Erdoes #850 - working report - Erdős Problem a Day

Erdős problem #850 — wave 6n report

Date: 2026-07-27 UTC

Disposition: PARTIAL. (d, live-page observation) The problem remains

open. The outputs are (i) an (a) elementary necessary-condition

reduction, (ii) a (d) from-scratch exact exclusion for the larger member

$y\le 10^9$, and (iii) a (b) stronger exclusion

$y<1.4\cdot 10^{12}$ modulo a directly relevant published computational

theorem that is not cited on the live problem page.

Claim labels

The requested labels are used on every substantive mathematical conclusion:

(a) elementary-rigorous;
(b) rigorous modulo the explicitly named theorem;
(c) plausible/structural-unverified or a cost estimate;
(d) computational-only.

Unlabelled material below is a definition, verbatim quotation, command,

bibliographic record, or description of what a source says.

0. Mandatory live-page check

(d, live-page observation) I accessed both the [live problem

page](https://www.erdosproblems.com/850) and its [discussion

thread](https://www.erdosproblems.com/forum/discuss/850) through the Bright

Data browser path on 2026-07-27. The problem page displayed:

OPEN;
0 claimed proofs for this problem;
Interested in collaborating None;
Currently working on this problem None;
Likes this problem 0xashrk;
This problem looks difficult aminb_el, Dogmachine;
This problem looks tractable None;
The results on this problem could be formalisable None;
I am working on formalising the results on this problem None.

Thus no mandatory stop condition was present.

Verbatim live statement

> Can there exist two distinct integers $x$ and $y$ such that $x,y$ have

> the same prime factors, $x+1,y+1$ have the same prime factors, and

> $x+2,y+2$ also have the same prime factors?

(d, live-page observation) The four comments, all dated 19 January 2026,

were also read. Neel Somani asked whether positive integers should be made

explicit; Thomas Bloom replied that the intended restriction is

$x,y\ge 1$; Somani acknowledged this; Terence Tao suggested making such

contextual hypotheses explicit, especially for automated attacks, and noted

that the Lean formalisation has the intended domain. None of the comments

claims a proof or says that anyone is currently working on the problem.

Accordingly, this report uses positive integers.

(d, live-page observation) The page's listed mathematical context is:

the name “Erdős-Woods conjecture”;
for only the first two comparisons, the family

$x=2(2^r-1)$, $y=x(x+2)$;

Makowski's additional first-two-term pair $(75,1215)$;
no other first-two-term examples known, with OEIS A343101 cited;
Shorey and Tijdeman's negative answer to the three-term question under

Baker's strong explicit $abc$ conjecture;

cross-reference #677, the 2011 Benelux olympiad problem, and Guy B19.

The page said it was last edited 28 September 2025.

1. Primary-source literature audit

These are the specific hits I opened and checked; none claims an unconditional

solution of #850.

(d, search observation) Exact-title/statement searches over arXiv and the

web, including 2025–2026 date terms, found no later primary source claiming a

resolution. The newest directly relevant primary items I found were Hercher

and Lebowitz-Lockard below. This is a search report, not a theorem that no

other literature exists.

1. P. Erdős, [*Quelques problèmes de la théorie des

nombres*](https://static.renyi.hu/~p_erdos/1963-14.pdf), Enseignement

Mathématique 6 (1963), Problem 60. The source records the standard

two-term family and asks for other two-term examples.

2. A. Makowski, [*On a problem of

Erdős*](https://www.e-periodica.ch/cntmng?pid=ens-001%3A1968%3A14%3A%3A87),

Enseignement Mathématique 14 (1968), p. 193. The paper explicitly gives

$75,1215$ for the two-term problem.

3. R. Balasubramanian, T. N. Shorey, and M. Waldschmidt,

[*On the maximal length of two sequences of consecutive integers with the

same prime

divisors*](https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/Balasubramanian1989_Article_OnTheMaximalLengthOfTwoSequenc.pdf),

Acta Math. Hung. 54 (1989), 225–236, DOI

10.1007/BF01952052. It gives

unconditional bounds on the possible length, not a resolution at length

three.

4. T. N. Shorey and R. Tijdeman, [*Arithmetic properties of blocks of

consecutive integers*](https://arxiv.org/abs/1612.05438),

arXiv:1612.05438, Theorem 9.1. I checked the source text of the theorem and

proof: Baker's explicit $abc$ conjecture implies that no positive

\(n_1

5. N. Lebowitz-Lockard, [*On pairs of consecutive sequences with the same

radicals*](https://arxiv.org/abs/2507.09899), arXiv:2507.09899

(14 July 2025). This preprint explicitly says the three-term problem is

unresolved and proves broader length/counting bounds. It does not settle

existence at length three.

6. Most importantly, C. Hercher, [*On One of Erdős' Problems—An Efficient

Search for Benelux

Pairs*](https://cs.uwaterloo.ca/journals/JIS/VOL28/Hercher3/hercher15.pdf),

Journal of Integer Sequences 28 (2025), Article 25.8.5, published 12

December 2025; see also arXiv:2506.01099.

Theorem 1 classifies every two-term Benelux pair

\(m

repository](https://github.com/ChristianHercher/Erdos850), whose current

commit I checked as

54b80aca0fd9b7bc33b13f65cc00b50824e90082. The repository contains the

program but no completion log or independently checkable certificate for

the month-long search. I therefore use the $1.4\cdot10^{12}$ result only

with label (b), and do not present it as my own computational verification.

2. Elementary reduction

Write

\[ \operatorname{rad}(n)=\prod_{p\mid n}p \]

and suppose, without loss of generality, that \(1\le x

$d=y-x$, and for $i=0,1,2$ put

\[ r_i=\operatorname{rad}(x+i)=\operatorname{rad}(y+i). \]

Define

\[ R=\operatorname{lcm}(r_0,r_1,r_2) =\operatorname{rad}\bigl(y(y+1)(y+2)\bigr). \]

Divisibility lemma

(a) Every putative solution satisfies

\[ R\mid d \quad\text{and hence}\quad R\le d=y-xProof. If $p\mid y+i$, equality of prime supports gives $p\mid x+i$.

Therefore $p\mid(y+i)-(x+i)=d$. Thus $r_i\mid d$ for each $i$, so their

least common multiple $R$ divides $d$. Positivity gives \(d

This yields a one-variable obstruction:

(a) If

\[ \operatorname{rad}\bigl(n(n+1)(n+2)\bigr)\ge n \tag{2} \]

for every $n\le Y$, then #850 has no solution whose larger member is at most

$Y$. This implication loses information, so a value violating (2) would

only be a candidate, not a counterexample.

Two further consequences are useful:

(a) Each of $y,y+1,y+2$ must be non-squarefree. If $y+i$ were

squarefree, then $r_i=y+i>d$, contradicting $r_i\mid d$.

(a) The gap has the exact form

\[ y-x=tR\qquad(t\in\mathbb Z_{\ge1}). \tag{3} \]

Fixed-support formulation

Let $S_i$ be the prime support of $x+i$ (equivalently $y+i$).

(a) Apart from the prime $2$, the three sets $S_0,S_1,S_2$ are

disjoint. The only possible overlap is $2\in S_0\cap S_2$, when $x,y$

are even. Both $x$ and $y$ obey

\[ n\equiv0\pmod p\ (p\in S_0),\qquad n\equiv-1\pmod p\ (p\in S_1),\qquad n\equiv-2\pmod p\ (p\in S_2). \tag{4} \]

The generalized Chinese remainder theorem therefore puts both numbers in one

residue class modulo $R$; equation (3) is the resulting gap.

(b, Størmer's theorem) For a fixed finite union

$S=S_0\cup S_1\cup S_2$, there are only finitely many consecutive

$S$-smooth pairs $n,n+1$, and they can be enumerated via finitely many Pell

equations. See D. H. Lehmer, On a problem of Størmer, Illinois J. Math. 8

(1964), 57–79

(Project Euclid,

DOI 10.1215/ijm/1256067456).

