ERDŐS/DAILY

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ERDőS #672 · PARTIAL

Erdős problem #672 — wave 7f

Date of live check: 2026-07-27 (UTC).

Claim labels used below are exactly those requested:

0. Mandatory live-page gate

(d) I accessed the live problem page, its LaTeX-source view, and the four-comment discussion through the Bright Data browser path. Direct-page tracker metadata was not used as a substitute.

(d) The page says its status is open, has 0 claimed proofs, and lists Currently working on this problem: None, Interested in collaborating: None. It was last edited 01 February 2026. Thus the mandatory collision/claimed-proof skip was not triggered.

Verbatim live statement

(d) The following is copied verbatim from the page's LaTeX-source view (only Markdown math delimiters have been added):

> Can the product of an arithmetic progression of positive integers \(n,n+d,\ldots,n+(k-1)d\) of length \(k\geq 4\) (with \((n,d)=1\)) be a perfect power?

Results listed on the live page

These are page reports, not new claims of this run.

\[ (-6)(-1)4\cdot9=6^3, \]

showing why positivity matters.

All four live comments

\[ (-3)(-1)(1)(3)=3^2 \]

and its evident symmetric generalization when negative terms are allowed.

1. Primary-source literature audit

(b) The following theorem ranges were checked in the papers themselves, rather than inferred from search snippets.

1. Győry, Hajdu and Pintér, Perfect powers from products of consecutive terms in arithmetic progression, Compositio Math. 145 (2009), Theorem 1.1, proves the full conjecture for \(3DOI and paper

2. For squares, Theorem B of that same paper synthesizes Hirata-Kohno–Laishram–Shorey–Tijdeman and Tengely: for positive initial term, \(d>1\), and \(b=1\), there is no square for \(5\leq k\leq109\). The \(d=1\) case is Erdős–Selfridge, and \(k=4\) is Euler. The original square paper is An extension of a theorem of Euler, Acta Arith. 129 (2007). Original DOI, Tengely note DOI

3. Hajdu, Tengely and Tijdeman, Cubes in products of terms in arithmetic progression, proves that a primitive positive progression cannot have cubic product when \(2DOI and publisher page

4. Hajdu and Kovács, Almost fifth powers in arithmetic progression, Theorem 1, proves that the product of \(3\leq k\leq54\) consecutive nonzero terms in a primitive progression is never a fifth power. DOI and paper

5. Bennett, Powers from products of \(k\) terms in progression: finiteness for small \(k\), proves that only finitely many integer tuples can solve the equation with \(4\leq k\leq15177\). This is a finiteness result, not a list, effective height bound, or nonexistence theorem. DOI and publisher page

6. Bennett and Siksek, A conjecture of Erdős, supersingular primes and short character sums, Annals of Mathematics 191 (2020), Theorem 2, proves that for an effectively computable sufficiently large \(k\), a nontrivial solution with prime exponent \(\ell\) must satisfy

\[ \ell\leq \exp(10k). \]

Faltings' theorem then gives finiteness for those sufficiently large \(k\), but not an explicit solution list. DOI and Annals paper

(d) There is a transcription discrepancy worth reporting: the main live page's LaTeX says \(\ell>e^{10^k}\), while its newest comment and Bennett–Siksek's actual Theorem 2 use \(\ell>\exp(10k)=e^{10k}\). The primary paper is unambiguous.

(c) Searches by the exact equation, paper titles, citations, and author publication lists found no later primary paper claiming a positive counterexample or a complete proof. This is an honest search miss, not a theorem that no such paper exists. A 2023 paper of Hajdu still summarizes the universal solved range as \(k\leq34\) and Bennett's result as finiteness through \(15177\). 2023 DOI

2. New reduction and certified height exclusion

Write

\[ F_{n,d,k}=\prod_{i=0}^{k-1}(n+id),\qquad M=n+(k-1)d. \]

(a) I use \(d\geq1\), as in every cited formulation. A decreasing positive progression can be reversed, preserving its product and primitivity, to make its common difference positive; the constant \(d=0\) case is implicitly excluded by the nontrivial-progressions formulation and the cited equation.

