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Claim. Samuel Korsky, Improved Bounds for Cubical Decompositions, manuscript dated 5 August 2026, posted the same day as a partial proof claim on the site's proof-claims tab of Problem 769; the tab records the result as obtained using GPT-5.6 Pro, and the manuscript's acknowledgments say that the author contributed the initial discussion of quadratic nonresidues in the upper bound and that almost all of the remaining work, including the proofs and the writing, was carried out by GPT-5.6 Pro in an extended interaction, the author reviewing the manuscript and accepting responsibility for it. Let $P(n)$ be the largest prime $p$ with $p-1\mid n$ and let $\theta=1/(4\sqrt e)$. Theorem 1.1 states that for every $\varepsilon>0$ there is $C_\varepsilon$ with

$$c(n)\le\bigl(C_\varepsilon\max\{P(n),\,n^{\theta+\varepsilon}\}\bigr)^n$$

for all sufficiently large $n$, so that $c(n)\le n^{(\theta+\varepsilon)n}$ for large odd $n$, where $P(n)=2$; and that under the generalized Riemann hypothesis for Dirichlet $L$-functions there is an absolute $C_0$ with $c(n)\le(C_0\max\{P(n),(\log n)^2\})^n$ for large $n$, so $c(n)\le(C_0\log^2n)^n$ for large odd $n$. Theorem 1.2 states that there is an absolute $C_1$ such that for all large $n$ and every odd prime $p$ with $p-1\mid n$, a tiling of the unit $n$-cube by $N\not\equiv1\pmod p$ cubes has

$$N\ge\Bigl(\Bigl(1-\frac1{\log_23}\Bigr)n-\log_2n-C_1\Bigr)2^n,$$

and hence that this lower bound holds for $c(n)$ when $n$ is even (take $p=3$). The upper bound reduces, through a numerical-semigroup lemma, to finding a short range $2\le m\le M$ of grid sizes for which the increments $m^n-1$ have no common prime factor; a common prime $q$ puts $1,\ldots,M$ into the subgroup of $n$-th roots of unity of $\mathbf F_q^\times$, a nonresidue estimate of Pollack at the Burgess scale (or the conditional estimate of Lamzouri, Li and Soundararajan) supplies a prime outside a proper such subgroup, and products of a fixed number of small primes then give more than $n$ distinct elements of it. The lower bound doubles a tiling and deletes the unit cubes: when every side length is a $p$-adic unit the tile count is $1$ modulo $p$, so some tile has a side of nonzero $p$-adic valuation, and an entropy estimate against a lower-dimensional reduction gives the constant $1-1/\log_23$. The manuscript's notes on the tab record the consequence for the threshold $h(n)$ of Problem 770: for large $n$, $P(n)\le h(n)\le\max\{P(n),n^{\theta+\varepsilon}\}$, and under GRH $h(n)\le\max\{P(n),C(\log n)^2\}$.

Submission note. Posted to erdosproblems.com as a proof claim by Samuel Korsky (account SamKorsky) on 5 August 2026, giving "GPT-5.6 Pro" as the AI used:

We prove for odd $n$ that for every $\varepsilon>0

$$\$ c(n)\le > \bigl(C_\varepsilon n^{1/(4\sqrt e) + \varepsilon}\bigr)^n,$$

with

$n^{1/(4\sqrt e)+\varepsilon}$ replaced by $C(\log n)^2$ under GRH. For even $n$, we prove that

$$c(n)\ge\left(\left(1-\frac1{\log_2 3}\right)n-\log_2 > n-O(1)\right)2^n.$$

The upper bounds combine a numerical-semigroup reduction

with character-nonresidue estimates and a product amplification inside the subgroup ${x^n=1}\subset\mathbb F_q^\times$. The lower bound uses a $p$-adic congruence obstruction preserved by doubling and deleting unit cubes. Much of the work for the upper bound and all of the work for the lower bound was performed by GPT - I think the lower bound argument is especially interesting! Notes: There is application to Erdős Problem 770; if $P(n)=\max{p\text{ prime},,p-1\mid n}$ and

$$h(n)=\min\left\{M>2:\gcd_{2\le m\le > M}(m^n-1)=1\right\},$$

the argument gives, for every $\varepsilon>0$ and all

sufficiently large $n$,

$$P(n)\le h(n)\le\max\{P(n),n^{1/(4\sqrt > e)+\varepsilon}\},$$

and under GRH,

$$h(n)\le\max\{P(n),C(\log n)^2\}. >$$

Consequently $h(n)=P(n)$ whenever $P(n)$ exceeds the corresponding scale;

this answers the large-$P(n)$ part of Problem 770 unconditionally above the Burgess exponent and, under GRH, for every fixed positive-power threshold. The density and liminf questions remain open.

Covers. The upper bound on $c(n)$ for all large $n$ in terms of $P(n)$ and the Burgess exponent, which gives $c(n)=o(n^n)$ along the odd integers and so a negative answer to the uniform question $c(n)\gg n^n$; the lower bound of order $n2^n$ for even $n$, above the bound $2^{n+1}-1$ of Connor and Marmorino; and, conditionally on the generalized Riemann hypothesis, the polylogarithmic form of the upper bound. The order of $c(n)$ is not determined, and the case $n+1$ prime, in which $P(n)=n+1$ and the stated upper bound has the shape $(Cn)^n$, is not improved.

Standing. The manuscript is unpublished and unrefereed; the site's label is unchanged, its page was last edited on 1 October 2025, and no reviewer was recorded as of 2026-10-07. The one comment on the entry is the author's own, of 21 September 2026, announcing further results obtained with the AI system it names as Astra, a lower bound $c(n)\ge n2^n$ for all large $n$ and an unconditional upper bound $((4/e^2+o(1))\log^2n)^n$ for almost all $n$, with details to follow; a thread comment is not a dated manuscript and gets no page, and nothing of it is adopted here. The corpus has not checked the proofs. The claim is therefore claimed. The earlier listing of Jeffrey Zeng, on Zeng's claim page, gives the weaker exponent $1/2$ for odd $n$ by an elementary route; the manuscript cites it and does not use it.