Erdoes #969 - working report - Erdős Problem a Day

Erdős problem #969 — wave w026

Date: 2026-07-29 UTC

Result in one paragraph

The problem remains open. I did not find a construction or a new asymptotic bound. I did obtain three verifiable outputs:

[d, computational-only] An independent exact calculation corrects an

off-by-one in the prose surrounding the negative extremizer in Mossinghoff--Oliveira e Silva--Trudgian (2021): at $m=154953313738409$, \[ Q(m-1)=94200318939698,\qquad Q(m)=94200318939699. \] The paper's reported normalized value $-1.1254291388$ uses the first (pre-jump) count, although its prose assigns the second count to $x=m-\varepsilon$. The theorem and normalized extremum are unaffected.

[d] A dependency-free exhaustive verifier gives sharp signed and

$n^{1/4}$-normalized extrema for every integer $1\leq n\leq10^7$.

[a/b] An exact decomposition shows that, under RH, attaining the

conjectural exponent is equivalent to one concrete missing estimate: square-root cancellation in a Möbius-weighted reciprocal-square sawtooth sum. This identifies what Liu's $11/35$ result still does not supply.

Here and below:

[a] means elementary-rigorous;
[b] means rigorous modulo the explicitly named theorem;
[c] means plausible/structural-unverified;
[d] means computational-only, with an exact reproducible finite check.

0. Mandatory live-page gate

The live page was loaded on 2026-07-29 through the Bright Data browser path, not datacenter curl: <https://www.erdosproblems.com/969>. A full-page screenshot and DOM text were inspected.

Live state:

status: OPEN;
page last edited: 19 October 2025;
comments: 0;
claimed proofs: 0;
“Interested in collaborating”: None;
“Currently working on this problem”: None;
all other interest/difficulty/formalisation markers: None.

Thus neither stop condition (claimed proof/solution/falsification, or current worker) was present.

Verbatim current statement

Let $Q(x)$ count the number of squarefree integers in $[1,x]$. Determine the order of magnitude in the error term in the asymptotic \[ > Q(x)=\frac{6}{\pi^2}x+E(x). > \]

The page then records the following known results. [b] The elementary bound is $E(x)\ll x^{1/2}$, and the prime number theorem gives $E(x)=o(x^{1/2})$. Walfisz gives an unconditional bound of shape $x^{1/2-o(1)}$. Evelyn and Linfoot give an $x^{1/4}$-scale lower bound. A bound $E(x)\ll x^{1/4}$ would imply RH. Even assuming RH the order is unknown; the page's best conditional upper bound is

E(x)\ll x^{11/35+o(1)}

by Liu.

The page's LaTeX source, including bibliography, is <https://www.erdosproblems.com/latex/969>.

1. Source and literature audit

Original Erdős sources

The two originals linked by the live page were located and inspected.

P. Erdős, Some Recent Advances and Current Problems in Number Theory,

in Lectures on Modern Mathematics, vol. III, Wiley, 1965, pp. 196--244; the squarefree-error question is on p. 206. A scan is available at <https://combinatorica.hu/~p_erdos/1965-17.pdf>.

P. Erdős, *Some Problems and Results on Additive and Multiplicative Number

Theory, in Analytic Number Theory* (Temple University, 1980), Lecture Notes in Mathematics 899, Springer, 1981, pp. 171--182; the update is on p. 176. A scan is available at <https://users.renyi.hu/~p_erdos/1981-33.pdf>.

The 1981 source explicitly says that the error cannot be $o(x^{1/4})$ and mentions the then-new RH-conditional $x^{1/3+o(1)}$ result. [b]

Page-cited benchmarks

C. J. A. Evelyn and E. H. Linfoot, *On a Problem in the Additive Theory of

Numbers IV*, Ann. of Math. (2) 32 (1931), 261--270, <https://doi.org/10.2307/1968190>. Later primary literature states their result as $E(x)=\Omega(x^{1/4})$. [b]

A. Walfisz, Weylsche Exponentialsummen in der neueren Zahlentheorie,

Mathematische Forschungsberichte XV, 1963. The bound is commonly written \[ E(x)=O\!\left(x^{1/2} \exp\!\left\{-c(\log x)^{3/5}(\log\log x)^{-1/5}\right\}\right). \] This exact attribution and form are also recorded in the 2021 primary paper below. [b]

H.-Q. Liu, On the distribution of squarefree numbers, J. Number Theory

159 (2016), 202--222, <https://doi.org/10.1016/j.jnt.2015.07.013>. The publisher metadata and abstract exist and say that the 1993 RH-conditional result is improved. The full publisher text was paywalled in this environment; the exact $11/35$ exponent was cross-checked in the live page and in the Mossinghoff--Oliveira e Silva--Trudgian paper, not inferred from the abstract.

