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Statement

For every red-blue coloring of the Euclidean plane, either two red points have distance $1$, or there are blue points

$$x,\ x+d,\ x+2d,\ x+3d,\ x+4d, \qquad \|d\|=1.$$

Equivalently, $\mathbb E^2\to(\ell_2,\ell_5)$. There is no measurability or other regularity assumption on the coloring. If $K$ is the least positive integer admitting a coloring that avoids both a red unit pair and a blue unit-step $\ell_K$, then $K\geq6$.

Proof

Suppose for a contradiction that neither configuration occurs. There is a red point $A$, because an entirely blue plane contains a blue $\ell_5$. On the circle of radius $5$ about $A$, choose points $B,C$ with $|BC|=1$. For example, two radii making angle $2\arcsin(1/10)$ give such a chord. Since there is no red unit pair, at least one of these points is blue; call it $B$.

Set $u=(B-A)/5$, and let $v$ be the counterclockwise $60^\circ$ rotation of $u$. The unit triangular lattice

$$L=A+\mathbb Zu+\mathbb Zv$$

contains both $A$ and $B=A+5u$. If $L$ contains a red $T_3$, then Lemma 6 gives a pattern invariant under translation by every vector in $5(\mathbb Zu+\mathbb Zv)$. If it contains no red $T_3$, the same invariance follows from Lemma 7. Their normalizing lattice rotations and translations do not change this period subgroup. In either case translation by $5u$ preserves color, contradicting the fact that $A$ is red and $B$ is blue.

Thus a blue unit-step $\ell_5$ exists whenever no red unit pair does. It contains a blue unit-step $\ell_k$ for every $k\leq5$, so no such $k$ can be an avoiding length. This proves the bound on $K$.

Source and proof scope

Theorem 1 on published p. 2, with its concluding proof on p. 8; Theorem 1.1 in arXiv v2. The two linked lattice lemmas include all earlier same-paper dependencies. The complete chain uses finite forced-color configurations, elementary Euclidean rotations, and lattice arithmetic. No external Ramsey result is an input. Choosing the lattice basis along $AB$ makes the final period argument explicit; no assertion about arbitrary distance-$5$ lattice vectors is needed.

This is a lower bound for the least avoiding length, not a determination of that length. The original question remains separate from this proved theorem.

Bears on. #188.