# Coloring of $K_{17}$

For any 3-coloring of $K_{17}$ I have to show there exists either a red, blue or green triangle. To start, can I use proof by contradiction with color red, blue, green? So $(0,0,136)$ means all 136 edges are green. Clearly this has green triangle. If we color the "outside" of the graph all one color, say red, then we have 17 edges that are red and no edge of the interior can be colored red in order to avoid red triangles. Then I try to obtain a contradiction?

This number is also known as $R(3,3,3)$. And $R(3,3)=6$. If you allowed an extra color then the complete number would be $R(3,3,3,3)$. Fore more info on this notation see ramsey number in wikipedia.
Hint: Pick a vertex $v$. Consider the $16$ edges incident with $v$. By the pigeonhole principle, $6$ of those those edges will have the same color, say red. So there are $6$ vertices joined to $v$ by red edges. If two of those vertices are joined to each other by a red edge, there's your red triangle. If not . . .