Attack log — Chowla's cosine problem

2026-08-17 — start

• Folder empty except PROBLEM.md. House rules: write only here; no git; cite what we beat.
• Green 100 #81 (Dec 2025 update of open-problems.pdf, Harmonic analysis, p. 38): for |A|=n, is there θ with ∑_{a∈A} cos(aθ) ≤ −c √n? Construction-sharp upper barrier K(n) ≪ √n. Pre-2025 record Ruzsa: K(n) ≥ exp(c √log n). Update 2025 records Bedert and Jin–Milojević–Tomon–Zhang polynomial bounds; Green quotes Bedert as any exponent < 1/7.
• Current arXiv (fetched tonight):
  • Bedert, arXiv:2509.05260v3 (24 Jul 2026): K(n) ≫ n^{1/5−o(1)}. Streamlined §5 gives K(n) ≫ n^{1/12}. Mid-argument in §7 gives K(n) ≫ n^{1/7} with no o(1), then Lemma 7.4 upgrades the |B_t| lower bound and produces the 1/5−o(1). No numerical c.
  • Jin–Milojević–Tomon–Zhang, arXiv:2509.03490v2: min ∑ cos(ax) ≤ −|A|^{1/10−o(1)}.
• Neither paper, nor Green, nor Open Problem Garden, states an explicit numerical c>0 with K(n) ≥ c n^α for a fixed α. An SDP float is not a dent. The square-root conjecture is not claimed.

2026-08-17 — first false start: extract c from n^{1/5−o(1)}

• The o(1) in Bedert Theorem 1.1 is a real (log n)^{4/5} loss (Lemma 7.4, M ≈ (log n)^2, π(M) ≈ M/log M). An explicit c for n^{1/5} does not fall out. A bound K(n) ≥ c n^{1/5}/(log n)^{4/5} is possible for large n but the n0 where it beats n^{1/7} depends on a large Cstar and is not the clean dent.
• Abandoned as the main claim. Kept n^{1/7} (no logs) as the target.

2026-08-17 — second false start: enumerate small A

• Exact K(n) for small n would give a table, not a universal constant. Affine symmetry does not bound max A. Residue, not a dent. Not run as a claim.

2026-08-17 — third false start: replace 1+sin by a better auxiliary

• Bedert §7 uses two nonnegative auxiliaries: 1+sin(2π t x) and 1−cos(2π t x + π/4). The isolation gap after multiplying is √2/2, coming from a 5-bit window (32 configurations).
• Searched (α, β, φ) for f = 2(1+α sin) hat{1}_A and r = (1−β cos(θ+φ))(f*f) on a 11 × 11 × 17 grid plus a local refinement (optimize_aux.py). Bedert's (1, 1, π/4) already minimises the implied Cstar on that family. No new auxiliary.

2026-08-17 — the click

• Section 7 of Bedert already proves |B_t| ≪ K^4 by testing |φ_t| ≤ ψ_t + 32 K^2 against |Q1|*|Q1|. Combined with Roth+AP |B_t| ≫ N/K^3 this is K ≫ N^{1/7} with no o(1). The implied constant is uncomputed in the paper but every piece is finite:
  1. Lemma 7.2 is 32 exact algebraic numbers.
  2. Lemma 7.1 is Young + an overflow split; the additions give |Q1| ≤ 3(hat{1}_A+K) and ||Q2||_2 ≤ 14 K^2.
  3. The quadratic z^2 ≤ A z + B with A, B ∈ Q(√2) has a rational enclosure.

2026-08-17 — certificate

• verify_lemma72.py: all 32 windows match Bedert's table exactly in Q(√2). Gap √2/2.

• track_constants.py / verify_certificate.py: Cstar ≤ 215816, 2048 · 215816 = 441991168 ≤ 18^7 = 612220032.

• Conversion through the symmetric formulation (hat{1}_{B∪−B} = 2 ∑ cos(2π b x)) yields, for every n ≥ 1,

  K(n) ≥ n^{1/7} / 18.

• Replay: compute/run_all.sh (exit 0).

• What this does not do: beat Bedert's exponent 1/5−o(1); prove the √n conjecture; produce an SDP dual without a sum-of-squares certificate.

Published record we compare against

• Best published exponent: Bedert v3, n^{1/5−o(1)}. We do not beat the exponent.
• Best published explicit numerical c in a bound K(n) ≥ c n^α (α > 0 fixed): none found in Green #81, Bedert v3, JMTZ v2, Ruzsa 2004, Open Problem Garden, or the May 2026 conjugate-function paper arXiv:2605.18087 (which cites 1/7−o(1) / 1/10−o(1) and does not give a numerical c).
• Best published explicit-form weaker bound: Littlewood/L^1 ⇒ K(n) ≫ log n (McGehee–Pigno–Smith / Konyagin). Steen–Cambridge notes record that A = 128 works in the L^1 inequality; that is a logarithmic bound, beaten by n^{1/7}/18 for all large n.

The dent is the constant 1/18 in a fully polynomial bound, independently replayable, not a new exponent.