# Homogeneous van der Waerden

The Erdős Discrepancy Problem is whether in any two-coloring of the naturals for any $$C$$ there is a sequence $$d, 2d, \ldots nd$$ such that the difference of red and blue numbers in it is more than $$C$$. This was recently shown to be true by Tao (building on Polymath5).

Now consider the following stronger conjecture, which also generalizes van der Waerden.

In any $$k$$-coloring of the naturals for any $$n$$ there is a monochromatic sequence $$(i+1)d, (i+2)d, \ldots (i+n)d$$.

If true, this would of course be quite a strong result, so I more expect that someone might be able to show a simple counterexample to it. What about the even stronger density version?

This is false already for $$k=2,n=4$$. Color an integer $$m$$ according to the parity of the exponent of $$2$$ in the prime factorization. Among $$i+1,i+2,i+3,i+4$$ at least one number is odd, and at least one is divisible by $$2$$ and not by $$4$$. Those two numbers have the parity of the exponent different, hence so do the corresponding two among $$(i+1)d,(i+2)d,(i+3)d,(i+4)d$$. Hence those four numbers can't have the same color.
Edit: for completeness, it's false for $$k=2,n=3$$ as well. For integer $$m$$, write it as $$3^i\cdot j,3\nmid j$$ and color $$m$$ according to $$j\mod 3$$. Then among $$i+1,i+2,i+3$$ the two which are not divisible by $$3$$ will have different remainders, do multiplying them by $$d$$ gives numbers of different colors. Hence your conjecture only holds in trivial cases $$k=1$$ and $$n\leq 2$$.