Important later primary result absent from the live-page notes

M. J. Mossinghoff, T. Oliveira e Silva, and T. S. Trudgian, The distribution of $k$-free numbers, Math. Comp. 90 (2021), 907--929, <https://doi.org/10.1090/mcom/3581>, author manuscript <https://arxiv.org/abs/1912.04972>.

Their Theorem 1 proves, in particular,

\liminf_{x\to\infty}\frac{E(x)}{x^{1/4}}<-3,\qquad \limsup_{x\to\infty}\frac{E(x)}{x^{1/4}}>3. \tag{1}

[b] Their Theorem 2 and exhaustive computation prove for real

$0<x\leq10^{18}$

-1.12543x^{1/4}<E(x)<1.11653x^{1/4}. \tag{2}

The computation reportedly took about six core-years and was independently double-checked. This means the $10^7$ computation below is a transparent local audit/table, not a new range record.

Searches by exact title, DOI, exponent $11/35$, arXiv, and the OpenAlex citation graph for Liu located no later paper claiming a better pointwise exponent for this same $Q(x)$. This is only a documented search miss, not a proof of absence. The live page, last edited in October 2025, also still calls Liu's exponent best.

2. A one-unit correction at the published negative extremizer

Let

\[ m=154953313738409. \]

The proof of Theorem 2 in the 2021 paper describes the negative extremum at $x=m-\varepsilon$, prints $Q_2(x)=94200318939699$, and prints normalized value approximately $-1.1254291388$. Table 2 similarly places the integer count $Q_2(m)$ beside a column headed by the left-limit value $R_2(m-\varepsilon)/m^{1/4}$.

The standalone verifier computes from scratch

Q(N)=\sum_{d\leq\sqrt N}\mu(d)\left\lfloor\frac{N}{d^2}\right\rfloor. \tag{3}

It sieves all $\mu(d)$ through $\lfloor\sqrt m\rfloor=12448024$, then obtains

\boxed{Q(m-1)=94200318939698},\qquad \boxed{Q(m)=94200318939699}. \tag{4}

[d] These are exact integer computations.

Independently,

m=120199\cdot1289139791,

and trial division certifies both factors prime. Hence $m$ is squarefree and the unit jump in (4) is also elementary-certified. [a]

With

\frac{607927101854026628663276779258}{10^{30}} <\frac6{\pi^2}< \frac{607927101854026628663276779259}{10^{30}}, \tag{5}

the verifier proves by integer fourth-power comparisons that the pre-jump count satisfies

1.1254291387699 < \frac{(6/\pi^2)m-Q(m-1)}{m^{1/4}} < 1.1254291387700, \tag{6}

whereas inserting the paper's displayed post-jump count gives

1.1251457061815 < \frac{(6/\pi^2)m-Q(m)}{m^{1/4}} < 1.1251457061816. \tag{7}

[d]

Thus the printed normalized value is correct and uses $Q(m-1)$; the one-unit issue is confined to the prose assigning $Q(m)$ to the left-hand point $m-\varepsilon$. In the table, the $Q(m)$ column can be understood as the count after the squarefree jump, while the normalized column explicitly uses the pre-jump value.

Sharp constants through $10^{18}$

Combining the exact arithmetic above with the paper's exhaustive global extremizer assertion gives the following sharpened formulation of its rounded Theorem 2. [b, modulo that theorem]

Let $c=6/\pi^2$. Then

\sup_{0<x\leq10^{18}}\frac{E(x)}{x^{1/4}} =\frac{29-43c}{43^{1/4}} \in(1.1165225284069,1.1165225284070), \tag{8}

attained at $x=43$, while

\inf_{0<x\leq10^{18}}\frac{E(x)}{x^{1/4}} =-\frac{cm-94200318939698}{m^{1/4}} \in(-1.1254291387700,-1.1254291387699). \tag{9}

The latter is a left-limit infimum at $m$, rather than a value attained at the integer $m$.

