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Claim. For every ϵ\epsilon with 0<ϵ≤10<\epsilon\le1 there is a constant c(ϵ)c(\epsilon) such that the eventual-time thresholds of the Legendre-symbol partial sums have mean value c(ϵ)c(\epsilon) over the primes: π(x)−1∑p≤xg(ϵ,p)→c(ϵ)\pi(x)^{-1}\sum_{p\le x}g(\epsilon,p)\to c(\epsilon) as x→∞x\to\infty, where g(ϵ,p)g(\epsilon,p) is the least tt such that ∣∑n≤m(n/p)∣<ϵm\bigl|\sum_{n\le m}(n/p)\bigr|<\epsilon m for every m≥tm\ge t. The proof ends with the quantitative form ∑p≤xg(ϵ,p)=c(ϵ)π(x)(1+O((log⁡log⁡x)−1/8))\sum_{p\le x}g(\epsilon,p)=c(\epsilon)\pi(x)(1+O((\log\log x)^{-1/8})) and identifies c(ϵ)=∑μ≥1μdμc(\epsilon)=\sum_{\mu\ge1}\mu d_\mu, where dμd_\mu is the limiting frequency of the primes with g(ϵ,p)=μg(\epsilon,p)=\mu. Since π(x)∼x/log⁡x\pi(x)\sim x/\log x, this is the asymptotic ∑p<xfϵ(p)∼cϵx/log⁡x\sum_{p<x}f_\epsilon(p)\sim c_\epsilon x/\log x asked by Problem 981, Erdős's display (80) of 1965, with the two-sided threshold gg in place of the one-sided threshold ff of Erdős and of the problem's statement, and with c(ϵ)≥1c(\epsilon)\ge1 because every g(ϵ,p)≥1g(\epsilon,p)\ge1. The theorem is quoted from the printed page on the result page theorem_p165; the digest is on the source card elliott_1969_conjecture_erdos_concerning_character_sums.

The one-sided threshold. The paper opens with Erdős's one-sided threshold f(ϵ,p)f(\epsilon,p) and the conjecture as its display (1), notes f(1,p)=n2(p)f(1,p)=n_2(p), the least quadratic nonresidue, and then proves the theorem for the two-sided g(ϵ,p)g(\epsilon,p), saying of the original definition only that simple changes in the argument give a similar result (p. 164); no adaptation is printed. Termwise f(ϵ,p)≤g(ϵ,p)f(\epsilon,p)\le g(\epsilon,p), which does not by itself transfer the asymptotic. The problem's wording, the one-sided form, therefore rests on the printed theorem together with the author's remark, as the problem page's Status records. For ϵ≥1\epsilon\ge1 the one-sided form is settled without this paper: the problem page's Formulation shows that for ϵ>1\epsilon>1 both thresholds equal 11 for every odd prime, and records the identity f(1,p)=n2(p)f(1,p)=n_2(p) that makes the instance ϵ=1\epsilon=1 Erdős's theorem (78) under its Origin.

Argument, in outline. The primes p≤xp\le x whose two-sided sums reach ϵm\epsilon m for some m≥Nm\ge N are few by a fourth-moment large-sieve bound (Lemma 7); for the rest, g(ϵ,p)=μ≤Ng(\epsilon,p)=\mu\le N is decided by the symbols (n/p)(n/p) for n≤Nn\le N, so by the residue of pp modulo N!N!, and with N=[log⁡log⁡x]N=[\sqrt{\log\log x}] the Siegel--Walfisz theorem gives each μ\mu its limiting frequency dμd_\mu. The proof was not checked here beyond this outline.

Depends on. Nothing in this wiki.

Acceptance. Refereed: P. D. T. A. Elliott, A conjecture of Erdős concerning character sums, Indagationes Mathematicae (Proceedings) 72 (1969), no. 2, 164--171 (Nederl. Akad. Wetensch. Proc. Ser. A 72 = Indag. Math. 31), communicated by N. G. de Bruijn at the meeting of 25 January 1969, the date this page is named by. Reviewed: the site's curator, Thomas F. Bloom, marks the problem PROVED and credits the proof to this paper in the problem's commentary (page last edited 27 December 2025); the curator neither wrote nor submitted the result. Independently of the site, Tang and Zhang, who found the paper while writing on the first-passage variant, restate Erdős's (80) as their Conjecture 1.1 and write that Elliott proved it (arXiv:2512.24631v2, p. 2; card tang_2025_average_first_passage_times_character_sums), and the site's thread of 27 December 2025 records the same reading. Not formalized: no Lean statement of the problem exists, and nothing was built or audited here. The one-sided qualification above is disclosed and is not a dispute; no dispute of the theorem was found in the searches dated on the problem page.