OPEN
This is open, and cannot be resolved with a finite computation.
- $500
Let $f(n)$ be maximal such that there exists a set $A$ of $n$ points in $\mathbb{R}^2$ in which every $x\in A$ has at least $f(n)$ points in $A$ equidistant from $x$.
Is it true that $f(n)\leq n^{o(1)}$? Or even $f(n) < n^{O(1/\log\log n)}$?
This is a stronger form of the unit distance conjecture (see
[90]).
The set of lattice points imply $f(n) > n^{c/\log\log n}$ for some constant $c>0$. Erdős offered \$500 for a proof that $f(n) \leq n^{o(1)}$ but only \$100 for a counterexample. This latter prize is downgraded to \$50 in
[ErFi97].
It is trivial that $f(n) \ll n^{1/2}$. A result of Pach and Sharir (Theorem 4 of
[PaSh92]) implies $f(n) \ll n^{2/5}$. Hunter has observed that the circle-point incidence bound of Janzer, Janzer, Methuku, and Tardos
[JJMT24] implies\[f(n) \ll n^{4/11}.\]Fishburn (personal communication to Erdős, later published in
[ErFi97]) proved that $6$ is the smallest $n$ such that $f(n)=3$ and $8$ is the smallest $n$ such that $f(n)=4$, and suggested that the lattice points may not be best example.
See also
[754].
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This page was last edited 28 December 2025.
Additional thanks to: Zach Hunter and Desmond Weisenberg
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 #92, https://www.erdosproblems.com/92, accessed 2026-02-13
