GPT 6 Astra Ultra attempt on Math-5389

Erdős Problem 45: a divisor construction and its growth scale

1) Problem and known input

For every $k\geq2$, does there exist an integer $n_k$ such that every $k$-colouring of \[ D(n_k)=\{d:1<d<n_k,\ d\mid n_k\} \] contains a monochromatic subset $S$ with $\sum_{d\in S}1/d=1$? The current problem page, checked on 23 September 2026, records an affirmative answer from Croot's unit-fraction colouring theorem.

The precise external input is the following finite statement from Croot's 2003 paper: there is an absolute constant $b>1$ such that every $k$-colouring of the integers in $[2,b^k]$ has a monochromatic set of distinct denominators whose reciprocals sum to $1$. The analytic proof of that theorem is not reproduced here. We give the full transfer to proper divisors and a matching growth scale, with each further input identified. This is a reconstruction of known results, not a claim of a new solution.

2) Explicit transfer from an interval to proper divisors

Choose an integer $L\geq\max(3,b^k)$ and put \[ n_k=\operatorname{lcm}(1,2,\ldots,L). \] Every integer from $2$ through $L$ divides $n_k$. These divisors really are proper: since $L$ and $L-1$ are coprime, \[ n_k\geq L(L-1)>L. \] Restrict any given colouring of $D(n_k)$ to $\{2,\ldots,L\}$. Croot's theorem supplies a monochromatic $S\subseteq\{2,\ldots,L\}$ with reciprocal sum $1$. Because this entire interval lies in $D(n_k)$, this is exactly the subset required in the problem. No repeated denominator or the excluded divisor $1$ has been used.

Even the elementary estimate $n_k\leq L!\leq L^L$ gives a doubly exponential bound. With $L=\lceil\max(3,b^k)\rceil$, \[ \log n_k\leq L\log L\leq C^k \] for a suitable absolute $C>1$ and all $k\geq2$, enlarging $C$ to cover the finitely many initial cases. Thus $n_k\leq\exp(C^k)$. One can sharpen the estimate on this particular least common multiple, but that is unnecessary for the stated growth scale.

3) A rigorous greedy obstruction for small reciprocal mass

Here is a useful sufficient condition for a colouring with no solution. If a finite set of denominators all at least $2$ has total reciprocal mass less than $k/2$, assign its elements successively to $k$ colour classes, keeping the sum in each class strictly below $1$.

To justify that the assignment never gets stuck, let the next weight be $w=1/d\leq1/2$. If every class rejected it, each existing class would have mass at least $1-w\geq1/2$. The already assigned mass would then be at least $k/2$, contradicting the assumed total mass. Every class consequently has mass less than $1$, and no subset of one class can have mass exactly $1$.

Applying this to $D(n)$ shows that any integer having the required Ramsey property must satisfy \[ \sum_{d\in D(n)}\frac1d\geq\frac k2. \] This deliberately coarse necessary bound is enough below. We do not need a sharper packing assertion with threshold $k$.

4) A doubly exponential necessary size

We use the classical Mertens product estimate \[ \prod_{p\leq y}(1-1/p)^{-1}\ll\log y\qquad(y\geq3) \] as a second external theorem. For $n$ sufficiently large, put $y=\log n$. Factorization of the divisor sum gives \[ \sum_{d\mid n}\frac1d =\prod_{p^a\parallel n}(1+p^{-1}+\cdots+p^{-a}) \leq\prod_{p\mid n}(1-1/p)^{-1}. \] The contribution of $p\leq y$ is $O(\log y)$ by Mertens. If there are $r$ distinct prime factors exceeding $y$, then $y^r<n$, so $r\leq\log n/\log y$. Since $-\log(1-1/p)\leq2/p$, \[ \log\prod_{\substack{p\mid n\\p>y}}(1-1/p)^{-1} \leq\frac{2r}{y} \leq\frac{2}{\log\log n}. \] The large-prime contribution is bounded by an absolute constant. Consequently \[ \sum_{d\mid n}\frac1d\leq C_0\log\log n \] for all sufficiently large $n$, with an absolute $C_0$. Combining this with the necessary reciprocal mass gives \[ \log\log n_k\geq\frac{k}{2C_0},\qquad n_k\geq\exp\left(\exp(c k)\right) \] for an absolute $c>0$ and sufficiently large $k$. Together with the construction, this identifies the doubly exponential scale. It does not identify the least possible integer $n_k$ or the optimal constants in the exponent.

5) Verification and outcome

The proper-divisor check $n_k>L$ prevents an endpoint error in the interval transfer. The lower bound uses only positive weights, so a class of mass less than $1$ cannot contain a hidden exact solution. Both uses of asymptotic constants are uniform in $k$; Croot's $b$ and Mertens' constant are absolute.

Outcome: SOLVED using Croot's finite colouring theorem. The Ramsey transfer is complete. The additional lower bound uses the explicitly stated Mertens estimate. Neither analytic theorem is asserted to have been independently reproved or formally checked here.

6) Sources

• E. S. Croot III, On a coloring conjecture about unit fractions, Annals of Mathematics 157 $2003$, 545--556, introductory corollary, https://annals.math.princeton.edu/wp-content/uploads/annals-v157-n2-p04.pdf.
• T. F. Bloom, Problem 45, https://www.erdosproblems.com/45, checked 23 September 2026. The page credits the lower-bound observation to Mehtaab Sawhney; the argument above uses a weaker elementary packing threshold which already gives the same scale.
• Mertens' product estimate is also stated as equation $2$ in T. F. Bloom, On a density conjecture about unit fractions, https://arxiv.org/html/2112.03726v2.

The estimate records full coverage with the named external theorems accepted; it is not a novelty claim.

7) COMPLETION ESTIMATE

COMPLETION: 100%