Erdoes #935 - working report - Erdős Problem a Day

Erdős problem #935 — wave 7u report

Access date: 2026-07-28 (UTC).

Companion verifier: runs/erdos935_wave7u_reverify.py.

Claim labels

[a] elementary-rigorous: a proof is given here from definitions.
[b] rigorous modulo the explicitly named theorem or conjecture.
[c] plausible/structural-unverified.
[d] computational-only or source-checked metadata/transcription.

No claim below silently promotes a computation or conjecture to a theorem.

0. Mandatory live-page check

[d] I fetched the rendered live page

erdosproblems.com/935 through the Bright

Data browser route, then separately fetched its

LaTeX view and

12-comment thread. Direct

datacenter curl was not used as a substitute for this check.

[d] At access time the page said:

status OPEN;
last edited 08 February 2026;
12 comments;
0 claimed proofs for this problem;
“Currently working on this problem: None”;
“Interested in collaborating: Woett”;
“Likes this problem: Woett”;
the difficult, tractable, formalisable, and working-on-formalisation markers

were all None.

[d] Thus the literal stop trigger on page 935 did not fire. The linked,

essentially equivalent Problem 367 page

now separately lists sproutseeds as a current worker; this is recorded as a

collaboration warning, but it was not the page-935 status specified by the

task. No external claim, message, or outreach was made in this run.

Verbatim current statement

The following is copied verbatim from the live LaTeX view (only placed in a

code block so that its source characters are preserved):

For any integer $n=\prod p^{k_p}$ let $Q_2(n)$ be the powerful part of $n$, so that\[Q_2(n) = \prod_{\substack{p\\ k_p\geq 2}}p^{k_p}.\]Is it true that, for every $\epsilon>0$ and $\ell\geq 1$, if $n$ is sufficiently large then\[Q_2(n(n+1)\cdots(n+\ell))<n^{2+\epsilon}?\]If $\ell\geq 2$ then is\[\limsup_{n\to \infty}\frac{Q_2(n(n+1)\cdots(n+\ell))}{n^2}\]infinite?

If $\ell\geq 2$ then is\[\lim_{n\to \infty}\frac{Q_2(n(n+1)\cdots(n+\ell))}{n^{\ell+1}}=0?\]

Results and discussion already on the page

[d] The page attributes the question to Erdős [Er76d], reports Erdős's

comment that the first assertion “seems very difficult to prove”, and states

that Mahler's result gives, for every fixed \(\ell\geq1\),

\[ \limsup_{n\to\infty}\frac{Q_2(n(n+1)\cdots(n+\ell))}{n^2}\geq1. \]

[d] The page now records the second question as answered affirmatively:

the Pell equation \(x^2-8y^2=1\), via the construction on Problem 367 and the

independent rediscovery in Feng et al. [Fe26], gives

\[ \limsup_{n\to\infty} \frac{Q_2(n(n+1)(n+2))}{n^2}=\infty. \]

It also records that the \(abc\) conjecture implies a positive answer to the

third question, and notes the analogous \(Q_r\) questions.

[d] I read all 12 comments. In chronological order they are by Woett,

Nat Sothanaphan, TerenceTao, Kevin Barreto, Woett, Thomas Bloom, fengt, Woett,

fengt, BorisAlexeev, fengt, and old-bielefelder (2–5 February 2026). Their

mathematical content concerns: recognizing the near-identity with Problem 367;

priority versus independent rediscovery of the Pell construction; the fact

that Pell equations are standard here; moving the Aletheia result to an

“Independent Rediscovery” category; and the fact that Aletheia incorrectly

described its response as solving the whole problem even though only its

second part survived review. None of the comments claims a proof of the

remaining first or third question.

1. Primary-source literature audit

[d] The original source exists and was inspected:

P. Erdős, [“Problems and results on number theoretic properties of consecutive

integers and related questions”](https://combinatorica.hu/~p_erdos/1976-39.pdf),

Proceedings of the Fifth Manitoba Conference on Numerical Mathematics

(1976), 25–44. The relevant discussion is on PDF page 8 (printed page 32).

It contains the three directions represented on the live page and the

“seems very difficult” sentence.

[d] Feng et al.,

[“Semi-Autonomous Mathematics Discovery with Gemini: A Case Study on the

Erdős Problems”](https://arxiv.org/abs/2601.22401), arXiv:2601.22401,

was downloaded and checked at §4.2, pp. 21–23. The reviewed paper explicitly

says that Aletheia solved the second question, gives the Pell proof, and

only remarks that \(abc\) implies the third. It does not publish an

unconditional proof of the first or third question.

