slug: zhang-character-sum-difference-question-d-refutation
bibkey: zhang2025problems
doi: 10.48550/arXiv.2506.17235
url: https://arxiv.org/abs/2506.17235v1
triage: theorem
motivation_gids:
  - D5/S3/ArithSums/ZhangCharacterSumDifferenceRefutation.result

Refutation of Zhang’s character-sum constant restriction

Problem

Wenpeng Zhang, Some interesting number theory problems, arXiv:2506.17235v1, printed page 3, states problem (B) and question (D) verbatim as:

(B). Whether there are infinitely many pairs of fundamentally different integer coefficients polynomials f(x) and g(x) (That is, (f(x)/p) ≠ (g(x)/p)) such that Σ_{x=1}^{p−1} (f(x)/p) − Σ_{x=1}^{p−1} (g(x)/p) = c, (2) where c is a fixed constant.

(D). Whether the values of c can only be 0 or 2?

Identity (2) compares the Legendre character sums of two fundamentally different integer-coefficient polynomials. Following the preceding corollary, the formal statement uses one constant for every odd prime.

Motivation

Issue #8626 records the source quotation, full quantifiers, literature-status scope, and the proposed quadratic counterexample. The result tests the literal restriction on c, including the paper’s stated meaning of fundamentally different polynomials.

Gap

The bounded source and literature searches found no earlier proof or refutation of Question (D) in the searched scope. Nica’s arXiv:2507.09991 addresses identity (1), not the constant restriction in Question (D). This dossier makes no claim of exhaustive literature coverage or publication priority.

Route

Set f(X) = X^2 and g(X) = (X+1)^2. For every odd prime p, all terms in the first character sum equal 1, so its value is p-1. In the second sum, the term at x=p-1 is 0 and all earlier terms equal 1, so its value is p-2. The two symbol-valued functions differ at x=p-1, and their character-sum difference is the fixed constant c=1.

Falsifier

A proof that either displayed sum has a different value for some odd prime, or that the two Legendre-symbol functions agree throughout the summation domain for some odd prime, would invalidate the counterexample.

Evidence

• Frozen result: D5/S3/ArithSums/ZhangCharacterSumDifferenceRefutation.result.
• The result has type Not claim and is kernel-checked in the pinned Lean toolchain.
• Its axiom closure is propext, Classical.choice, and Quot.sound.

Triage

theorem; resolution refuted for Question (D)’s literal restriction on the constant. The proof shape is bind-only, with no escape witness; admission is by the preregistered external-open-problem resolution in issue #8626.

ASSUMED-UNVERIFIED

The bounded literature search is not exhaustive and does not establish worldwide novelty or priority. Unpublished resolutions remain unverified.