GPT 6 Astra Ultra attempt on Math-5504

FORMAL RESTATEMENT

Erdős #214; finite geometric deduction from Juhász's lemma, 2026-09-23. If $S\subset\mathbb R^2$ contains no pair of points at distance one, must its complement contain the four vertices of a unit square?

QUICK LITERATURE AND CONTEXT CHECK

The answer is affirmative. Juhász proved a stronger theorem about arbitrary four-point configurations in a two-colouring with a forbidden distance in one colour. Here the only non-elementary input is his preliminary monochromatic rhombus lemma. The subsequent deduction for the unit square is fully reconstructed in vector coordinates.

OBJECTIVE AND ATTACK PLAN

Colour $S$ blue and its complement red. A blue point forces every point at distance one from it to be red. Starting from a red equilateral rhombus, apply this rule to two candidate squares and their translates.

WORK

Declared external input (Juhász, Lemma 3). In a red--blue colouring of the plane with no blue pair at distance $t>0$, there is a red rhombus of side $t$ whose acute angles are $60$ degrees. The lemma has no measurability or other regularity hypothesis. Its proof is not reproduced here.

Apply the lemma with $t=1$. After translation, the four red vertices can be written \[ 0,\quad u,\quad v,\quad u+v, \qquad |u|=|v|=|u-v|=1. \tag{1} \] Choose a unit vector $w$ perpendicular to $u$. The points $0,u,w,u+w$ form a unit square.

If $w$ and $u+w$ are both red, we are done. If $u+w$ is blue, then $v+w$ and $u+v+w$ are red, since their distances from $u+w$ are respectively $|u-v|=1$ and $|v|=1$. Together with the already red $v,u+v$, they form the translated square \[ \{v,u+v,v+w,u+v+w\}. \tag{2} \]

It remains to consider $w$ blue and $u+w$ red. Let $R$ be the rotation through angle $60$ or $-60$ degrees that carries $u$ to $v$, and set $w'=Rw$. Then $w'$ is a unit vector perpendicular to $v$, and \[ |w-w'|=2\sin(\pi/6)|w|=1. \] Hence $w'$ is red. If $v+w'$ is red, then $\{0,v,w',v+w'\}$ is a red unit square. If $v+w'$ is blue, the two points $u+w'$ and $u+v+w'$ are red: their distances from it are $|u-v|=1$ and $|u|=1$. The four points \[ \{u,u+v,u+w',u+v+w'\} \tag{3} \] then form a red unit square. These cases exhaust the possible colours.

VERIFICATION AND SCOPE

Every forced colour follows from an explicitly checked unit distance. Every displayed square has two perpendicular unit sides, so none is a degenerate parallelogram. No assumption about interiors, measurability, density, or closedness of $S$ is used. The result is complete relative to Juhász's rhombus lemma; this file does not claim to replace its proof or to establish the full arbitrary-four-point theorem independently.

SOURCES

• Current problem: https://www.erdosproblems.com/214.
• R. Juhász, Ramsey type theorems in the plane, Journal of Combinatorial Theory, Series A 27 $1979$, 152--160, Lemma 3 and Theorem 1: https://www.cs.umd.edu/~gasarch/TOPICS/ERT/JuhERT.pdf.

CONCLUSION

Attempt status: solved using a named external lemma. The complement necessarily contains a unit square. The geometric deduction reconstructs the cited method, with no novelty claim. Completion rate estimate: 100%.