[d] I also read the newer

Problem 367 discussion,

because page 935 itself says the questions are equivalent up to fixed

constants. A 10 June 2026 comment by ScottHughes claims a stronger Pell lower

bound for the already-solved second direction, an \(r\)-full extension, and

the \(abc\)-conditional first direction. The site labels comments as

unverified; I used none of those claims as a theorem and did not duplicate

the Pell construction.

[d] A dangerous false lead was rejected. The repository's

raw, unreviewed model response

claims a 2008 paper by J. Cilleruelo, *The powerful part of the product of

consecutive integers*, Bull. London Math. Soc. 40 (2008), 873–877.

Exact-title searches found no such paper. More decisively, a Crossref query

for ISSN 0024-6093, volume 40 (2008), shows that pages 863–875 are

Deshouillers' A refined bound for sum-free sets in groups of prime order and

pages 876–886 are Ortega-Cerdà's *Interpolating and sampling sequences in

finite Riemann surfaces*. The alleged 873–877 article cannot occupy that page

range. Nothing in this report uses it.

[d] Terence Tao,

“Products of consecutive integers with unusual anatomy”,

arXiv:2603.27990v2 (22 April 2026), was checked because it postdates the last

edit of page 935. Its “very bad” intervals are intervals whose whole product

is powerful. Theorem 1.8 proves a counting asymptotic for integers lying in

such intervals, but the paper explicitly says the conjecture that a very bad

interval has length at most two remains open. It does not give the uniform

lower bound on exponent-one primes needed below.

[d] Carlo Sanna,

[“On the exponents in the factorizations of \(r\) consecutive

numbers”](https://doi.org/10.2989/16073606.2021.1938277),

Quaestiones Mathematicae 45 (2022), 1221–1228, was checked. Its Lemma 3.3

gives the same basic \(O_r(x/y^{1/2})\) tail estimate for large powerful parts

that appears in the elementary density argument below. I reprove the needed

estimate rather than importing it.

[d] For the conditional radical bound I checked P. Ribenboim,

“Finite sets of binary forms”,

Publ. Math. Debrecen 68 (2006), 261–282. Theorem 4.1 attributes the binary

form radical theorem to Granville (earlier, with extra endpoint hypotheses,

to Langevin), and Lemma 4.2 gives the \(n-1-\epsilon\) exponent after fixing

the second variable to \(1\).

[d] Searches for the exact problem, “powerful part” plus consecutive

integers, the alleged Cilleruelo title, and later citations found no verified

primary source proving the remaining uniform assertions. This is a reported

search miss, not a proof that no such literature exists.

2. Exact reformulation

For fixed \(\ell\), put

\[ P_\ell(n)=\prod_{i=0}^{\ell}(n+i),\qquad w(m)=\frac{m}{Q_2(m)} =\prod_{v_p(m)=1}p, \]

and

\[ W_\ell(n)=\frac{P_\ell(n)}{Q_2(P_\ell(n))} =\prod_{v_p(P_\ell(n))=1}p. \]

[a] These are exact integer identities, directly from the definition of

\(Q_2\). Consequently

\[ \frac{Q_2(P_\ell(n))}{P_\ell(n)}=\frac1{W_\ell(n)}. \tag{2.1} \]

The third question is therefore exactly \(W_\ell(n)\to\infty\). The first

question is exactly the eventual inequality

\[ W_\ell(n)> n^{\ell-1-\epsilon}\prod_{i=1}^{\ell}\left(1+\frac{i}{n}\right). \tag{2.2} \]

Thus its asymptotic content is the uniform lower bound

\(W_\ell(n)\ge n^{\ell-1-o(1)}\).

Whole-product versus individual powerful parts

Define the fixed constant

\[ C_\ell=\prod_{p\leq\ell} p^{\lceil(\ell+1)/p\rceil}. \tag{2.3} \]

[a] Lemma.

\[ Q_2(P_\ell(n)) \leq C_\ell\prod_{i=0}^{\ell}Q_2(n+i). \tag{2.4} \]

[a] Proof. If \(p>\ell\), then \(p\) divides at most one of the

\(n+i\), so the \(p\)-contribution agrees on the two sides before the factor

\(C_\ell\). Let \(p\leq\ell\), write \(e_i=v_p(n+i)\), let

\(E=\sum e_i\), and let \(s=\#\{i:e_i=1\}\). If \(E<2\), there is no missing

factor. If \(E\geq2\), the whole-product powerful part contains \(p^E\),

whereas the product of the individual powerful parts contains

\(p^{E-s}\); the ratio is \(p^s\). An interval of \(\ell+1\) integers

contains at most \(\lceil(\ell+1)/p\rceil\) multiples of \(p\). Multiplying

over \(p\leq\ell\) proves (2.4). \(\square\)

[a] This constant is precisely why computing

\(\prod_i Q_2(n+i)\) alone would not be a valid checker for the stated

problem.

