AI Math

Landau–Siegel zeros and a Stepanov-type nonvanishing lemma

GPT-6 Pro and Claude Opus 5.5 · posted and maintained by Jacob Pescador · v3, 22 October 2026 · unrefereed · PDF · TeX source

Let χ be a primitive real character of conductor q≥3. We prove Barreto’s bound

∑p≤Xχ(p)=−1logpp≤12logX+14logq+loglogX+O(1)∑p≤Xχ(p)=−1logpp≤12logX+14logq+loglogX+O(1)

uniformly in X≥3 and q, with an effective implied constant, and hence that every zero β∈(0,1) of L(s,χ) satisfies (1−β)logq≥1−O(loglogq/logq), so there is no Landau–Siegel zero. As in OpenAI’s proof, the norm of an interpolation determinant is bounded above by the size of its entries and below by its divisibility by high powers of the odd inert primes. Here the points are conjugate pairs (α,α′) of quadratic integers in Barreto’s rectangular box, which gives the 14. In our proof a single congruence, that conjugation is the Frobenius modulo an odd inert prime, gives both the divisibility and the nonvanishing. Modulo a large inert prime p0 every point lies on the graph w=zp0, and as zp0 has derivative zero modulo p0, a linear relation among the rows of the interpolation matrix would give a polynomial in z with more roots, counted with multiplicity, than its degree, as in Stepanov’s method.

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Cite

GPT-6 Pro and Claude Opus 5.5 (posted by J. Pescador), “Landau–Siegel zeros and a Stepanov-type nonvanishing lemma,” preprint, version 3, 22 October 2026, https://jacobpescador.github.io/ai-math/papers/landau-siegel-stepanov/.

BibTeX
@misc{aimath-landau-siegel-stepanov,
  author       = {{GPT-6 Pro} and {Claude Opus 5.5}},
  title        = {{Landau--Siegel zeros and a Stepanov-type nonvanishing lemma}},
  howpublished = {Preprint, posted by J.~Pescador, \url{https://jacobpescador.github.io/ai-math/papers/landau-siegel-stepanov/}},
  note         = {Version 3, 22 October 2026},
  year         = {2026}
}

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