I was wondering if every 2-coloring of the integer set would result in some number of equidistant (equally-spaced) points. I proved that there will always be 3 equally-spaced points of the same color, and I wonder if I could locate a bound for the largest number of equidistant points I could find. (If it exists, of course.)
I tried induction, but it does not seem to work when parity comes in place.
Any help is appreciated!
Edit: by "equally-spaced" I meant occurring in an arithmetic sequence. An equivalent form of the question is: if we partition the integer set into two sets, do we always get an $n$-term arithmetic sequence whose terms belong to exactly one of the sets? I already showed that we do when $n=3$, and I hope to further generalize it to larger $n$.
This follows from Szemerédi's theorem, I believe. Let $C_i$ denote the vertices with colour $i$ for $i=1,2$. Since these sets partition $\mathbb{N} \subseteq \mathbb{Z}$, at least one of them has positive upper density. Therefore, by Szemerédi, at least one of the sets contains a $k$-term arithmetic progression for every $k$.