Consequently each fixed support partition is decidable. The unresolved

uniform step is that $S$, $R$, and the exponents are all unbounded.

3. Independent exact computation

The standalone checker is

runs/erdos850_wave6n_reverify.py, SHA-256

23860a2cc711bed3458d14c9d29eea904444068cdfc790151f656e6be141cd84

It has no factoring/database dependency; NumPy is used only for exact

fixed-width array operations.

Algorithm and correctness

(a) The radical sieve starts with r[n] = n. For every prime $p$ and

every power $p^e\le Y+2$, $e\ge2$, it divides all multiples of $p^e$ by

$p$. If $v_p(n)=a$, the entry is divided once for each

$e=2,\ldots,a$, leaving exactly one factor $p$. Hence the resulting

array is exactly $\operatorname{rad}(n)$.

For

\[ a=\operatorname{rad}(n),\quad b=\operatorname{rad}(n+1),\quad c=\operatorname{rad}(n+2), \]

consecutive-number coprimality gives

\[ \operatorname{rad}(n(n+1)(n+2)) =ab\,\frac{c}{\gcd(a,c)}. \tag{5} \]

(a) For $Y\le2^{32}-2$, $ab\le n(n+1)<2^{64}$, so the first product

is exact in uint64. The final product is not formed in fixed width:

\[ ab\,fwhich makes the comparison overflow-free.

The code independently audits the fast sieve against:

1. a differently ordered prime-multiple radical sieve for every

$0\le n\le200000$; and

2. ordinary trial division at 64 deterministic checkpoints spread through

the full range.

It also trial-factors every candidate appearing in Hercher's classification;

this arithmetic audit is separate from the billion-value necessary-condition

scan.

Reproduction command

python3 runs/erdos850_wave6n_reverify.py \
  --limit 1000000000 --audit-limit 200000

The run made on this VM used NumPy 2.4.4. Its essential output was:

building exact radicals through 1,000,000,002 ...
radical sieve seconds: 76.006
independent radical-sieve audit: 0..200000 OK
full-range trial-factor checkpoints: 64 OK
necessary-condition scan: no y <= 1,000,000,000 has rad(y(y+1)(y+2)) < y
preliminary rad(y)*rad(y+1) < y cases: 845
global minimum R/y in the scanned range: y=2, R=6,
    reduced_ratio=3/1, decimal=3
total seconds: 94.100
RESULT no_necessary_condition_hits=True limit=1000000000

/usr/bin/time -v reported 94.49 seconds wall time, 55.14 seconds user time,

40.93 seconds system time, and peak RSS 8,004,124 KiB.

(d) For every $1\le n\le10^9$, the computation found the stronger

finite-range inequality

\[ \operatorname{rad}(n(n+1)(n+2))\ge3n, \]

and $n=2$ attains equality. The checker did not count all equality cases,

so no uniqueness claim is made.

(a)+(d) Combining this exact scan with (1) proves that no #850

counterexample has larger member $y\le10^9$. The word “proves” here is

finite/computational, not a uniform theorem.

4. The stronger literature corollary

Hercher's Theorem 1 says that every first-kind Benelux pair

\(m \[ (75,1215) \]

\[ m=2^k-2,\qquad n=2^k(2^k-2),\qquad 2\le k\le20. \tag{6} \]

Every #850 pair is in particular a first-kind Benelux pair, so only these

listed candidates need their third terms checked.

(a) The exceptional candidate fails because

\[ \operatorname{rad}(75+2)=\operatorname{rad}(77)=77,\qquad \operatorname{rad}(1215+2)=\operatorname{rad}(1217)=1217. \]

Trial division by the primes at most $\sqrt{1217}<35$ verifies that 1217 is

prime.

(a) Every member of (6) fails uniformly:

\[ m+2=2^k,\qquad n+2=2\bigl(2^{2k-1}-2^k+1\bigr). \]

The parenthesized factor is odd and at least $5$ for $k\ge2$. Hence

$\operatorname{rad}(m+2)=2$, while $\operatorname{rad}(n+2)$ contains an

odd prime.

