Erdős problem 891 — wave w021

Date checked: 2026-07-28 UTC.

Claim labels used throughout:

• (a) elementary-rigorous — proved here from elementary facts.
• (b) rigorous-modulo-named-theorem — dependent on the cited theorem or primary source.
• (c) plausible/structural-unverified — a heuristic or an unclosed route.
• (d) computational-only — an exhaustive finite calculation. Here “computational-only” can still mean an exact certificate check; it never means a uniform theorem.

0. Mandatory live-page gate

(b; live-page observation) I fetched the live page through the Bright Data browser path, including its discussion thread, before doing mathematics. On 2026-07-28 the page was marked OPEN, showed 0 claimed proofs, listed Currently working on this problem: None, and listed Interested in collaborating: None. Thus the stop/skip condition was not triggered. The other visible markers were “Likes this problem: Dogmachine”, “This problem looks difficult: Dogmachine”, and no marker for tractable/formalisable/formalising.

The live statement, copied verbatim from the page's LaTeX view, is:

Let $2=p_1<p_2<\cdots$ be the primes and $k\geq 2$. Is it true that, for all sufficiently large $n$, there must exist an integer in $[n,n+p_1\cdots p_k)$ with $>k$ many prime factors?

(b; interpretation fixed by the live page and original sources) The January 2026 live-page comment explicitly says that the factors are required to be distinct. This also agrees with Schinzel's 1959 formulation (“diviseurs premiers distincts”) and Erdős–Selfridge's definition of their prime-factor-counting function. I therefore write

Results and comments actually listed on the live page

(b) The page attributes to Schinzel, via Pólya's theorem on gaps between integers supported on a fixed finite prime set, the result with interval length

in place of \(p_1\cdots p_k\).

(b) The page says the problem is unknown even for \(k=2\), namely whether every sufficiently late block of six integers contains an integer with at least three distinct prime factors.

(b) The page records Weisenberg's observation that Dickson's conjecture conditionally makes the answer negative when the interval length is shortened to \(p_1\cdots p_k-1\). Its displayed construction uses \(L_k=\operatorname{lcm}(1,\ldots,p_1\cdots p_k)\) and the linear forms \((L_k/m)n'+1\), \(1\leq m<p_1\cdots p_k\).

(c; unverified live comment, StijnC, 2025-08-24) A bad interval must contain a number \(p_1^{e_1}\cdots p_k^{e_k}\); the comment gives probabilistic heuristics near such a number, guesses \(n>4372\) for \(k=2\), and guesses \(n>8611\) for \(k=3\), while explicitly noting that a rigorous argument looks hard.

(c; unverified live comment, Dogmachine, 2026-01-08) Mihăilescu/Catalan is invoked to say that every three consecutive integers starting at 8 contain a number with at least two distinct prime divisors. The same comment clarifies “distinct”.

(b; cited theorem in a live comment, Dogmachine, 2026-06-13) Eggleton and MacDougall are cited for the fact that ten consecutive integers cannot all have exactly two distinct prime factors. This concerns \(\omega(m)=2\) for every member, not the present mixed condition \(\omega(m)\leq2\).

Live sources: problem 891, discussion thread, and LaTeX view.

1. Primary-source literature check

(b) A. Schinzel, “Unsolved problem 31,” Elemente der Mathematik 14 (1959), 82–83, is present in the ETH Zürich scan. It defines the corresponding shortest eventual length \(g(k)\), derives the Pólya upper bound used on the live page, and states in particular that among every ten consecutive integers greater than 92 one has three distinct prime divisors. It conjectures the primorial-sized improvement. Primary scan, pp. 82–83.

(b) P. Erdős and J. L. Selfridge, “Some problems on the prime factors of consecutive integers,” Illinois Journal of Mathematics 11 (1967), 428–430, explicitly restate on p. 430 that Schinzel obtains \(p_1\cdots p_{k-1}p_{k+1}\), conjecture that \(p_1\cdots p_k\) is right, and say they cannot improve Schinzel even for \(k=2\). Primary Erdős archive PDF.

(b) G. Pólya's cited paper really exists: “Zur arithmetischen Untersuchung der Polynome,” Mathematische Zeitschrift 1 (1918), 143–148. The Göttingen archive supplies the paper and metadata. The precise consequence used here—gaps between consecutive integers whose prime factors lie in a fixed finite set tend to infinity—is stated explicitly in the two primary sources above and in R. Tijdeman's later primary paper. Pólya archive, Tijdeman, Compositio Mathematica 26 (1973), 319–330.

(b) R. B. Eggleton and J. A. MacDougall, “Consecutive Integers with Equally Many Principal Divisors,” Mathematics Magazine 81 (2008), 235–248, Theorem 1, proves that there is no run of ten integers all satisfying \(\omega(m)=2\). Their set \(P_2\) means exactly two distinct prime divisors, so this does not rule out a six-term run mixing primes, prime powers, and two-prime products as problem 891 requires. Publisher record and DOI.

