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Problem 969

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Statement. Let Q(x)Q(x) count the number of squarefree integers in [1,x][1,x]. Determine the order of magnitude in the error term in the asymptotic

Q(x)=6π2x+E(x).Q(x)=\frac{6}{\pi^2}x+E(x).

Status. Open. The site's label is OPEN (the page was last edited on 2025-10-19, and its proof-claims tab was empty on 2026-10-06), and no claim page is recorded.

Source. erdosproblems.com/969, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #969, https://www.erdosproblems.com/969.

References.

  • [EvLi31] Evelyn, C. J. A. and Linfoot, E. H., On a problem in the additive theory of numbers. Ann. of Math. (2) (1931), 261-270.
  • [Li16] Liu, H.-Q., On the distribution of squarefree numbers. J. Number Theory (2016), 202-222.
  • [Wa63] Walfisz, Arnold, Weylsche Exponentialsummen in der neueren Zahlentheorie. (1963), 231.

Formalization. None recorded.

Current assessment

Site formulation(last edit): determine the order of magnitude of E(x)=Q(x)−6π2xE(x)=Q(x)-\frac{6}{\pi^2}x. The question is open on the site and no claim about it is on record. The site's summary of the literature leaves a gap between an omega result and the upper bounds. Evelyn and Linfoot [EvLi31] proved E(x)=Ω(x1/4)E(x)=\Omega(x^{1/4}): ∣E(x)∣≥c x1/4|E(x)|\ge c\,x^{1/4} for some c>0c>0 and arbitrarily large xx (the site writes this as E(x)≫x1/4E(x)\gg x^{1/4}; E(x)E(x) changes sign, so it is not a lower bound at every xx), and x1/4x^{1/4} is the expected true order. From above, no unconditional bound saves a power over the elementary E(x)≪x1/2E(x)\ll x^{1/2}, the prime number theorem giving o(x1/2)o(x^{1/2}) and Walfisz [Wa63] only x1/2x^{1/2} divided by a subpolynomial factor. The gap persists under the Riemann Hypothesis: that hypothesis gives E(x)≪x11/35+o(1)E(x)\ll x^{11/35+o(1)} (Liu [Li16]), the best conditional bound, and the site records that the true order of magnitude is unknown even assuming it; in the other direction, E(x)≪x1/4E(x)\ll x^{1/4} would imply the Riemann Hypothesis. The literature is not compiled on this page, and no status search beyond the site record is recorded.

The OpenAI Math Release holds two manuscripts titled The Quasi-Riemann Hypothesis. The first (OpenAI, 2026-09-30, subtitled A Zero-Free Half-Plane Re⁡(s)>7/8\operatorname{Re}(s)>7/8; intake card openai_2026_quasi_riemann_hypothesis_zero_free_half_plane_7_8) asserts that the Riemann zeta function, every Dirichlet LL-function and every finite-order Hecke LL-function over Q(−3)\mathbb{Q}(\sqrt{-3}) has no zero in the half-plane Re⁡s>7/8\operatorname{Re}s>7/8; the second (OpenAI, 2026-10-05; intake card openai_2026_quasi_riemann_hypothesis_zero_free_half_plane_11_12) asserts, by a different argument, the smaller half-plane Re⁡s>11/12\operatorname{Re}s>11/12 for the same families, a weaker zero-free region that the manuscript presents as an intermediate step toward 7/87/8. A third manuscript, Uniform exclusion of Landau–Siegel zeros (OpenAI, 2026-10-01; intake card openai_2026_uniform_exclusion_landau_siegel_zeros), asserts a gap 1−β≥c/log⁡q1-\beta\ge c/\log q, with an absolute c>0c>0, for every real zero β\beta of every primitive nonprincipal real Dirichlet LL-function of conductor q≥3q\ge3; it concerns real zeros of Dirichlet LL-functions and not ζ\zeta, so it bears on nothing here. The release's Lean catalog lists comparator statements for the 7/87/8 half-plane (for ζ\zeta, for every Dirichlet LL-function and for the Hecke family) and for the Siegel-zero gap, and none for the 11/1211/12 manuscript (manuscripts at https://github.com/openai/math/blob/adc7f1241/preprints/The-Quasi-Riemann-Hypothesis-September-30-2026/paper.pdf, https://github.com/openai/math/blob/adc7f1241/preprints/The-Quasi-Riemann-Hypothesis-October-5-2026/paper2.pdf and https://github.com/openai/math/blob/adc7f1241/preprints/Uniform-exclusion-of-Landau-Siegel-zeros-October-1-2026/paper.pdf, Lean under https://github.com/openai/math/tree/adc7f1241/lean). The release names no Erdős problem, and none of the three manuscripts mentions Erdős. A zero-free half-plane Re⁡s>θ\operatorname{Re}s>\theta for ζ\zeta gives, by the classical Perron argument, M(x)=∑n≤xμ(n)≪xθ+εM(x)=\sum_{n\le x}\mu(n)\ll x^{\theta+\varepsilon}, and since Q(x)=∑dμ(d)⌊x/d2⌋Q(x)=\sum_d\mu(d)\lfloor x/d^2\rfloor, the standard split of this sum gives E(x)≪x1/(3−θ)+εE(x)\ll x^{1/(3-\theta)+\varepsilon}: that is x8/17+εx^{8/17+\varepsilon} for θ=7/8\theta=7/8 and x12/25+εx^{12/25+\varepsilon} for θ=11/12\theta=11/12, a power saving the unconditional literature above does not have. That deduction is this repository's own, not the release's, and the release's theorems are recorded on the cards at claims level only. No claim page is written for the release: the release asserts nothing about Problem 969, and even a verified power saving E(x)≪x1/2−δE(x)\ll x^{1/2-\delta} would not determine the order of magnitude, so it would settle no instance of the question. The connection is recorded as context.

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