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Claim. Jeffrey Zeng posted a partial proof claim on the site's proof-claims tab of Problem 769 on 24 July 2026 (claim 132, the first discussion link), and the same day posted the $h(n)$ part of the result on the tab of Problem 770 (claim 133, the second); the tab records the result as obtained using an OpenAI internal model. There is no manuscript: the listing's summary and notes carry the whole argument. The source card zeng_2026_collective_coprimality_threshold records the cross-listed claim 133, with its date, attribution and evidence limits, and the corpus's own-words reconstruction of its $h(n)$ bounds; the step from $h(n)$ to $c(n)$ is not reconstructed there. Let $P(n)$ be the largest prime $p$ with $p-1\mid n$, let $S(n)=\lfloor\sqrt{4n}\rfloor+1$, and let $h(n)$ be the least $M>2$ for which the integers $2^n-1,3^n-1,\ldots,M^n-1$ have no common factor, the gcd taken over the whole list. The claim is that for every $n\ge1$

$$P(n)\le h(n)\le\max\{P(n),S(n)\}, \qquad P(n)>2\sqrt n\ \Longrightarrow\ h(n)=P(n),$$

so that odd $n$, where $P(n)=2$, have $h(n)\le S(n)$. The refinement that replaces one cube by $m^n$ cubes exists for every $m\ge2$ and raises the tile count by $m^n-1$; for odd $n$ the increments $m^n-1$ with $2\le m\le S(n)$ have no common factor, because $h(n)\le S(n)$, and a Frobenius-type bound for the numerical semigroup they generate gives

$$c(n)\le(S(n)-1)\,(2^n-2)\,(S(n)^n-1)+1=o(n^n).$$

The listing concludes that the uniform lower bound $c(n)\gg n^n$ asked about in the problem is false. The argument puts $1,\ldots,M$ into the subgroup of $n$-th roots of unity of $\mathbf F_q^\times$ for a prime $q$ dividing every listed power difference, which has at most $n$ elements; reduced fractions with numerator and denominator at most $M$ rule out $q>M^2$, a signed pigeonhole representation rules out $M<q\le M^2$ for even $n$, and the least quadratic nonresidue does so for odd $n$. The threshold $h(n)$ is the quantity of Problem 770, whose page records the same listing as a partial result for that problem.

Submission note. Posted to erdosproblems.com as a proof claim by Jeffrey Zeng (account jeffzeng) on 24 July 2026, giving "an OpenAI internal model" as the AI used:

AI-assisted, Lean-formalized partial result for #769/#770. All gcds are collective, not pairwise. Let $P(n)=\max\{p\text{ prime}:p-1\mid n\}$ and $S(n)=\lfloor\sqrt{4n}\rfloor+1$. For all $n\ge1$,

$$P(n)\le > h(n)\le\max\{P(n),S(n)\},\quad P(n)>2\sqrt n\Rightarrow h(n)=P(n).$$

For odd

$n$, $h(n)\le S(n)$ and

$$c(n)\le(S(n)-1)(2^n-2)(S(n)^n-1)+1=o(n^n). >$$

Thus #769’s proposed uniform lower bound $c(n)\gg n^n$ is false. This is

only a partial result for #770: its density, liminf, and general $\varepsilon>0$ questions remain unresolved. Novelty and priority have not been determined. AI disclosure: an OpenAI internal model. The central claims have a sorry-free Lean formalization. Notes: Let (G(n,M)=\gcd_{2\le a\le M}(a^n-1)), with $M=\max\{P(n),S(n)\}$. If $q\mid G(n,M)$, then $q>M$, and $H=\{x\in\mathbf F_q^*:x^n=1\}$ has at most $n$ elements. If $q>M^2$, reduced fractions $a/b$, $1\le a,b\le M$, inject into $H$, but their number is at least $M^2/4+M>n$. If $q\le M^2$ and $n$ is even, Dirichlet approximation gives $x=a/k$, $1\le k\le M$, $0<|a|<M$; hence every $x\ne0$ satisfies $x^n=1$, forcing $q-1\mid n$, impossible since $q>P(n)$. If $n$ is odd, $H$ consists of squares and the least quadratic nonresidue is at most $\lceil\sqrt q\rceil$, forcing $q>M^2$. Fermat gives $P(n)\le h(n)$. Cube refinements add $a^n-1$, so the gcd-one numerical-semigroup argument yields the stated cutoff.

Posted to erdosproblems.com as a proof claim by Jeffrey Zeng (account jeffzeng) on 24 July 2026, giving "an OpenAI internal model" as the AI used:

AI-assisted, Lean-formalized partial result for #770. Use the collective-gcd, strict-$M>2$ definition of $h(n)$. Let (P(n)=\max{p\text{ prime}:p-1\mid n}) and $S(n)=\lfloor\sqrt{4n}\rfloor+1$. For every positive $n$,

$$> P(n)\le h(n)\le\max\{P(n),S(n)\},\qquad P(n)>2\sqrt n\Rightarrow h(n)=P(n). >$$

If $n$ is odd, $h(n)\le S(n)$. More sharply, if $p=P(n)>2$ and

$C(p)=\#\{(a,b):1\le a,b\le p,\gcd(a,b)=1\}>n$, then $h(n)=p$, including cases with $p\le2\sqrt n$. These are partial results: the density, liminf, and general $\varepsilon>0$ questions remain unresolved. Novelty and priority have not been determined. AI disclosure: an OpenAI internal model. Notes: Put $G(n,M)=\gcd_{2\le a\le M}(a^n-1)$. If $q\mid G(n,M)$, then $q>M$, and $H=\{x\in\mathbf F_q^*:x^n=1\}$ has at most $n$ elements. For $q>M^2$, reduced fractions $a/b$, $1\le a,b\le M$, inject into $H$. Their count is $C(M)\ge M^2/4+M$, so $M\ge S(n)$ or $C(M)>n$ gives a contradiction. For $q\le M^2$ and even $n$, Dirichlet gives $x=a/k$ with $1\le k\le M$ and $0<|a|<M$, so every $x\ne0$ satisfies $x^n=1$; hence $q-1\mid n$, contradicting $q>P(n)$. For odd $n$, all elements of $H$ are squares and the least quadratic nonresidue is at most (\lceil\sqrt q\rceil), forcing $q>M^2$. Fermat gives $P(n)\le h(n)$. The restriction is necessary: $n=86$ has $P(n)=3$ but $h(n)=5$.

Covers. The bound $c(n)=o(n^n)$ along the odd integers, hence a negative answer to the question whether $c(n)\gg n^n$ holds uniformly in $n$, together with the stated bounds on $h(n)$. It does not give the order of $c(n)$ and says nothing about $c(n)$ for even $n$, in particular nothing about the case $n+1$ prime in which Erdős expected $c(n)>n^n$; the problem's request for good bounds is not settled by it.

Standing. The listing says the result is partial and that its novelty and priority are undetermined; it says the central claims have a Lean formalization without sorry, but links no Lean source, build record or certificate, so no formalization is linked here. The site's label is unchanged, its page was last edited on 1 October 2025 and the tab showed no comment on the entry as of 2026-10-07; no review by anyone outside the claimant is recorded, and the corpus's reconstruction on the source card is author-recorded compilation, not acceptance evidence. The claim is therefore claimed. Samuel Korsky's later listing, on Korsky's claim page, starts from the same subgroup observation and reaches a smaller exponent for odd $n$ by a different route; Korsky's manuscript cites this listing without using it. Star Fleet Math's Lean disproof of the $n^n$ bound, whose Star Fleet Math entry is dated ten days before this listing, is on its claim page; neither is known to cite the other.