(b; relevant but not resolving 891) Two very recent primary preprints were checked because their titles are close to this problem:

• Tao and Teräväinen, Quantitative correlations and some problems on prime factors of consecutive integers, arXiv:2512.01739v2 (25 April 2026), prove that infinitely many \(n\) satisfy \(\omega(n+j)\leq\Omega(n+j)\ll j\) for all positive \(j\).
• Cheuk Fung Lau, On the Number of Prime Factors of Consecutive Integers, arXiv:2604.15042v2 (24 June 2026), improves this to \(\omega(n+j)\leq\Omega(n+j)\leq C\log j\) for all \(j\geq2\), for infinitely many \(n\).

Their constants and quantifiers do not imply either side of problem 891: they do not give a pointwise lower bound in every fixed primorial-length interval, and they do not produce an interval in which every member has at most the prescribed fixed \(k\) distinct factors. Lau explicitly associates the applications in that paper with Erdős problems 248, 413, 826, and 679, not 891.

(b; honest search result) Exact-title, exact-phrase, and citation searches located the sources above but no primary source claiming a solution or a uniform \(k=2\) result for problem 891. This is only a report of the search, not a proof that no such paper exists.

2. Exact structural reduction

Put

Call \(n\) \(k\)-bad if

Smooth-centre lemma

(a) If \(n\) is \(k\)-bad, the unique multiple \(R\) of \(P_k\) in \(\{n,\ldots,n+P_k-1\}\) has the form

Proof. Any \(P_k\) consecutive integers contain exactly one multiple \(R\) of \(P_k\). Thus \(p_1,\ldots,p_k\) all divide \(R\), so \(\omega(R)\geq k\). Badness gives \(\omega(R)\leq k\). Equality follows, and no other prime can divide \(R\). ∎

(a) Consequently every bad start occurs uniquely as

for one of the smooth centres above. This turns a scan through \(X\) integers into a scan through only \(O_k((\log X)^k)\) centres. The converse is not asserted: most such centred windows are good.

Nearest-witness version

For such a centre \(R\), let \(L\) and \(U\) be the distances to the nearest integers on the left and right having at least \(k+1\) distinct prime factors (use \(+\infty\) if absent).

(a) A \(k\)-bad length-\(P_k\) window containing \(R\) exists if and only if

Indeed, the consecutive low-\(\omega\) block around \(R\) has length \(L+U-1\), and a length-\(P_k\) subblock containing \(R\) exists exactly under the displayed inequality. For \(k=2\), the whole open problem is therefore reduced to proving that, for every sufficiently large \(R=2^a3^b\), the nearest \(\omega\geq3\) witnesses on the two sides have distance sum at most six.

Certificate lemma

(a) If \(t\) has \(k+1\) pairwise-coprime divisors \(d_1,\ldots,d_{k+1}>1\), then \(\omega(t)\geq k+1\). Each \(d_i\) contains a prime factor, and pairwise coprimality makes these chosen primes distinct.

This is useful computationally because the \(d_i\) need not themselves be prime. No general-purpose factorisation or primality oracle is needed.

3. Exact finite computation

The standalone checker is runs/erdos891_wavew021_reverify.py. It uses only the Python standard library and performs:

1. an independent dense Eratosthenes sieve for exact \(\omega(m)\) through \(1{,}000{,}029\);
2. a second scan of the same range using only the smooth centres, asserting equality with the dense bad-start list;
3. exact trial-division factorizations around the final obstructions;
4. a sparse scan through the much larger bounds below;
5. for every candidate window beyond the last bad start, construction and direct validation of \(k+1\) pairwise-coprime divisors of some member of that window.

The sparse certificates strip all powers of independently sieved primes \(p\leq20{,}000\). A remaining cofactor is admitted as a certificate component only after the discovered prime powers have been completely removed; the script then directly checks nontriviality, divisibility of the product, and all pairwise gcds. It uses no sympy, no probable-prime test, no stored factor table, and no randomness.

The core of the sparse proof is:

def smooth_centres(base_primes, cap):
    def rec(i, value):
        if i == len(base_primes):
            yield value
            return
        value *= base_primes[i]          # exponent is at least one
        while value <= cap:
            yield from rec(i + 1, value)
            value *= base_primes[i]
    yield from rec(0, 1)

def certificate(t, need, trial_primes):
    remaining, pieces = t, []
    for p in trial_primes:
        if remaining % p:
            continue
        prime_power = 1
        while remaining % p == 0:
            remaining //= p
            prime_power *= p
        pieces.append(prime_power)
        if len(pieces) == need:
            break
    if len(pieces) < need and remaining > 1:
        pieces.append(remaining)
    if len(pieces) < need:
        return None
    pieces = tuple(pieces[:need])
    assert t % math.prod(pieces) == 0
    assert all(math.gcd(a, b) == 1
               for a, b in itertools.combinations(pieces, 2))
    return pieces

# For every R and every j with n=R-j in range, search
# d=-j,...,P_k-1-j for a certified value R+d.

The complete executable source, including both independent sieves, all range assertions, factorizations, and witness fingerprints, is the standalone file named above.