3. Unconditional almost-everywhere progress

[a] Powerful-number count. Every powerful integer has a unique

representation \(a^2b^3\) with \(b\) squarefree. Hence the number \(A(Y)\)

of powerful integers at most \(Y\) satisfies

\[ A(Y)\leq \sqrt{Y}\sum_{b\geq1}b^{-3/2} =\zeta(3/2)\sqrt{Y}. \tag{3.1} \]

Splitting into dyadic intervals then gives

\[ \sum_{\substack{d>T\\d\ {\rm powerful}}}\frac1d \ll T^{-1/2}. \tag{3.2} \]

[a] Tail lemma. Uniformly for \(1\leq T\ll X\),

\[ \#\{m\in[X,2X+O_\ell(1)]:Q_2(m)>T\} \ll_\ell \frac{X}{\sqrt T}. \tag{3.3} \]

[a] Proof. If \(Q_2(m)=d>T\), then \(d\) is a powerful divisor of

\(m\). A union bound over powerful \(d\) gives

\[ \sum_{\substack{Tby (3.1)–(3.2) and \(T\ll X\). \(\square\)

[a] Theorem (density version of the first question). Fix

\(\ell\geq2\) and \(0<\epsilon\leq\ell-1\). Then

\[ \#\left\{n\in[X,2X]: Q_2(P_\ell(n))\geq n^{2+\epsilon}\right\} \ll_{\ell,\epsilon} X^{\,1-\frac{2+\epsilon}{2(\ell+1)}}. \tag{3.4} \]

For \(\epsilon>\ell-1\), the desired inequality is elementary for every

sufficiently large \(n\), simply because \(Q_2(P_\ell(n))\leq P_\ell(n)\)

and the exponent \(2+\epsilon\) then exceeds \(\ell+1\). For \(\ell=1\),

the original first question is also elementary:

\(Q_2(n(n+1))\leq n(n+1)

[a] Proof of (3.4). A violation and (2.4) imply

\[ \prod_{i=0}^{\ell}Q_2(n+i)\geq X^{2+\epsilon}/C_\ell. \]

At least one factor is at least

\[ T=(X^{2+\epsilon}/C_\ell)^{1/(\ell+1)}. \]

Apply (3.3) to each of the \(\ell+1\) shifts. Since

\(\epsilon\leq\ell-1\), \(T\ll X\), and the stated exponent follows.

\(\square\)

[a] Theorem (density version of the third question). For every fixed

\(\ell\geq2\) and \(\eta>0\),

\[ \#\left\{n\in[X,2X]: \frac{Q_2(P_\ell(n))}{n^{\ell+1}}\geq\eta\right\} \ll_{\ell,\eta}X^{1/2}. \tag{3.5} \]

Thus the third limit is unconditionally true in natural density, with

a power-saving exceptional set.

[a] Proof. A member of the set in (3.5), together with (2.4), forces

some \(Q_2(n+i)\gg_{\ell,\eta}X\). Use (3.3) with

\(T\asymp_{\ell,\eta}X\), then sum over the fixed number of shifts.

\(\square\)

[a] Scope warning. Neither (3.4) nor (3.5) controls every \(n\).

An exceptional set of size \(O(\sqrt X)\) can still contain an infinite

sparse sequence, so these results do not settle either remaining uniform

question.

4. The \(abc\) route, checked precisely

Let \(R=\operatorname{rad}(P_\ell(n))\). Because the primes in

\(W_\ell(n)\) and \(Q_2(P_\ell(n))\) are disjoint,

\[ R=W_\ell(n)\operatorname{rad}(Q_2(P_\ell(n))). \]

[a] Since every exponent in \(Q_2\) is at least two,

\[ \operatorname{rad}(Q_2(P_\ell(n))) \leq\sqrt{Q_2(P_\ell(n))}. \]

Using \(P_\ell=W_\ell Q_2(P_\ell)\) gives the useful exact inequality

\[ R\leq\sqrt{W_\ell(n)P_\ell(n)},\qquad W_\ell(n)\geq\frac{R^2}{P_\ell(n)}. \tag{4.1} \]

[b] Assume the \(abc\) conjecture through the Granville–Langevin

radical consequence (Ribenboim, Lemma 4.2): for a squarefree polynomial

\(f\) of degree \(d\),

\[ \operatorname{rad}(f(n))\gg_{f,\delta}n^{d-1-\delta}. \tag{4.2} \]

