# A Congruence Condition For The Four-Distance Problem

## TL;DR

This paper establishes a 3-adic valuation condition for rational points equidistant from rectangle vertices, providing partial constraints on the long-standing four-distance problem related to the unit square.

## Contribution

It introduces a congruence condition involving 3-adic valuation that restricts potential solutions to the four-distance problem, advancing understanding of this open question.

## Key findings

- If $v_3(a)=0$, then rational distance points must have either $v_3(x)<0$ or $v_3(y)<0$.
- The result excludes one-fourth of potential solutions with bounded height for the four-distance problem.
- Provides a new algebraic constraint relevant to the existence of rational solutions in the four-distance problem.

## Abstract

Place the vertices of a rectangle at $\{(0, \pm 1/2), (a, \pm 1/2)\}$, where $a$ is rational. We show that if $v_3(a) = 0$, then any point $(x,y)$ that is rational distance from all four vertices of the rectangle has either $v_3(x) < 0$ or $v_3(y)<0$, where $v_3(\cdot)$ is the 3-adic valuation. The case of particular interest is the long-open four-distance problem, which asks whether such a rational distance point exists in the case $a=1$ of the unit square. For the four-distance problem, our result rules out one-fourth of all potential solutions with bounded height.