The standalone checker independently emitted the full arithmetic table. Its

endpoints and exceptional row are:

| kind/$k$ | $m$ | $n$ | $\operatorname{rad}(m+2)$ | $\operatorname{rad}(n+2)$ |

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

| exceptional | 75 | 1,215 | 77 | 1,217 |

| 2 | 2 | 8 | 2 | 10 |

| 20 | 1,048,574 | 1,099,509,530,624 | 2 | 84,577,656,202 |

(b, Hercher Theorem 1) There is no #850 counterexample with

\[ 1\le xThis is the strongest concrete exclusion found in the literature search. Its

classification premise is Hercher's published computation; my script verifies

the deduction and all listed candidates but does not reproduce that

month-long exhaustive search.

5. Why the argument does not become uniform

The elementary identity

\[ y(y+2)+1=(y+1)^2 \tag{8} \]

has

\[ \operatorname{rad}\bigl(y(y+1)(y+2)\bigr)=Rfor a putative solution.

(b, ordinary $abc$ conjecture) Applying $abc$ to (8), with any

$\varepsilon<1$, gives

\[ (y+1)^2\le C_\varepsilon R^{1+\varepsilon} Thus ordinary $abc$ leaves at most finitely many $y$, but its unspecified

constant does not eliminate the finite exceptional set.

(b, Shorey–Tijdeman Theorem 9.1) Baker's stronger explicit $abc$

conjecture yields

\[ y^2an immediate contradiction. This is exactly why the conditional result

closes the problem while ordinary $abc$ gives only finiteness.

(b, Stewart–Yu's unconditional $abc$-type bound) The available

unconditional shape

\[ (y+1)^2< \exp\!\left(CR^{1/3}(\log R)^3\right) \]

does not contradict \(R

large $y$. Hence this standard machinery stalls at the precise missing

power-law radical estimate.

An exact sufficient lemma would be

\[ \operatorname{rad}(n(n+1)(n+2))\ge n \quad\text{for every positive }n. \tag{9} \]

(c) Lemma (9) is structurally plausible from the computation but is

unproved here and must not be treated as a theorem. It is a special

high-quality $abc$ assertion for (8); proving it uniformly, or exploiting

the separate supports $S_0,S_1,S_2$ to bypass it, is the exact analytic

wall.

Cost of extending only the finite scan

The $10^9$ run consumed about 96 CPU-seconds. A full unsegmented array to

$1.4\cdot10^{12}$ would require about 11.2 TB, so scaling requires a

segmented implementation.

(c, measured scaling estimate) Ideal linear scaling of the observed run

to $1.4\cdot10^{12}$ is about 38 core-hours; allowing for segmentation and

prime-loop overhead gives roughly 50–80 core-hours, or about 2–3 wall hours on

32 cores. At a realistic \$0.05–\$0.10 per core-hour, compute cost is roughly

\$2.50–\$8, excluding engineering and storage. A scan to $10^{13}$ would

be roughly 350–600 core-hours. I did not run either job here. Even such a

scan would establish only a larger finite bound and could not close the

unbounded problem without an analytic finiteness/uniformity step.

6. Verified state

(a) Any solution forces

\(\operatorname{rad}(y(y+1)(y+2))\mid(y-x)

CRT class modulo that radical, and makes $y,y+1,y+2$ all non-squarefree.

(d) The necessary inequality fails for every $y\le10^9$; the exact

scan actually found $R(y)\ge3y$ throughout this range.

(b) Modulo Hercher's published Theorem 1, every

$y<1.4\cdot10^{12}$ is excluded after an elementary check of his complete

two-term candidate list.

(c) No construction, counterexample, or unconditional uniform proof was

found. The missing step is a uniform radical lower bound such as (9), or a

genuinely uniform use of the three separate prime-support sets.

PARTIAL: elementary reduction and an independent exact scan exclude y<=10^9; modulo Hercher's published exhaustive Theorem 1, all y<1.4e12 are excluded, but the unbounded problem remains open.

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