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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 no red $T_3$, its coloring, up to translation and rotation by a multiple of $60^\circ$, has

$$(a,b)\text{ red}\quad\Longleftrightarrow\quad2a+b\equiv0\pmod5. \tag{1}$$

All other lattice points are blue. In particular, both primitive unit directions are color periods after multiplication by $5$.

Obtaining the first red pair

Use the triangular coordinates and forcing rules. The lattice contains a red point $A$, since otherwise any five consecutive points along a unit lattice direction would be blue. Translate $A$ to $(0,0)$. The three lattice points $(-1,2),(2,-1),(-1,-1)$ form a side-$3$ equilateral triangle with center $A$. By Lemma 2, at least one is red. A rotation by a multiple of $60^\circ$ lets us take that red point to be $B=(-1,2)$.

The local pair rule: Figure 10

The following coordinates specify the entire forcing step:

The progression $D,E,F,G,B'$ has unit step $(1,0)$, so $B'$ is red. Then its unit neighbor $N$ is blue. The progression $C,H,I,G,N$ has unit step $(1,-1)$, forcing $C$ red, and $A',N,G,I,H$ has unit step $(-1,1)$, forcing $A'$ red.

Put $w=(-1,2)$ and $s=(3,-1)$ in this proof. We have established the following rule: if $P$ and $P+w$ are red, then

$$P+2w,\qquad P+s,\qquad P+s+w \quad\text{are red}. \tag{2}$$

The rule holds at every translate of the diagram. Applying its half-turned version to $P,P-w$ gives the corresponding rule with $w,s$ replaced by $-w,-s$. These operations preserve the hypotheses, including the absence of a red $T_3$ anywhere in $L$.

From the pair rule to all parallel lines

Starting with $0,w$, rule (2) repeatedly extends the red line to $2w,3w,\ldots$. Applying the reversed rule to $w,0$ gives $-w$, and repetition gives all negative multiples too. Thus $\mathbb Zw$ is red. Since $w=(-1,2)$ has coprime coordinates, these are exactly the lattice points on that line.

Rule (2) at $0,w$ gives the pair $s,s+w$, which similarly extends to the entire line $s+\mathbb Zw$. The reversed rule applied to $w,0$ gives $w-s$ and $-s$, so $-s+\mathbb Zw$ is red as well. Apply the same two operations on a red line $ms+\mathbb Zw$: they produce the lines $(m+1)s+\mathbb Zw$ and $(m-1)s+\mathbb Zw$. Induction proves that every point of

$$H=\mathbb Zw+\mathbb Zs$$

is red. This explicitly supplies both directions of the line propagation implicit in Figure 9.

The subgroup $H$ consists exactly of the integer pairs in (1). Indeed, $2a+b$ vanishes modulo $5$ on both generators. Conversely, if $2a+b\equiv0\pmod5$, then

$$(a,b)=\frac{a+3b}{5}\,w+\frac{2a+b}{5}\,s,$$

and both coefficients are integers. To determine every remaining color, observe that the homomorphism $(a,b)\mapsto2a+b\pmod5$ takes the unit directions $v,u,-u,-v$ to $1,2,3,4$, respectively. For any point outside $H$, subtract the unit direction with its nonzero residue. The result is a red point of $H$, forcing the original point blue. This proves (1) and its periods.

Source and dependencies

Lemma 7, Figures 9–10, published pp. 7–8; Lemma 2.6 in arXiv v2. The source's reference to three points at distance $\sqrt3$ from a lattice point is made precise by selecting three alternating points from its six such neighbors. The finite progressions, backward propagation, parallel-line induction and remaining residue classes are all explicit above. No external theorem is used beyond elementary lattice arithmetic and Lemma 2.

Bears on. #188.