Unrolling residues to avoid progressions

Mathematics is often described as the study of patterns, some of the most beautiful of which are periodic. In the integers these periodic patterns take the form of arithmetic progressions, i.e., a set of integers which are equally spaced, or have the same common difference. For example, 5, 11, 17, 23, 29 is an arithmetic progression with 5 terms where consecutive terms have a difference of 6. Suppose we take the numbers 1, 2, . . . , 28 and we divide these numbers into two sets, which we can represent by “coloring” the numbers red (r) and blue (B). We can then look at how many times we have three equally spaced terms all from the same set. In other words how many times can we find three equally spaced numbers which are monochromatic in either red or blue. If we want to maximize the number of such occurrences then the obvious thing to do is to paint 1, 2, . . . , 28 either all red or all blue. Then every three equally spaced numbers is monochromatic, and in this case there are 182 such sets of these triples (see Observation 1). We can also easily find the average number of such occurrences. This is the same as asking how many of these monochromatic triples would we expect if we colored by flipping a coin with one side marked r and the other B. If this is a fair coin then given any three equally spaced numbers the probability the coin would always land on the same side is 2 8 . By linearity of expectation we can conclude that on average we have 1 4 · 182 = 45.5 of these equally spaced triples will be monochromatic. What if we want to find the minimum number of these equally spaced monochromatic triples? By the random argument we know we can at least find one coloring with ≤ 45 of these, but we would hope to do better. For the case of 1, 2, . . . , 28 it turns out there is a unique(!) way to color these to minimize the number, namely: