GPT 6 Astra Ultra attempt on Math-5390

Erdős Problem 46: monochromatic unit fractions and rational targets

1) Problem and the theorem being used

The question asks whether every finite colouring of the positive integers has distinct monochromatic denominators $2\leq n_1<\cdots<n_t$ with \[ \frac1{n_1}+\cdots+\frac1{n_t}=1. \] The problem page checked on 23 September 2026 records the affirmative answer due to Croot. His finite theorem states that some absolute $b>1$ has the following property: every $r$-colouring of $\{2,\ldots,\lfloor b^r\rfloor\}$ contains a monochromatic finite set $S$ whose reciprocal sum is $1$.

This note takes that precisely stated theorem as an external input. It proves the passage to an arbitrary colouring, the infinitely many disjoint representations, and the rational-target strengthening with the colour and distinctness quantifiers checked. The substantial analytic proof in Croot's paper is not repeated, and no new solution or independent formal verification is claimed.

2) The original assertion

Let $\chi:\{2,3,\ldots\}\to\{1,\ldots,r\}$ be the given colouring. Its restriction to Croot's finite interval supplies the required set $S$. Ordering its members increasingly gives the denominators in the question. As $S$ is a set, every denominator appears once. It contains at least two elements because $1/n<1$ for every $n\geq2$.

3) Moving every denominator beyond any fixed cutoff

Fix an integer $T\geq2$. Give the integers $2,\ldots,T$ individual fresh colours, all different from one another and from the original $r$ colours. Retain $\chi$ on the integers larger than $T$. The resulting colouring has $r+T-1$ colours, so Croot's theorem applies again. A monochromatic reciprocal sum equal to $1$ cannot use a fresh colour: its entire colour class is a singleton $\{n\}$ with $n\geq2$. Therefore the resulting unit representation lies entirely beyond $T$ and is monochromatic for the original colouring.

Beginning with one such representation, choose the next cutoff larger than all denominators used so far, and repeat. This produces infinitely many pairwise disjoint finite sets $S_1,S_2,\ldots$, each of reciprocal sum $1$, each monochromatic. Some fixed original colour occurs for infinitely many of them, by the infinite pigeonhole principle. Thus there is one colour containing infinitely many pairwise disjoint unit representations. This last pigeonhole step is needed before combining several representations into a larger rational target.

4) Every positive rational has a monochromatic representation

Let $a,b$ be positive integers. Colour $m\geq2$ by \[ \chi_b(m)=\chi(bm). \] The preceding argument applied to $\chi_b$ gives infinitely many pairwise disjoint unit representations in a common induced colour. Choose any $a$ of their denominator sets, say $S_1,\ldots,S_a$. For each of them, \[ \sum_{m\in S_j}\frac1{bm}=\frac1b. \] The union \[ U=\bigcup_{j=1}^a bS_j \] therefore consists of distinct denominators, all of the same original colour, and \[ \sum_{n\in U}\frac1n=\frac ab. \] Multiplication by $b$ is injective, and the sets $S_j$ are disjoint, so denominators from different representations cannot collide. Choosing the $S_j$ far enough along the construction makes every denominator in $U$ exceed any prescribed cutoff.

In fact, for each fixed $b$ the same induced colour works for every positive numerator $a$, because the infinite collection of unit representations was chosen before specifying how many to combine. We do not assert that the colour can be fixed independently of $b$.

5) Checks and outcome

Simply scaling a single unit representation by $b$ gives $1/b$, not $a/b$. Taking $a$ copies with the same denominators would violate distinctness, and choosing $a$ arbitrary monochromatic solutions would not guarantee a common colour. The disjoint infinite family and subsequent pigeonhole selection address both issues.

Outcome: SOLVED, as a reconstruction using Croot's theorem. The original assertion and the stated disjointness and rational-target consequences are fully deduced. The finite theorem remains the explicitly cited external mathematical dependency.

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 46, https://www.erdosproblems.com/46, checked 23 September 2026. This page records both the disjoint representations and the rational extension, which the above argument spells out.

The estimate measures coverage using the named published theorem, not the originality of the result or an independent reproof of Croot's work.

7) COMPLETION ESTIMATE

COMPLETION: 100%