2.1 Reduction of the exponent

(a) If \(F_{n,d,k}=Y^e\) for some \(e\geq2\), and \(q\mid e\) is prime, then

\[ F_{n,d,k}=\left(Y^{e/q}\right)^q. \]

(b) For \(35\leq k\leq38\), the square, cube, and fifth-power results above rule out \(q=2,3,5\). Hence every counterexample at one of these first four unresolved lengths would give a prime exponent

\[ q\geq7. \]

2.2 Elementary interval-prime lemma

Lemma (a). Suppose \(F_{n,d,k}=Y^q\), where \(q\) is prime. Let \(p\) be prime with

\[ p\leq k\leq2p. \]

If

\[ Mthen \(p\mid d\).

Proof (a). Assume \(p\nmid d\). The indices \(i\) for which \(p\mid n+id\) form one residue class modulo \(p\). An interval of \(k\) indices with \(p\leq k\leq2p\) contains one or two representatives of this class.

If there is one representative, its contribution to \(v_p(F_{n,d,k})\) is at most \(q-2\), because every term is at most \(M

If there are two representatives \(i \[ (n+jd)-(n+id)=pd, \]

which has \(p\)-adic valuation exactly \(1\). Consequently the two terms cannot both be divisible by \(p^2\); one has valuation exactly \(1\), and the other has valuation at most \(q-2\). Thus in either case

\[ 0But \(F_{n,d,k}=Y^q\) requires \(q\mid v_p(F_{n,d,k})\), a contradiction. Therefore \(p\mid d\). \(\square\)

2.3 Four certificates

(a) If the lemma forces distinct primes \(p_1,\ldots,p_r\) to divide \(d\), then

\[ M=n+(k-1)d\geq1+(k-1)\prod_{j=1}^r p_j. \]

(b) Combining this observation with \(q\geq7\) proves the following:

| \(k\) | interval primes \(p\leq k\leq2p\) | their product | certified bound on \(M\) |

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

| 35 | \(19,23,29,31\) | \(392863\) | \(M\geq 13357343\) |

| 36 | \(19,23,29,31\) | \(392863\) | \(M\geq 13750206\) |

| 37 | \(19,23,29,31,37\) | \(14535931\) | \(M\geq 47045881=19^6\) |

| 38 | \(19,23,29,31,37\) | \(14535931\) | \(M\geq 47045881=19^6\) |

Proof of the certificates (a/b).

  • For \(k=35\), assume \(M<13357343\). This is less than \(19^6\), hence less than every \(p^6\) in the row. The lemma with \(q\geq7\) forces all four primes into \(d\), giving

\[ M\geq1+34(392863)=13357343, \]

contradiction.

  • For \(k=36\), the identical argument gives

\[ M\geq1+35(392863)=13750206. \]

  • For \(k=37\), assume \(M<19^6\). All five primes are forced into \(d\), whence

\[ M\geq1+36(14535931)=523293517>19^6, \]

contradiction.

  • For \(k=38\), the corresponding forced lower bound is

\[ 1+37(14535931)=537829448>19^6. \]

Thus the inequalities in the table are uniform in both \(n,d\), not outcomes of sampling.

2.4 Exact size of the excluded finite regimes

(a) For step \(s=k-1\), the number of all positive pairs \((n,d)\) with \(n+sd \[ R_s(B)=D(B-1)-s\frac{D(D+1)}2,\qquad D=\left\lfloor\frac{B-2}{s}\right\rfloor. \]

Möbius inversion gives the exact number of primitive pairs:

\[ C_k(B)= \sum_{g=1}^{\lfloor(B-1)/k\rfloor} \mu(g)\, R_{k-1}\!\left(\left\lfloor\frac{B-1}{g}\right\rfloor+1\right). \]

(d) A from-scratch linear Möbius sieve evaluated this formula. The certificates therefore exclude exactly the following many primitive \((n,d)\) pairs in their respective triangular height regimes:

| \(k\) | \(B\) (all \(M

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

| 35 | 13,357,343 | 1,595,080,842,488 |

| 36 | 13,750,206 | 1,641,995,050,554 |

| 37 | 47,045,881 | 18,687,973,255,143 |

| 38 | 47,045,881 | 18,182,892,896,208 |

The pair counts are computational-only; the exclusion itself is (b) because it is an elementary proof conditional only on the three named exponent theorems.

