The Welch function is not third-order sum-free

A mix of human-written and AI-generated textHuman understanding: some partsmath.NT — Number Theorycs.IT — Information Theory

Contributed by Deep Bhattacharjee ↗

Version 1 / Oct 22, 2026 / CC BY 4.0

Abstract

Let n=2m+1n=2m+1. Hou and Zhao conjectured that for n≥7n\ge7 the Welch function X2m+3X^{2^m+3} on $\F_{2^n}$ sums to zero over some three-dimensional affine subspace, that is, it is not third-order sum-free. We prove this. The proof counts exactly the elements xx for which some subspace spanned by 11, xx and a third element gives a zero sum: there are (2n+1−3Kn)/4(2^n+1-3K_n)/4 of them, where KnK_n is a binary Kloosterman sum.

Provenance statement

Deep Bhattacharjee chose the problem, Conjecture 4.10 of Hou and Zhao (arXiv:2609.31489), and directed the work. Claude (Anthropic), through Claude Code, searched the literature, found and checked the proofs, wrote the verification programs and drafted the paper. The proof is complete by hand in eight pages: the kernel Z(x) of the Hou–Zhao linearized polynomial has dimension 3 exactly when two traces equal 1, the number of such x is (2^n+1−3K_n)/4 with K_n the binary Kloosterman sum, and Weil's bound makes it positive for every n ≥ 7. C and Python programs check the criterion and every lemma at every element of F_(2^n) (n ≤ 21 and n ≤ 13), C++ counts the zero-sum subspaces from the definition for n ≤ 11, a Lean kernel certificate settles n = 5 and n = 7, and Lean 4 with Mathlib proves the algebraic identities of the proof. Code and proofs: release v1.1.0, https://github.com/creelie/eConjectureWelchSumFree/releases/tag/v1.1.0.

Tools used

Anthropic
Claude CodeVersion Opus 5.5 Ultracode

References

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Version history

  1. v1Submitted by Deep BhattacharjeeInitial depositCurrentOct 22, 2026