Reproduction command:

python runs/erdos891_wavew021_reverify.py

The complete run passed all assertions in about 16.5 seconds on this VM.

Exhaustive results

(d) The following are exact finite classifications, not asymptotic conclusions:

| \(k\) | \(P_k\) | all starts verified through \(X\) | smooth centres through \(X+P_k-1\) | relevant centres | candidate starts above last bad | certificate failures | last bad start | |---:|---:|---:|---:|---:|---:|---:|---:| | 2 | 6 | \(10^{30}\) | 3052 | 3016 | 18092 | 0 | 4372 | | 3 | 30 | \(10^{12}\) | 2369 | 2316 | 69475 | 0 | 8615 |

(d) For \(k=2\), the complete set of bad starts up to \(10^{30}\) consists of 69 integers:

The deterministic witness digest is 00de7e30ba8627c65062bf7b4d525761446e2beead0f7fcba80486163148ff57.

(a) The final bad block is certified directly by

Every line has at most two distinct prime factors, while \(4378=2\cdot11\cdot199\), so the immediately following start is good.

(d) For \(k=3\), the complete set of bad starts up to \(10^{12}\) consists of 464 integers:

The deterministic witness digest is d667b1603e8b3097e40cb02d6517566a337a1632e46f1ed1904023b0501ecead.

(a) The final five bad starts arise from the 34-term low-\(\omega\) block \(8611,\ldots,8644\). Exact trial division gives:

8611=79*109             8612=2^2*2153
8613=3^3*11*29         8614=2*59*73
8615=5*1723            8616=2^3*3*359
8617=7*1231            8618=2*31*139
8619=3*13^2*17         8620=2^2*5*431
8621=37*233            8622=2*3^2*479
8623=8623              8624=2^4*7^2*11
8625=3*5^3*23          8626=2*19*227
8627=8627              8628=2^2*3*719
8629=8629              8630=2*5*863
8631=3^2*7*137         8632=2^3*13*83
8633=89*97             8634=2*3*1439
8635=5*11*157          8636=2^2*17*127
8637=3*2879            8638=2*7*617
8639=53*163            8640=2^6*3^3*5
8641=8641              8642=2*29*149
8643=3*43*67           8644=2^2*2161

Each has at most three distinct prime factors. The next endpoint is

so the start 8616 is good.

(d; correction to an unverified comment) Thus the live comment's suggested \(k=3\) cutoff “\(n>8611\)” is false: starts 8612, 8613, 8614, and 8615 are also bad. The sharp statement supported by this computation is only that every \(8615<n\leq10^{12}\) is good.

4. What remains, exactly

(a) The finite checks do not settle “for all sufficiently large \(n\)”. A new bad smooth centre beyond \(10^{30}\) for \(k=2\), or beyond \(10^{12}\) for \(k=3\), would not contradict anything proved here.

(a) The exact missing lemma is:

For every fixed \(k\geq2\), for all sufficiently large exponent vectors \(a_i\geq1\), and for every \(0\leq j<P_k\), some \[ > p_1^{a_1}\cdots p_k^{a_k}+d,\qquad -j\leq d<P_k-j, > \] has at least \(k+1\) distinct prime factors.

For \(k=2\), the nearest-witness formulation above is equivalent and especially sharp.

(b; why Pólya does not close it) Pólya's theorem controls the gaps in the integers supported on one fixed finite set of primes. Here each of the \(P_k-1\) neighbours may use a different and arbitrarily large set of at most \(k\) primes. There is no fixed finite support to which Pólya can be applied. This is precisely why Schinzel's extra \(p_{k+1}/p_k\) of interval length is useful and why the cited argument stops short of the primorial.

(c; structural obstruction) Boundary families such as \(R=3\cdot2^a\) or \(R=2\cdot3^b\) lead to simultaneous prime-power/almost-prime conditions on several exponential shifts \(R+d\). For example, when \(a\) is odd, \(\gcd(3,2^a-1)=1\), so if \(R-3=3(2^a-1)\) has at most two distinct prime factors then \(2^a-1\) has at most one; in the nontrivial exponent range Catalan-type reasoning reduces this to a Mersenne-prime-type condition. A bad window needs several such conditions simultaneously. No theorem found in the literature search rules out all exponent values in these systems.

(d; computational wall) Sparse enumeration is cheap—the number of \(k=2\) centres below \(10^D\) is only quadratic in \(D\)—but witness discovery is not uniform. Raising a finite cutoff eventually produces neighbours for which small-prime stripping finds too few coprime components; deciding those cases requires factoring increasingly large exponential shifts. A scan to a larger finite \(X\) could extend the table, but no finite amount of core-hours proves the required all-exponent lemma. Factoring generic 100-digit and larger residuals can range from seconds to core-days per residual depending on their factor structure, and the size is unbounded. The real bottleneck is therefore the uniform Diophantine/sieve lemma above, not the enumeration cost.

PARTIAL: Exact smooth-centre reduction; all k=2 starts through 10^30 and k=3 starts through 10^12 classified, with last bad starts 4372 and 8615 respectively, but no uniform all-large-n lemma.