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Source. Theorem 12, printed p. 348, physical p. 8 of the published paper.

Let $L_k$ be the configuration of $k$ collinear points separated successively by unit distance. For every integer $n\ge1$,

$$R(L_3,n,4),\qquad R(L_4,n,3),\qquad R(L_6,n,2) \quad\text{are false}. \tag{1}$$

Avoiding $L_3$ with four colors

Color $x\in\mathbb R^n$ by

$$\lfloor |x|^2\rfloor\pmod4. \tag{2}$$

Suppose $x-u,x,x+u$, where $|u|=1$, had the same color. Put $t_-=|x-u|^2$, $t_0=|x|^2$, and $t_+=|x+u|^2$. Then

$$t_++t_--2t_0=2. \tag{3}$$

Write $t_\sigma=4k_\sigma+r+\theta_\sigma$, with a common $r\in\{0,1,2,3\}$ and $0\le\theta_\sigma<1$. Equation (3) would give

$$2=4(k_++k_--2k_0)+\theta_++\theta_--2\theta_0. \tag{4}$$

The final term lies strictly between $-2$ and $2$. If the integer in parentheses is $0$, it would have to equal $2$; if it is $1$, it would have to equal $-2$; every other value is farther outside the interval. This is impossible, so (2) contains no monochromatic $L_3$.

Avoiding $L_4$ with three colors

Color $x\in\mathbb R^n$ by

$$\lfloor 2|x|^2\rfloor\pmod3. \tag{5}$$

Suppose $x+iu$, $1\le i\le4$, with $|u|=1$, had the same color. Put $y_i=2|x+iu|^2$ and let $f_i=y_i-\lfloor y_i\rfloor\in[0,1)$. The quadratic sequence $y_i$ satisfies

$$y_1+y_3=2y_2+4, \qquad y_2+y_4=2y_3+4. \tag{6}$$

All four integer parts are congruent modulo $3$. On separating integer and fractional parts in the first equation, one gets

$$4=3M+f_1+f_3-2f_2$$

for an integer $M$. Since the fractional expression lies strictly between $-2$ and $2$, necessarily $M=1$, and hence

$$f_1+f_3-2f_2=1. \tag{7}$$

The second equation similarly gives

$$f_2+f_4-2f_3=1. \tag{8}$$

Adding (7) and (8) yields

$$f_1+f_4=f_2+f_3+2,$$

whose left side is strictly below $2$ while its right side is at least $2$. Thus (5) contains no monochromatic $L_4$.

Avoiding $L_6$ with two colors

Color $x\in\mathbb R^n$ by the parity of

$$\left\lfloor\frac{|x|^2}{6}\right\rfloor. \tag{9}$$

Suppose $x+iu$, $1\le i\le6$, with $|u|=1$, had the same color, and set

$$a_i=\frac{|x+iu|^2}{6}.$$

Then

$$a_{i+1}+a_{i-1}=2a_i+\frac13\qquad(2\le i\le5), \tag{10}$$

and all $\lfloor a_i\rfloor$ have the same parity. Put

$$b_i=a_i+(i-4)\lfloor a_3\rfloor+(3-i)\lfloor a_4\rfloor. \tag{11}$$

Adding an integer affine function of $i$ preserves (10) and every fractional part. Moreover,

$$0\le b_3,b_4<1,$$

and every $\lfloor b_i\rfloor$ is even, since modulo $2$ the three coefficients in (11) sum to $1+(i-4)+(3-i)=0$.

The first two recurrences give

$$b_2=2b_3-b_4+\frac13, \qquad b_5=2b_4-b_3+\frac13. \tag{12}$$

Thus $-2/3<b_2,b_5<7/3$. Their even integer parts imply

$$b_2,b_5\in[0,1)\cup[2,7/3). \tag{13}$$

But (12) also gives

$$2b_2+b_5=3b_3+1, \qquad b_2+2b_5=3b_4+1. \tag{14}$$

If $b_2\ge2$, the first left side is at least $4$ while its right side is strictly below $4$; the second identity rules out $b_5\ge2$ in the same way. Hence $b_2,b_5\in[0,1)$.

The remaining recurrences give

$$b_1=2b_2-b_3+\frac13, \qquad b_6=2b_5-b_4+\frac13. \tag{15}$$

Again $-2/3<b_1,b_6<7/3$, so their even integer parts put them in the union in (13). The identities

$$2b_1+b_4=3b_2+1, \qquad b_3+2b_6=3b_5+1 \tag{16}$$

then exclude the second interval, just as in (14). Therefore every $b_i$ lies in $[0,1)$. Finally, the quadratic sequence satisfies

$$b_1+b_6=b_3+b_4+2. \tag{17}$$

The left side of (17) is strictly below $2$, while the right side is at least $2$, a contradiction. Thus (9) contains no monochromatic $L_6$, and all three assertions in (1) follow.