Erdős problem 368 — live-page audit, exact smoothness table, and the uniform wall

Access date: 2026-07-28 UTC.

Claim labels used below:

• (a) elementary-rigorous
• (b) rigorous modulo a named theorem
• (c) plausible/structural-unverified
• (d) computational-only

0. Mandatory live-page gate

I fetched the live problem page, its LaTeX view, and the complete discussion thread through a Bright Data browser. Direct datacenter curl was not used as a substitute.

The rendered live state is:

• Status: OPEN.
• Comments: 4.
• Claimed proofs: 0.
• Likes this problem: ebarschkis.
• Interested in collaborating: None.
• Currently working on this problem: None.
• Looks difficult: None.
• Looks tractable: None.
• Results could be formalisable: None.
• Working on formalising the results: None.

Thus neither mandatory stop condition fires.

Verbatim live statement

The following is copied verbatim from the site's LaTeX view (including its known-results discussion):

How large is the largest prime factor of $n(n+1)$?

Let $F(n)$ be the prime in question. P\'{o}lya \cite{Po18} proved that $F(n)\to \infty$ as $n\to\infty$. Mahler \cite{Ma35} showed that $F(n)\gg \log\log n$. Schinzel \cite{Sc67b} observed that for infinitely many $n$ we have $F(n)\leq n^{O(1/\log\log\log n)}$.

The truth is probably $F(n)\gg (\log n)^2$ for all $n$. Erd\H{o}s \cite{Er76d} conjectured that, for every $\epsilon>0$, there are infinitely many $n$ such that $F(n) <(\log n)^{2+\epsilon}$.

Pasten \cite{Pa24b} has proved that\[F(n) \gg \frac{(\log\log n)^2}{\log\log\log n}.\]The largest prime factors of $n(n+1)$ are listed as A074399 in the OEIS.

The citation header is #368: [Er65b,p.218][Er76d,p.27][ErGr80,p.69], and the tag is number theory.

All four live comments

In chronological order:

1. TongLL, 11:09 on 09 Jan 2026: says Størmer appears to have proved

\(F(n)\to\infty\) in 1897, points to Mahler (1935) and Lehmer (1964), and sketches the Pell/primitive-divisor reduction using \(u=2n+1\).

1. Wouter CvB, 16:05 on 10 Jan 2026: asks whether Pasten's paper, whose

title concerns \(n^2+1\), really implies the displayed result for \(n(n+1)\).

1. BorisAlexeev, 19:17 on 10 Jan 2026: answers by specializing Pasten's

Corollary 1.5 to \(x=1,y=n\). The comment text says “smallest prime factor,” but the primary paper's Corollary 1.5 says largest prime factor \(\mathscr P(xy(x+y))\); I use the primary paper.

1. BorisAlexeev, 06:09 on 17 Feb 2026: notes an Aristotle formalisation,

using Pell's equation, of the qualitative fact \(F(n)\to\infty\). This is not a claim to solve the quantitative open problem.

The page warns that comments are unverified. None is a claimed proof of the open quantitative question and none declares a current worker.

1. Primary-source literature check

The displayed asymptotic state

• Erdős's primary scan,

Problems and results on number theoretic properties of consecutive integers and related questions, exists and on printed page 27 states \(P(n(n+1))>c\log\log n\), followed by the conjectural upper exponent \(2+\epsilon\) infinitely often and lower exponent \(2-\epsilon\) eventually. (a: source/content check) The downloaded scan had SHA-256 2c098f2c3e5079ef09a44f693c7db43a29428bb1c8bf7df7ab57b020306a9ad0.

• Pasten's paper is genuinely

arXiv:2312.03566, later Inventiones mathematicae 236 (2024), 373–385. Its Corollary 1.5 states \[ \mathscr P(xy(x+y)) \geq \kappa\,\frac{(\log_2 y)^2}{\log_3 y} \] for coprime positive \(x<y\). Setting \(x=1,y=n\) gives exactly the live page's bound. (b, Pasten Corollary 1.5)

I did not independently re-audit scans of Pólya [Po18], Mahler [Ma35], or Schinzel [Sc67b]; their precise claims above are recorded as live-page ground truth, not reconstructed from memory.

