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Claim. There are no integer $x\ge2$, primes $p,q$ and integers $a,b,y,z$ with either $x^3-1=p^2a^3$ and $x^3+1=q^3z^2$, or $x^3-1=p^3y^2$ and $x^3+1=q^2b^3$. So no triple $(x^3-1,x^3,x^3+1)$ of consecutive powerful numbers has outer members of these two mixed shapes. This is Theorem 1 of B. Y. Sayim, Nonexistence of consecutive powerful triplets around cubes with mixed prime factorizations, a Zenodo preprint under the concept DOI 10.5281/zenodo.20654529, which resolves to the newest version; the version records are linked separately, version 1.0 published 2026-06-12 and version 1.3 dated 2026-09-11 (Creative Commons Attribution 4.0, by the record). The author announced it in the problem's discussion thread and on its proof-claims tab on 2026-10-04 as a partial result. The argument factors $x^3\mp1=(x\mp1)(x^2\pm x+1)$, follows the $3$-adic valuations and the placement of $p$ and $q$ through eighteen branches, closes sixteen of them by the known integral points on five Mordell curves $Y^2=X^3+k$ and the cubic lemmas of Tzanakis and of She, and reduces the two remaining branches to the single equation $t^6+t^3+1=3w^2$, with $t\ge7$ in the first family and $t\le-5$ in the second. Its Theorem 2 finds that this equation has only the rational solutions $(t,w)=(1,\pm1)$: the genus-two curve has the palindromic symmetry $t\mapsto1/t$, and one of its elliptic quotients has Mordell–Weil rank $0$ and trivial torsion (LMFDB curve 1296.k1), which pins down the rational points. The curve computations were done in SageMath and checked against the LMFDB, by the paper's own account. The paper also refines Beckon's congruence constraint on the smallest member of any consecutive powerful triple (Proposition 3) and shows that no finite congruence sieve of that type can settle the conjecture. The statements above are those of version 1.3. The repository linked as code, an identical copy of versions 1.0 and 1.3 by the Zenodo records, holds the LaTeX source and the SymPy and SageMath verification scripts.

Submission note. Posted to erdosproblems.com as a proof claim by Berkay Yüksel Sayim (account Berkay_Yueksel_Sayim) on 4 October 2026, giving "Fable 5 by Claude, inside Claude code. doublechecked with chat gpt another model" as the AI used:

I claim that two mixed forms are impossible for the neighbours of a cube. Powerful: every prime dividing the number divides it at least twice (72 = 2³·3²). For x ≥ 2 I look at x³−1, x³, x³+1, with primes p, q: (M1) x³−1 = p²·a³ and x³+1 = q³·z² (M2) x³−1 = p³·y² and x³+1 = q²·b³ "Mixed": left and right have different shapes. Chan and She excluded equal shapes. Idea: split x³∓1 = (x∓1)(x²±x+1); look at x mod 3 and where p, q sit: 18 cases. 16 are impossible by known results (e.g. v² = u³+2). In the last case of each form x−1 = t³ and x²−x+1 = 3w², so t⁶+t³+1 = 3w². A rank-0 elliptic curve (SageMath) gives only t = 1, but the cases need t ≥ 7 (M1), t ≤ −5 (M2). So no x works. Partial result: other shapes are not covered; #364 is not settled. Notes: Partial result: only these two mixed forms are excluded, other factorizations of the neighbours of a cube are not covered, and #364 is not settled. The key step uses a rank-0 elliptic curve and is computer-assisted (SageMath). I also posted this as a comment in this thread.

Covers. The two shape combinations $(p^2a^3,q^3z^2)$ and $(p^3y^2,q^2b^3)$ for the outer members of a powerful triple centered at a cube, with $p$ and $q$ prime. With Chan's exclusion of $(p^3y^2,q^3z^2)$ (claim page, card) and She's exclusion of $(p^2a^3,q^2b^3)$ (claim page, card), every combination in which each outer member is either a prime squared times a cube or a prime cubed times a square (the four combinations of $\{p^3\square,\,p^2\,\text{cube}\}$) is excluded. It says nothing about triples whose middle member is not a cube, or whose outer members factor otherwise, so the question of Problem 364 stays open, as the claim itself says. The result is a negative answer for the shapes it covers, a partial no to the question whether such triples exist, so its claim value is disproved.

Standing. The claim is pending. The site's proof-claims tab lists it as a partial proof claim, the site's label is unchanged, and the curator has not credited the result, so there is no acceptance evidence. The claim's tools field names the AI system Fable 5 by Claude, used inside Claude Code, with a second check by a ChatGPT model. The paper's section on its use of AI tools says the case analysis and the descent to the rank-zero quotient were produced by the AI system (Claude's Fable 5 model, inside Claude Code) under the author's direction, with verification passes and cross-checks by a model of another company; the disclosure was added in version 1.2 by the record's version notes, replacing the shorter acknowledgment of versions 1.0 and 1.1. No comment had been posted on the claim.