For \(f(x)=P_\ell(x)\), \(d=\ell+1\). Equations (4.1)–(4.2) yield

\[ W_\ell(n)\gg_{\ell,\delta}n^{\ell-1-2\delta}, \qquad Q_2(P_\ell(n))\ll_{\ell,\delta}n^{2+2\delta}. \tag{4.3} \]

Choosing \(\delta\) arbitrarily small proves both the first question and,

for \(\ell\geq2\), the third question conditionally on \(abc\). This

derivation identifies the exact exponent supplied by the named theorem;

it is not an unconditional claim.

5. Exact finite computation

What was computed

[d] The standalone verifier exhaustively scans every

\(1\leq n\leq10^9\) for \(\ell=2\) and \(\ell=3\), computes the integer

\(W_\ell(n)\), and finds its exact minimum in each displayed range. All

minimizers in the table are unique.

|---:|---:|---:|---:|---:|

| \(1\)–\(9\) | 3 | 2 | 3 | 1 |

| \(10\)–\(99\) | 3 | 48 | 13 | 24 |

| \(100\)–\(999\) | 61 | 242 | 305 | 242 |

| \(10^3\)–\(9{,}999\) | 29 | 9,800 | 6,545 | 1,680 |

| \(10^4\)–\(99{,}999\) | 515 | 59,534 | 1,306,855 | 23,760 |

| \(10^5\)–\(999{,}999\) | 985 | 332,928 | 58,997,939 | 530,450 |

| \(10^6\)–\(9{,}999{,}999\) | 8,113 | 1,294,298 | 128,066,455 | 2,328,480 |

| \(10^7\)–\(99{,}999{,}999\) | 36,395 | 96,549,408 | 15,586,040,119 | 63,101,375 |

| \(10^8\)–\(10^9\) | 460,357 | 128,961,799 | 9,377,106,367,033 | 129,800,448 |

[d] In particular, the following are sharp over the stated finite

range:

\[ \frac{Q_2(n(n+1)(n+2))}{n(n+1)(n+2)} \leq\frac1{460357} \quad(10^8\leq n\leq10^9), \tag{5.1} \]

with equality only at \(n=128961799\), and

\[ \frac{Q_2(n(n+1)(n+2)(n+3))} {n(n+1)(n+2)(n+3)} \leq\frac1{9377106367033}, \tag{5.2} \]

with equality only at \(n=129800448\).

[d] Combining the exact finite minima with the elementary inequalities

\(\prod_{i=1}^{\ell}(1+i/n)\leq \prod_{i=1}^{\ell}(1+i/10^8)\) gives, throughout the same range,

\[ \frac{Q_2(P_2(n))}{n^3} \leq 2.1722272714436844\times10^{-6}, \tag{5.3} \] \[ \frac{Q_2(P_3(n))}{n^4} \leq 1.0664271267260991\times10^{-13}. \tag{5.4} \]

These decimal bounds are corollaries of the exact rational bounds, not

floating-point search criteria.

Independent extremizer factorizations

[d] For the final \(\ell=2\) range, scalar trial division independently

gave

\[ \begin{aligned} 128961799&=137^2\cdot6871,\\ 128961800&=2^3\cdot5^2\cdot11^2\cdot73^2,\\ 128961801&=3^5\cdot67\cdot89^2. \end{aligned} \]

Thus the exponent-one primes of the whole product are exactly \(6871\)

and \(67\), and \(6871\cdot67=460357\).

[d] For the final \(\ell=3\) range, the independent factorization was

\[ \begin{aligned} 129800448&=2^8\cdot3^3\cdot89\cdot211,\\ 129800449&=11393^2,\\ 129800450&=2\cdot5^2\cdot13^2\cdot15361,\\ 129800451&=3\cdot11^3\cdot32507. \end{aligned} \]

The \(2\)- and \(3\)-valuations aggregate across terms, so the exponent-one

primes of the whole product are \(89,211,15361,32507\), whose product is

\(9377106367033\).

Algorithm and reproducibility

[a] For a segment \([L,R]\), the verifier initializes an array with

the integers themselves. For every prime \(p\leq\sqrt R\), it divides the

\(p^2\)-progression by \(p^2\), then the \(p^j\)-progression by \(p\) for

each \(j\geq3\). A number with \(v_p(m)=e\geq2\) is therefore divided by

\(p^2p^{e-2}=p^e\), while a prime occurring exactly once remains. This

computes every individual \(w(m)\) exactly.

