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Statement

Suppose the plane has no red unit-distance pair and no blue $\ell_5$. If a unit triangular lattice $L$ contains a red $T_3$, then, up to translation and rotation by a multiple of $60^\circ$, its red points have triangular coordinates

$$S+5\mathbb Z^2,\qquad S=\{(0,0),(1,1),(2,2),(3,0),(4,-2),(2,-1)\}. \tag{1}$$

Every other lattice point is blue. In particular, translation by five times either primitive unit basis vector preserves every color. This describes the restriction of the hypothetical plane coloring; it is not a construction of such a coloring of the plane.

Propagating one six-point configuration

Use the triangular coordinates and configurations. By Lemma 5, the lattice contains a red $T_6$. Normalize it to $S$ and label

$$A=(0,0),\ B=(1,1),\ C=(2,2),\ D=(3,0),\ E=(4,-2),\ F=(2,-1).$$

We first force its translate by $(5,0)$. Points $I=(3,-3)$ and $J=(5,-1)$ are blue by Lemma 3: the triangles $A,D,I$ and $C,F,J$ have side $3$ and red centers $F$ and $D$, respectively. The following points are blue unit neighbors:

If $R=(5,-4)$ were red, its unit neighbors $P=(5,-5)$ and $Q=(4,-4)$ would be blue. Then $K,L,I,Q,P$ would be a blue progression with step $(1,-1)$. Thus $R$ is blue. The progression $A',J,N,M,R$, where $A'=(5,0)$, has unit step $(0,-1)$, forcing $A'$ red.

The points $S_1=(2,1)$ and $S_2=(3,1)$ are unit neighbors of $D$, while $S_3=(4,1)$ and $S_4=(5,1)$ are unit neighbors of $A'$. All four are blue, so the unit horizontal progression $S_1,S_2,S_3,S_4,B'$ forces $B'=(6,1)$ red.

Reflection in the horizontal line $AD$ acts as $(a,b)\mapsto(a+b,-b)$. It preserves the red seed $S$, exchanging $B,F$ and $C,E$, and fixes $A'$. Apply the preceding forcing argument to this reflected configuration: its image of $B'$ is $F'=(7,-1)$, which is therefore red. This uses the same red seed, not a symmetry assumption on the coloring.

Next $U=(0,3)$ is blue by Lemma 3, using triangle $U,A,D$ with red center $B$. Points $V=(1,3)$ and $W=(2,3)$ are blue unit neighbors of $C$. If $X=(4,2)$ were red, its unit neighbors $X_1=(3,3)$ and $X_2=(4,3)$ would be blue, making $U,V,W,X_1,X_2$ a blue horizontal $\ell_5$. Thus $X$ is blue. Reflecting this entire argument in $AD$ also forces $Y=(6,-2)$ blue.

The red triangle $A',B',F'$ has side $\sqrt3$ and must extend to a red $T_6$ by Lemma 5. Relative to $A'$, its vertices are $0,w,t$, where $w=(1,1)$ and $t=(2,-1)$; the blue points $X,Y$ are $w-t,t-w$. The four-completion enumeration on the configuration page leaves only $S$ itself: $S-w$ contains $t-w$, $S-t$ contains $w-t$, and $-S+w+t$ contains both. Therefore all of $S+(5,0)$ is red.

Propagation in every lattice direction

The equilateral six-point set $S$ is invariant under $120^\circ$ rotation about its center $(2,0)$. The linear part of this rotation is $(a,b)\mapsto(-a-b,a)$; it takes $(1,0)$ successively to $(-1,1)$ and $(0,-1)$. Applying the proved translation rule to the rotated seed consequently forces the red translates by

$$d_1=(5,0),\qquad d_2=(-5,5),\qquad d_3=(0,-5).$$

The rule applies again to every new red translate. Induction therefore forces $S+n_1d_1+n_2d_2+n_3d_3$ red for all nonnegative integers $n_i$. Since $d_1+d_2+d_3=0$, the negative of each generator is a sum of the other two. These nonnegative combinations thus generate the entire group $5\mathbb Z^2$. This proves that every point in (1) is red, with no unproved passage from a finite diagram to the whole lattice.

Forcing all other residues blue

Modulo $5$, the six red residues are $(0,0),(1,1),(2,2),(3,0),(4,3),(2,4)$. Each of the remaining nineteen residues has a red unit neighbor. The following table lists one such neighbor for every remaining residue; coordinate differences are understood modulo $5$ and are one of the six unit directions.

For an arbitrary lattice point in one of these residues, choose the corresponding unit difference and the appropriate $5\mathbb Z^2$ translate of the red representative. That neighbor is red by (1), so the point is blue. This proves the exact coloring and its periods.

Source and corrections

Lemma 6, Figures 7–8, published pp. 5–7; Lemma 2.5 in arXiv v2. Both versions end the extension step by calling $A',B',C',D',E',F'$ blue; the preceding argument and the claimed red translate require red. The arXiv text additionally calls the conditionally forbidden progression $A'JNMR$ red; the published text correctly says blue. The rewrite expands the reflection arguments, enumerates the alternative completions, proves the propagation induction, and supplies the residue check. No external theorem is used.

Bears on. #188.