Exact-title, formula, Google Scholar-style, and arXiv searches for P(n(n+1)), P(xy(x+y)), and “largest prime factor of consecutive integers” found no post-Pasten primary paper with a stronger uniform lower bound for this problem. A very recent hit, Yang, arXiv:2607.16032, concerns positive density and the relative order of \(P(n)\) and \(P(n+1)\), not a uniform lower bound on their maximum. This is an honest search result, not proof that no uncatalogued improvement exists.

A directly relevant exact theorem missing from the live page

Luca and Najman, On the largest prime factor of \(x^2-1\), Math. Comp. 80 (2011), 429–435, DOI 10.1090/S0025-5718-2010-02381-6, prove an unconditional complete enumeration of the nontrivial \(x\) for which \(P(x^2-1)<100\). Their Section 3 explains the completeness check that removes the apparent GRH dependence of the regulator computation.

There is a published 2014 erratum. The arXiv v1 HTML still displays a stale total of 16,167. The author-hosted corrected paper states 16,223; its SHA-256 in this run was a90b2eb77018d9392609160c40b69869819666be44e0c083c793eef395dc6ee6. The current author-hosted tables linked from the publication page contain 13,374 odd and 2,849 even values, again totalling 16,223. I use only this corrected version and hash-pin both tables below.

Corollary 5 of the corrected paper explicitly says that if

then \((t-1)/2,(t+1)/2\) is the largest pair of consecutive integers whose prime factors are all below 100. Thus this paper is not merely adjacent literature; it gives a complete finite regime of Problem 368.

2. Exact reduction from the corrected enumeration

Write \(P(m)\) for the largest prime factor of \(m\), with \(P(1)=1\), and

For \(n\geq1\), put \(x=2n+1\). Then

Multiplication by 4 introduces no new prime beyond 2, which already divides \(n(n+1)\). Therefore

Conversely, every odd \(x\geq3\) is uniquely \(2n+1\). This is an elementary bijection between indices with \(F(n)<100\) and odd nontrivial solutions of \(P(x^2-1)<100\). (a)

For each prime \(p\leq97\), define

Because Luca–Najman's list is complete through 97, filtering its odd part by largest prime factor and applying (1) gives every \(M(p)\) and \(C(p)\). Maximality and completeness of the following table are (b), modulo the corrected Luca–Najman theorem; the table extraction and all 16,223 smooth factorisations were independently recomputed (d).

The row \(p=41\) recovers the 869 odd solutions reported historically by Lehmer and quoted in Luca–Najman's introduction. This is a useful external consistency check, not an input to the computation.

Sharp endpoint certificate

Let

Direct trial division gives

Hence \(F(N)=79<100\). (a) An independent sympy.factorint call agreed with both complete factorizations.

Luca–Najman completeness plus (1) proves

Thus \(N\) is exactly the last index with \(F(n)<100\). (b, Luca–Najman) This is a sharp, global finite-regime theorem, not evidence that the unresolved asymptotic conjecture is proved.

3. Exact formulation of what remains

A positive integer is \(Y\)-smooth if all its prime factors are at most \(Y\). The desired uniform lower bound

is equivalent, after absorbing finitely many cases into the constant, to the following uniform smooth-neighbour bound:

There is an absolute \(C\) such that every consecutive \(Y\)-smooth pair \(n,n+1\) satisfies \[ > \log n\leq C\sqrt Y. \tag{3} > \]

Indeed, the conjectural lower bound and \(F(n)\leq Y\) imply (3). Conversely, apply (3) with \(Y=F(n)\); both \(n\) and \(n+1\) are \(F(n)\)-smooth, so

This equivalence is (a) and isolates a precise missing lemma rather than the vague instruction “improve the smooth-number estimates.”

