Erdős Problem #605

Erdős Problem #605: many equal distances on a sphere

Explicit incidence construction, 24 September 2026.

Statement

We construct, for every sufficiently large $n$, $n$ distinct points on the unit sphere in $\mathbb R^3$ with $\gg n^{4/3}$ pairs at distance $\sqrt2$. This supplies the requested $f(n)\to\infty$, with $f(n)$ a positive constant times $n^{1/3}$. The argument reconstructs the point--line incidence mechanism behind the known Erdős--Hickerson--Pach result.

A planar incidence system

Fix an integer $k\geq1$. Take \[P=\{(x,y):1\leq x\leq k, 1\leq y\leq2k^2, x,y\in\mathbb Z\} \] and the lines \[\mathcal L=\{\ell_{m,b}:y=mx+b, 1\leq m\leq k, 1\leq b\leq k^2\}. \] There are $2k^3$ points and $k^3$ distinct lines. Every line contains exactly $k$ of the listed points: for each integer $x\in[1,k]$, its value $mx+b$ lies between $2$ and $2k^2$, and no other horizontal coordinates occur in $P$. Thus there are exactly $k^4$ incidences between these two families.

Mapping both families to the same sphere

To $(x,y)\in P$ assign \[u_{x,y}=\frac{(x,y,1)}{\sqrt{x^2+y^2+1}}, \] and to $\ell_{m,b}$ assign \[v_{m,b}=\frac{(m,-1,b)}{\sqrt{m^2+1+b^2}}. \] Both are unit vectors. The first assignment is injective because the ratios of its first two coordinates to its third recover $(x,y)$. The second is injective because division by the negative second coordinate recovers $m,b$. Their images are disjoint: all $u$ vectors have positive second coordinate and all $v$ vectors have negative second coordinate. Hence we have exactly $3k^3$ distinct points on one unit sphere.

Their cross dot product is \[u_{x,y}\cdot v_{m,b} =\frac{mx-y+b}{\sqrt{x^2+y^2+1}\sqrt{m^2+1+b^2}}. \] It vanishes exactly when the point lies on the line. Two unit vectors have squared distance $2-2u\cdot v$, so every one of the $k^4$ incidences gives a pair at distance $\sqrt2$. These unordered pairs are all different, because one endpoint lies in each of two disjoint injectively parameterized families. Any further equal distances inside either family can only increase the count.

Every sufficiently large cardinality

For $n\geq24$, set $k=\lfloor(n/3)^{1/3}\rfloor$. Then $3k^3\leq n$ and $k\geq\tfrac12(n/3)^{1/3}$. Add arbitrary distinct sphere points outside the finite constructed set until the total is exactly $n$. The original pairs remain, yielding \[\#\{\{u,v\}:|u-v|=\sqrt2\} \geq k^4\geq\frac{n^{4/3}}{16\,3^{4/3}}. \] Thus $f(n)=n^{1/3}/(16\,3^{4/3})$ works for large $n$, and its initial values may be chosen smaller if needed. If unit distance is preferred, scale the sphere by $1/\sqrt2$; its radius becomes $1/\sqrt2$. This explicitly fixes radius versus diameter conventions.

Outcome and attribution

Attempt status: solved. The construction, injectivity, fixed common distance and all-$n$ passage are complete, without an external incidence theorem. It proves the existence question for one common distance; the stronger known assertion for every prescribed distance is not asserted here. Sources: https://www.erdosproblems.com/605; P. Erdős, D. Hickerson and J. Pach, A problem of Leo Moser about repeated distances on the sphere, Theorem 2(ii), https://users.renyi.hu/~p_erdos/1989-02.pdf. No novelty or formal verification is claimed.

COMPLETION ESTIMATE

COMPLETION: 100%