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Source. Published p. 354, Theorem 17 (published scan).

Statement. In every finite coloring of $\mathbb Q$ there are rationals $x_1,y_1,x_2,y_2$ such that each pair $x_i,y_i$ has one color and

$$(x_1-y_1)(x_2-y_2)=1.$$

The two pairs need not have the same color as one another.

Complete proof relative to van der Waerden. Suppose there are $k\ge1$ colors. Put $M=k!(2k+1)^2$. The exact arithmetic-progression theorem in external_inputs gives a monochromatic progression $a,a+d,\ldots,a+(M-1)d$ in the positive integers, with $d>0$. In particular every difference $dn$, $1\le n<M$, occurs between two points of one color.

Among the $k+1$ rationals

$$\frac1{d\,k!(k+i)},\qquad 1\le i\le k+1,$$

two, with indices $i<j$, have the same color. In that order their difference is

$$\frac{j-i}{d\,k!(k+i)(k+j)}=\frac1{dn},\qquad n=\frac{k!(k+i)(k+j)}{j-i}.$$

Since $1\le j-i\le k$, the denominator $j-i$ divides $k!$, so $n$ is a positive integer. Also $i\le k$ and $j\le k+1$ give $n\le k!(2k)(2k+1)<M$. Choose the first pair from the progression with difference $dn$ and the second pair as above. Their product is one. $\square$

The strict $n<M$ is what a progression of $M$ points supplies. The source briefly includes the unused endpoint $n=M$ among its available differences; the actual selected integer satisfies the strict bound, so the proof closes without an extra progression term. This illustrates why the linear obstruction of Theorem 16 does not extend to arbitrary homogeneous polynomials in the differences.

Bears on. #174.