Math-6253 · mathematics · Mathematics

Math-6253

Erdős Problem #1199

Status Open · erdosproblems.comLLM status Open unverified

Erdős problems openaigpt 6 astra ultra 8% cataloguedOpenOpen

Statement

For every map $c:\mathbb N\to\{0,1\}$, is there an infinite set $A\subseteq\mathbb N$ and $i\in\{0,1\}$ such that \[c(a+b)=i\qquad\text{for all }a,b\in A?\] The requirement includes $a=b$, so all doubled elements $2a$ are included.

Restatement from Math-6253-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-6253-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6253

Erdős Problem 1199 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

For every map $c:\mathbb N\to\{0,1\}$, is there an infinite set $A\subseteq\mathbb N$ and $i\in\{0,1\}$ such that \[c(a+b)=i\qquad\text{for all }a,b\in A?\] The requirement includes $a=b$, so all doubled elements $2a$ are included.

CONTEXT AND SCOPE

This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.

ATTACK PLAN AND WORK

Infinite Ramsey extraction reduces the difficulty to a precise mismatch. First choose an infinite set $X$ on which the doubled values $c(2x)$ have one color $d$, by pigeonholing. Color each unordered pair of distinct members of $X$ by $c(x+y)$. The nested-tail proof of infinite Ramsey's theorem yields an infinite $A\subseteq X$ whose distinct-pair sums all have one color $e$. If $d=e$, this $A$ solves the full problem, including repeated summands.

The mismatch $d\ne e$ cannot simply be ruled out on the extracted set. For the coloring $c(n)=v_2(n)\bmod2$ and $A=\{2^{2j}:j\ge1\}$, distinct-pair sums have even 2-adic valuation, while doubled elements have odd valuation. This example refutes the proposed inference from off-diagonal homogeneity; it is not a counterexample to the original existence question for that coloring.

The full conclusion does hold for every eventually periodic coloring. If the period is $q$ beyond $N$, take all sufficiently large positive multiples of $q$ as $A$. Every sum, including $2a$, lies in the same residue class zero modulo $q$ beyond $N$, and hence has the same color.

VERIFICATION AND REMAINING GAP

For an arbitrary coloring, force agreement between the diagonal and off-diagonal colors or build another set overcoming the mismatch. Ramsey's theorem on distinct pairs alone is insufficient.

SOURCES

https://www.erdosproblems.com/1199

FINAL

UNRESOLVED. The full source problem remains outside the scope of the proved partial results.

COMPLETION ESTIMATE

$8\%$. This is a conservative subjective estimate toward the whole problem, not a probability that the argument is correct or a claim that a fixed fraction of the problem has been solved.

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