Monochromatic solution to $x + qy = z$ for $q \in \mathbb{Q}$

Question

I want to find a proof that for any given $r \in \mathbb{N}$ and $q \in \mathbb{Q}$ every $r$-coloring of the integers shows a monochromatic solution for $x + qy = z$.

I am aware that if the equation is $x + y = z$ we have Shur's Theorem and if $x + y = 2z$ the problem is equivalent to finding a monochromatic arithmetic progression of length 3 $\{x, x + d, x + 2d\}$ (and we can prove it finding a necessary amount of monochromatic pairs of numbers with the same sum).

This problem is in a book I'm studying and I was trying to adapt the proof for the other problem involving $x+y = 2z$ (even though the former doesn't try to find arithmetic progressions) but couldn't solve it yet. I also thought about trying to use some probabilistic argument; don't know if it is of any use here.

Any help is much appreciated, thanks in advance!

Answer 1

Writing $q=a/b$ and multiplying by $b$ your equation is equivalent to the integer coefficients equation $bx+ay-bz=0$, which is partition regular by Rado's theorem since the coefficents of $x$ and $z$ sum to zero.

Answer 2

Shur's theorem states that there is always a + y = b, so there is always (aq) + qy = (bq), let x = aq and z = bq, so there is always x + qy = z