[a] A prime \(p>\ell\) cannot divide two entries of an

\((\ell+1)\)-term block. For \(p\leq\ell\), whether it must be removed from

the product of individual \(w(n+i)\) depends only on \(n\bmod p^2\). The

verifier constructs this periodic correction from the valuations, rather

than assuming multiplicativity. Its maximum correction is \(2\) for

\(\ell=2\) and \(18\) for \(\ell=3\).

[a] Values are saturated at \(10^{15}\) only after a guarded

calculation: the uncorrected product is first capped at

\(10^{15}C_{\max}\). Since the eventual correction is at most

\(C_{\max}\), a capped value cannot become a false value below \(10^{15}\).

All certified minima are below \(10^{15}\).

[d] Before the full scan, the script compares every individual sieve

value through 10,003 and every block value for both \(\ell=2,3\) through

\(n=10,000\) against independent scalar trial division. After the scan it

scalar-refactors every decade extremizer, reconstructs \(P_\ell=W_\ell Q_2\),

and asserts the hard-coded table.

Run:

cd /home/exedev/MathDyad
/usr/bin/time -f 'wall=%E maxrss=%MKB' \
  python3 runs/erdos935_wave7u_reverify.py

[d] The clean default run ended with:

scalar/vector cross-check: PASS (all starts n<=10,000)
FULL CERTIFICATE: PASS
elapsed_seconds=183.113
wall=3:03.30 maxrss=439812KB

[d] Verifier SHA-256:

d7fe8a5cbe418d87cfe9e28b5ff5fdbe211604d64bb865314a8bc2a4983793bf.

The complete 484-line code is in the companion .py; it contains the

prime sieve, segmented computation, periodic correction, independent

trial division, range assertions, and command-line options. A quick audit

can use --limit 10000000, while the default is required to reproduce the

full certificate.

[d] Scope warning. The finite table proves no eventual asymptotic. Its

early minima are not even monotone by decade (for example \(61\) is followed

by \(29\) for \(\ell=2\)), so no monotonic extrapolation was made.

6. Exact wall for a uniform proof

[a] By (2.1)–(2.2), the two remaining targets are precisely:

1. \(W_\ell(n)\geq n^{\ell-1-o(1)}\) uniformly in \(n\);

2. \(W_\ell(n)\to\infty\) uniformly in \(n\).

Average estimates such as (3.3) allow an \(O(\sqrt X)\) exceptional set and

therefore do not reach either target.

[a] The obstruction is already concrete for \(\ell=2\). Write

\[ n+i=a_i u_i,\qquad a_i=w(n+i),\quad u_i=Q_2(n+i). \]

Every \(a_i\) is squarefree, \(\gcd(a_i,u_i)=1\), and every \(u_i\) is

powerful. The only prime that can occur exactly once in one term but be

absorbed by another term of the three-term block is \(2\). Hence

\[ a_0a_1a_2\leq2W_2(n). \tag{6.1} \]

If \(W_2(n)\leq B\) infinitely often, one of finitely many squarefree

coefficient triples with \(a_0a_1a_2\leq2B\) must support infinitely many

powerful solutions of

\[ a_1u_1-a_0u_0=1,\qquad a_2u_2-a_1u_1=1. \tag{6.2} \]

[a] A theorem giving finiteness of (6.2) for every fixed bounded

coefficient triple would prove the third question for \(\ell=2\). Its

special coefficient choice \(a_0=a_1=a_2=1\) is the problem of three

consecutive powerful integers.

[d] Tao's 2026 paper explicitly records the corresponding

finiteness/no-triple phenomenon as still open unconditionally (and finite

under \(abc\)). Thus the checked literature supplies no generic Pell,

\(S\)-unit, or powerful-number-count theorem that proves the missing

simultaneous finiteness lemma.

[b] Alternatively, the near-optimal polynomial radical estimate

\[ \operatorname{rad}(P_\ell(n))\geq n^{\ell-o(1)} \]

would close the first question through (4.1), but the verified source gives

this strength conditional on \(abc\), not unconditionally. This is the exact

named-theorem gap in the radical approach.

[d] Extending the same exhaustive scan from \(10^9\) to \(10^{10}\)

would cost roughly 30–40 core-minutes at the measured throughput, with

about the same 440 MB segmented memory footprint. I did not run it because

it exceeds the requested few-minute budget and, regardless of outcome,

cannot provide the required uniform finiteness step.

PARTIAL: elementary density-one bounds and exact \(n\le10^9\) tables for \(\ell=2,3\) are verified; the uniform first and third questions remain open at the explicit radical/simultaneous-powerful finiteness wall.

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