3. Exact finite census for integer arguments

The following locations are exhaustive and sharp for integer $n$ in each listed block. [d] Displayed real values are rounded to 12 decimals; the comparisons deciding their locations use the rational enclosure (5), never floating point.

Write $F(n)=E(n)/n^{1/4}$.

| integer block | squarefree count in block | $\max E(n)$: $(n,Q(n),E)$ | $\min E(n)$: $(n,Q(n),E)$ | $\max F(n)$: $(n,F)$ | $\min F(n)$: $(n,F)$ | |---|---:|---:|---:|---:|---:| | $1\ldots9$ | 6 | $(7,6,1.744510287022)$ | $(1,1,0.392072898146)$ | $(7,1.072504257163)$ | $(9,0.305219732010)$ | | $10\ldots99$ | 55 | $(43,29,2.859134620277)$ | $(56,34,-0.043917703825)$ | $(43,1.116522528407)$ | $(56,-0.016054346009)$ | | $100\ldots999$ | 547 | $(719,441,3.900413766955)$ | $(380,229,-2.012298704530)$ | $(115,0.943097803683)$ | $(380,-0.455770853299)$ | | $1000\ldots9999$ | 5475 | $(9242,5626,7.537724665086)$ | $(5589,3394,-3.704572262155)$ | $(1663,0.942266003199)$ | $(1864,-0.483376054349)$ | | $10000\ldots99999$ | 54711 | $(47523,28905,14.480338591093)$ | $(80156,48715,-14.004776211358)$ | $(47523,0.980737439713)$ | $(80156,-0.832323514388)$ | | $100000\ldots999999$ | 547132 | $(351115,213474,21.675632523440)$ | $(436484,265330,-20.453125652959)$ | $(351115,0.890449785791)$ | $(436484,-0.795733430828)$ | | $1000000\ldots10000000$ | 5471365 | $(4026914,2448109,38.842564594213)$ | $(8771780,5332572,-30.793501113701)$ | $(2015403,0.899594897905)$ | $(1146476,-0.667194998374)$ |

Also,

\[ Q(10^7)=6079291. \]

Certification method:

Sieve a byte per integer, setting multiples of every prime square to zero.
Derive (5) from Machin's identity

\[ \pi=16\arctan(1/5)-4\arctan(1/239) \] with exact alternating-series remainder intervals. [a]

For raw errors, compare the resulting integer lower/upper numerators.
For normalized positive candidates, for example, certify

\[ \frac{E(n)}{n^{1/4}}< \frac{E(n_0)}{n_0^{1/4}} \] by the sufficient exact comparison \[ E_{\rm hi}(n)^4n_0<E_{\rm lo}(n_0)^4n. \] The negative case is analogous. Thus no computed fourth root proves an extremum.

Recompute $Q(n)$ at every reported location using (3), from a separately

built Möbius sieve.

The squarefree-indicator byte array through $10^7$ has SHA-256 14362c2c16ecee2d0db1ae18153ee236ff8c5c86b0b29910954ce008d21b28d4.

4. Exact reduction to the missing lemma

Let $\mu$ be the Möbius function,

y=\lfloor\sqrt x\rfloor,\qquad M(y)=\sum_{d\leq y}\mu(d),\qquad \psi(t)=\{t\}-\frac12,

and

S(x)=\sum_{d\leq y}\mu(d)\psi\!\left(\frac{x}{d^2}\right),\qquad T(y)=\sum_{d>y}\frac{\mu(d)}{d^2}.

The squarefree indicator identity

\mu^2(n)=\sum_{d^2\mid n}\mu(d)

gives, for every real $x\geq1$,

\begin{aligned} Q(x) &=\sum_{d\leq y}\mu(d)\left\lfloor\frac{x}{d^2}\right\rfloor\\ &=x\sum_{d\leq y}\frac{\mu(d)}{d^2} -\sum_{d\leq y}\mu(d)\left\{\frac{x}{d^2}\right\}. \end{aligned}

Since $\sum_{d\geq1}\mu(d)/d^2=1/\zeta(2)=6/\pi^2$, this is exactly

\boxed{E(x)=-S(x)-\frac12M(y)-xT(y).} \tag{10}

[a] The verifier checks the rational coefficients in (10) exactly at every tabulated extremizer.

