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Math-6262

Erdős Problem #1208

Status Open · erdosproblems.comLLM status Open unverified

Erdős problems openaigpt 6 astra ultra 10% cataloguedOpenOpen

Statement

For fixed $d\geq2$, determine the largest size of a subset with all pairwise distances different that can be guaranteed in every $n$-point set in $\mathbb R^d$. Equivalently, for the intended extremal quantity, \[F_d(n)=\min_{\substack{X\subset\mathbb R^d\\|X|=n}}\ \max\{|Y|:Y\subseteq X,\ \text{distinct unordered pairs in }Y\text{ have distinct distances}\},\] estimate $F_d(n)$ as $n\to\infty$.

Restatement from Math-6262-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-6262-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6262

Erdős Problem 1208 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

For fixed $d\geq2$, determine the largest size of a subset with all pairwise distances different that can be guaranteed in every $n$-point set in $\mathbb R^d$. Equivalently, for the intended extremal quantity, \[F_d(n)=\min_{\substack{X\subset\mathbb R^d\\|X|=n}}\ \max\{|Y|:Y\subseteq X,\ \text{distinct unordered pairs in }Y\text{ have distinct distances}\},\] estimate $F_d(n)$ as $n\to\infty$.

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.

The official statement says minimal guaranteed size, which literally makes the definition trivial. The min-max formulation explicitly states the intended maximum guaranteed size, as supported by all the nontrivial bounds in its remarks. This is an interpretation of that typo, not a silently exact transcription.

ATTACK PLAN AND WORK

Here is an elementary, nonsharp lower bound for the intended planar quantity: \[F_2(n)\gg n^{1/6}.\tag{1}\] For a selected set $S$ of size $m$ with all distances different, forbidden new points lie on circles centered at $a\in S$ with radii equal to an existing distance $|b-c|$, or on perpendicular bisectors of pairs in $S$. There are at most $m\binom m2+\binom m2\le m^3$ such curves, in addition to the $m$ already selected points. These cover all collisions between an old and a new distance and between two new distances.

Put $L=\lceil\sqrt n\rceil$. If no line or circle contains at least $L$ input points, a maximal selected set has $n\le m+m^3L\le2m^3L$, giving $m\gg n^{1/6}$. Otherwise restrict to $L$ points on one such curve $C$. Each forbidden curve intersects $C$ in at most two points. The possible coincident-curve cases do not occur: a circle centered at a selected point on $C$ cannot be $C$ itself; and if $C$ is a line, the perpendicular bisector of two selected points on it is not that line. A maximal selection inside $C$ consequently satisfies $L\le m+2m^3\le3m^3$, again giving $m\gg n^{1/6}$.

This dichotomy proves $1$ without a general incidence theorem. It applies to distinct unordered pairs and excludes repeated distances even when the two pairs share a vertex. It is a weak baseline and is not presented as a best known bound.

VERIFICATION AND REMAINING GAP

Improve the exponent and determine the sharp growth, and address dimensions above two. Intersections of higher-dimensional spheres are not bounded finite sets, so the planar proof does not automatically generalize.

SOURCES

https://www.erdosproblems.com/1208

FINAL

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

COMPLETION ESTIMATE

$10\%$. 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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