3. Reproducible checker and run

The standalone checker is runs/erdos672_wave7f_reverify.py. It uses only the Python standard library. It:

  • (a/d) generates the interval primes by trial division and recomputes every product and inequality;
  • (d) exhausts the normalized residues modulo \(p^2\) to verify the one-or-two-multiple structure used in the lemma;
  • (a/d) recomputes the primitive-pair counts by Möbius inversion and cross-checks the formula against literal enumeration on a small instance;
  • (d) factors any supplied progression term-by-term from scratch and takes the gcd of all prime exponents to decide whether its product is a perfect power;
  • (d) uses \(1,25,49\), whose product is \(35^2\), as a positive control.

Core certificate code:

from functools import reduce
from operator import mul

for p in interval_primes(k):
    assert bound <= p ** (q - 1)
    verify_residue_structure(k, p)

prime_product = reduce(mul, interval_primes(k), 1)
forced_height = 1 + (k - 1) * prime_product
assert forced_height >= bound

Commands actually run:

$ python3 runs/erdos672_wave7f_reverify.py
...
k=35 ... certified M >= 13357343 ... primitive_pairs_below=1595080842488
k=36 ... certified M >= 13750206 ... primitive_pairs_below=1641995050554
k=37 ... certified M >= 47045881 ... primitive_pairs_below=18687973255143
k=38 ... certified M >= 47045881 ... primitive_pairs_below=18182892896208
PASS: all independent arithmetic and structural checks succeeded.

$ python3 runs/erdos672_wave7f_reverify.py --box 100 100
Finite box: checked=24348 primitive triples, hits=0
PASS: all independent arithmetic and structural checks succeeded.

The finite box is merely a (d) sanity check; it is not used in the proof.

4. Exact remaining wall

(b) What remains at the first unresolved lengths is now explicit:

  • \(k=35,36,37,38\);
  • a prime exponent \(q\geq7\);
  • largest term at least the corresponding bound in the table.

(a) The interval-prime lemma cannot automatically go farther. At or beyond \(M=p^{q-1}\), two terms divisible by \(p\) may contribute valuations \(1+(q-1)=q\), which is compatible with a \(q\)-th power. Even below that threshold, forcing the four primes \(19,23,29,31\) into \(d\) gives only \(d\geq392863\), which is compatible with \(M\geq13357343\); the divisibility argument has spent all of its contradiction.

(c) The missing lemma is therefore a way to exclude the surviving \(q=7\) valuation/kernel patterns (or to force additional independent primes into \(d\)). In the standard approach one writes

\[ n+id=a_i x_i^q \]

with the \(a_i\) supported on primes below \(k\), then must eliminate the resulting coupled ternary generalized-Fermat equations. Győry–Hajdu–Pintér report that their \(k=32,33,34\) cases already took about a week each after substantial Magma/Maple sieving; extending that system to \(k=35\) is the concrete missing computation/theorem, not a routine brute-force loop.

(c/d) For scale, between the new \(k=35\) bound and \(19^6\) there are exactly

\[ 18,192,184,956,623 \]

primitive parameter pairs. Even an optimistic direct tester sustaining \(10^6\) fully factored progressions per second would require about \(5053.38\) core-hours (roughly \$200–\$500 at \$0.04–\$0.10/core-hour, before overhead). The supplied pure-Python checker is much slower than that optimistic rate. Such a scan was not run; a modular/ternary sieve is required.

(c) No counterexample, uniform proof, or effective enumeration of Bennett's finite set was obtained. The global problem remains open.

PARTIAL: Modulo verified square/cube/fifth-power theorems, any counterexample at the first unresolved lengths k=35..38 has prime exponent at least 7 and largest term at least 13,357,343; 13,750,206; 47,045,881; 47,045,881 respectively, with all arithmetic and exact excluded-pair counts independently checked.

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