Refreshed
Claimed solved

A recent preprint claims to prove the conjecture, but the proof has not yet been independently verified.

The problem asks whether the stated parity formula always holds for odd kk under the given coprimality conditions. No source in the scan identifies its original proposer or date.

February 2026 proof claim

“Parity of kk-differentials in genus zero and one” states that Conjecture 1.1 is true. Its proof uses Nk(n)=Fk(n+1)−Fk(n)N_k(n)=F_k(n+1)-F_k(n) and Jacobi-symbol arguments, and claims to make earlier conditional spin-parity results unconditional. The source attributes the key observation and formal proof of a main lemma to AxiomProver, with Lean verification artifacts discussed in an appendix.

Current status (as of February 2026): The conjecture is claimed proved in a preprint, but independent mathematical verification is still absent.

Sources

Solutions 0

No solutions have been posted yet.