Coloring classes of $\{1,2,3,\dots,n\}$

Question

I'm trying to prove the following statement

There is an integer $n_0$ such that for any $n\ge n_0$, in every $9$-coloring of $\{1,2,3,\dots,n\}$, one of the $9$ color classes contains $4$ integers $a,b,c,d$ such that $a+b+c=d$.

I thought about taking $\{1,2,\dots,n\}$ to be vertices of a graph where edges connect vertices of the same color. Then I tried to use Ramsey theorem, but to no avail. I thought also about Schur theorem and tried to find some generalization of it, but got nothing.

How should one prove the statement?

Please help, thanks!

Answer 1

Let $n_0=R(4,\dots,4)$, where 4 appears $r-1$ times. Consider any $r$-colouring $c:\{1,\dots,n\}\to\{1,\dots, r\}$. Enumerate the vertices of the graph $K_n$ by $\{1,\dots,n\}$ and colour edge $ij$ with $c(|i-j|)$. This gives a $r-1$-colouring of $K_n$. Then by definition of $n_0$, we must have a $K_4$ with all edges monochromatic. If this $K_4$ has vertices $x\le y\le z \le w$ then $a=y-x,b=z-y,c=w-z,d=w-x$ gives a solution to your equation.

Answer 2

Isn't this a direct consequence of Rado's single equation theorem?

(taken from the book Ramsey Theory over the Integers by B. Landmann and A. Robertson)

where $c_1=c_2=c_3=1$, $c_4=-1$ for your statement and we can take e.g., $D=\{c_1, c_4\}$. By "regular", it means your statement is true for all $r$-colorings with $r\ge1$.