The classical Pell reduction explains why fixed \(Y\) is computable but does not supply (3). If \(x=2n+1\) and \(D\) is the squarefree part of \(x^2-1\), then

where \(D\) is one of the \(2^{\pi(Y)}-1\) nonempty squarefree products of primes at most \(Y\). For fixed \(D\), Pell solutions are powers of a fundamental unit. The primitive-divisor theorem bounds the relevant power index by \(O(Y)\) when the second coordinate is \(Y\)-smooth. (b, Carmichael or Bilu–Hanrot–Voutier) It does not give the required uniform control of the fundamental unit/regulator as \(D\) ranges over all those products. The exact missing input within this Pell approach is a height/regulator bound strong enough to force (3) using the smoothness of the Pell coordinate. (c: structural diagnosis; (3) itself is the rigorous missing target)

This also diagnoses the computational wall. Luca–Najman process \(2^{25}-1=33,554,431\) Pell equations. Their paper reports about 12 CPU-days for regulators and 3 more for the remaining work on a Xeon E5430. Raising the bound only to include 101 already doubles the outer family to \(2^{26}-1=67,108,863\) Pell equations, with potentially larger regulators. A specialized reproduction is therefore at least tens to hundreds of modern core-hours, rather than a few-minute check; at roughly \$0.05–\$0.20/core-hour, a realistic compute-only budget is order \$5–\$40, before implementation and validation. I did not launch it on this box. (c: engineering estimate based on the paper's measured run) More importantly, any fixed extension remains finite and cannot prove the uniform estimate (3).

4. Reproduction and independent audit

The complete standard-library checker is runs/erdos368_wave8x_reverify.py, SHA-256 804a03fccb529cf93e2177cc393e0769ca886c96f650128ea02c440b4e1d328c.

Run from the repository root:

python runs/erdos368_wave8x_reverify.py

It downloads and hash-pins the two current primary-source tables:

oddsort.txt  sha256=035f2e7342d11bb86d4b7dacad611e7adb5601f8e5352a1e525e0646ad0f4380
evensort.txt sha256=05dd3086ca24abf63f4ae1fa2f20957db6a32a5a5e3d341e6c495aeba637d5fc

It then checks strict ordering, parity, uniqueness, the corrected total 16,223, and fully trial-divides every \(x^2-1\) using a from-scratch list of the primes below 100. For every odd \(x\), it separately factors \((x-1)/2\) and \((x+1)/2\), checks identity (1), recomputes the 25 table rows, and asserts equality with the report. The core is:

for x in values:
    remainder = x*x - 1
    factors = {}
    for p in primes_below(100):
        while remainder % p == 0:
            factors[p] = factors.get(p, 0) + 1
            remainder //= p
    assert remainder == 1

for p in primes_below(100):
    eligible = [x for x in odd_values if largest_factor[x] <= p]
    x = eligible[0]                 # source list is strictly decreasing
    n = (x - 1) // 2
    row = (p, n, largest_factor[x], len(eligible))

Two timed runs completed in about 2.7–2.9 seconds with about 25 MiB maximum RSS and ended:

PASS: hashes, 16,223 entries, all factorizations, and table agree.

The checker validates every listed entry and every arithmetic consequence, but it deliberately does not pretend to rerun Luca–Najman's 15-CPU-day completeness computation. Accordingly, (2) and the word “exact” are always labelled modulo their named theorem.

PARTIAL: Modulo Luca–Najman’s corrected completeness theorem, the exact maxima and counts for every prime smoothness bound through 97 are verified, with last \(F(n)<100\) at \(n=9,591,468,737,351,909,375\); the open asymptotic bound is exactly the unresolved uniform estimate \(\log n=O(\sqrt Y)\) for consecutive \(Y\)-smooth integers.