Partial summation gives

T(y)=-\frac{M(y)}{y^2} +2\int_y^\infty\frac{M(t)}{t^3}\,dt. \tag{11}

[a]

Under RH, the standard Mertens estimate is

M(t)=O_\varepsilon(t^{1/2+\varepsilon}). \tag{12}

[b, standard RH theorem] Equations (11)--(12) imply

M(y)=O_\varepsilon(x^{1/4+\varepsilon}),\qquad xT(y)=O_\varepsilon(x^{1/4+\varepsilon}).

Consequently, under RH,

\boxed{ E(x)=O_\varepsilon(x^{1/4+\varepsilon}) \quad\Longleftrightarrow\quad S(x)=O_\varepsilon(x^{1/4+\varepsilon}). } \tag{13}

[b]

This is the clean remaining lemma:

Uniformly for real $x\geq2$, prove square-root cancellation \[ > \sum_{d\leq\sqrt x}\mu(d) > \psi\!\left(\frac{x}{d^2}\right) > \ll_\varepsilon x^{1/4+\varepsilon}. > \tag{14} > \]

The sum has length about $x^{1/2}$, so (14) asks for essentially square-root cancellation. [c, interpretation] RH controls the untwisted partial sums $M(t)$, but not the nonlinear reciprocal-square phases arising from a Fourier expansion of $\psi(x/d^2)$. One needs estimates uniform both in dyadic $d$-ranges and in the Fourier harmonic, while also controlling near-discontinuities where $x/d^2$ is close to an integer. This is the exact analytic wall.

Since the other two terms in (10) are already at the conjectural scale under RH, Liu's theorem is equivalently an

S(x)\ll_\varepsilon x^{11/35+\varepsilon}

bound in this formulation. The remaining exponent gap is

\frac{11}{35}-\frac14=\frac9{140}.

No manipulation of the untwisted Mertens bound alone removes that gap.

The Dirichlet-series identity

\sum_{n\geq1}\frac{\mu^2(n)}{n^s} =\frac{\zeta(s)}{\zeta(2s)}

also explains the $1/4$ barrier through the poles at halves of zeta zeros, but it does not by itself give the required pointwise uniform bound. [a/b]

5. Reproduction

Complete standalone verifier:

runs/erdos969_wavew026_reverify.py

SHA-256 of the verifier:

6e1d98f3a9fe6da8f8ce2eec7c931a92cd70052de3e90f86288fdc87aa906a64

Run from the repository root:

PYTHONUNBUFFERED=1 python runs/erdos969_wavew026_reverify.py

It uses only the Python standard library. The completed run took 37.118 seconds on this VM and ended with:

2021 extremizer audit: Q(154953313738408)=94200318939698, Q(154953313738409)=94200318939699; left normalized value in (-1.1254291387700,-1.1254291387699)
  The published 1.1254291388 magnitude uses the pre-jump count; using its displayed Q-at-jump gives magnitude in (1.1251457061815,1.1251457061816).
independent Möbius-floor checks completed in 37.118s
PASS: all claims certified from scratch in 37.118s

The key exact code paths are:

def q_via_mobius(n, mu):
    return sum(mu[d] * (n // (d*d))
               for d in range(1, isqrt(n) + 1))

# Positive normalized comparison; SCALE cancels after fourth powers.
assert ehi**4 * candidate_n < candidate_elo**4 * n

The complete file additionally derives the $\pi$ enclosure, constructs two independent sieves, certifies every candidate against every integer in its block, verifies the factorization of the published jump, and checks (10).

6. Honest wall and compute cost

The computation does not suggest a counterexample or close the asymptotic problem. In fact, the 2021 theorem (1) guarantees normalized excursions beyond $\pm3$ eventually, despite their staying below about $1.13$ through $10^{18}$. Finite searches therefore cannot determine the order without a uniform analytic step.

Repeating the 2021 exhaustive range is outside the allowed budget: its reported six core-years are about $52{,}600$ core-hours. At a representative $\$0.04$--$\$0.08$ per core-hour that is roughly $\$2{,}100$--$\$4{,}200$, before engineering and redundant verification. A naïve linear extension by one decade to $10^{19}$ would be about $526{,}000$ core-hours ($\$21{,}000$--$\$42{,}000$) on comparable throughput. [c, cost extrapolation] The needed progress is analytic estimate (14), not another small brute-force range.

PARTIAL: verified and corrected the pre-jump count at the published $10^{18}$-range negative extremizer, certified exact decade extrema through $10^7$, and reduced the RH-conditional $1/4+\varepsilon$ target to the single uniform sawtooth-sum bound (14); the asymptotic order remains open.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger