Putnam 2022, Problem B-3

Assign to each positive real number a color, either red or blue. Let $D$ be the set of all distances $d > 0$ such that there are two points of the same color at distance $d$ apart. Recolor the positive reals so that the numbers in $D$ are red and the numbers not in $D$ are blue. If we iterate this recoloring process, will we always end up with all the numbers red after a finite number of steps?

Solution:
The answer is yes. Let \( R_0, B_0 \subset \mathbb{R}^+ \) be the set of red and blue numbers at the start of the process, and let \( R_n, B_n \) be the set of red and blue numbers after \( n \) steps. We claim that \( R_2 = \mathbb{R}^+ \).  

We first note that if \( y \in B_1 \), then \( y/2 \in R_1 \). Namely, the numbers \( y \) and \( 2y \) must be of opposite colors in the original coloring, and then \( 3y/2 \) must be of the same color as one of \( y \) or \( 2y \).  

Now suppose by way of contradiction that \( x \in B_2 \). Then of the four numbers \( x, 2x, 3x, 4x \), every other number must be in \( R_1 \) and the other two must be in \( B_1 \). By the previous observation, \( 2x \) and \( 4x \) cannot both be in \( B_1 \); it follows that \( 2x, 4x \in R_1 \) and \( x, 3x \in B_1 \). By the previous observation again, \( x/2 \) and \( 3x/2 \) must both be in \( R_1 \), but then \( x = 3x/2 - x/2 \) is in \( R_2 \), contradiction. We conclude that \( R_2 = \mathbb{R}^+ \), as desired.