PROVED
This has been solved in the affirmative.
Let $x_1,\ldots,x_n\in \mathbb{R}^2$ and let $R(x_i)=\#\{ \lvert x_j-x_i\rvert : j\neq i\}$, where the points are ordered such that\[R(x_1)\leq \cdots \leq R(x_n).\]Let $\alpha_k$ be minimal such that, for all large enough $n$, there exists a set of $n$ points with $R(x_k)<\alpha_kn^{1/2}$. Is it true that $\alpha_k\to \infty$ as $k\to \infty$?
It is trivial that $R(x_1)=1$ is possible, and that $R(x_2) \ll n^{1/2}$ is also possible, but we always have\[R(x_1)R(x_2)\gg n.\]Erdős originally conjectured that $R(x_3)/n^{1/2}\to \infty$ as $n\to \infty$, but Elekes proved that for every $k$ and $n$ sufficiently large there exists some set of $n$ points with $R(x_k)\ll_k n^{1/2}$.
Mathialagan
[Ma21] proved that given a set $P$ of $k$ points and a set $Q$ of $n$ points, with $2\leq k\leq n^{1/3}$, there exists a point in $P$ which determines $\gg (kn)^{1/2}$ distances to points in $Q$. This immediately implies $R(x_k)\gg (kn)^{1/2}$ for $2\leq k\leq n^{1/3}$.
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This page was last edited 18 January 2026. View history
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Additional thanks to: Bois Alexeev and Terence Tao
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #652, https://www.erdosproblems.com/652, accessed 2026-04-11
