For what number of $n$ if we partition the edges of the complete graph $K_n$ into $4$ pieces then one of the pieces contains a $K_3$?
Suppose I start with one vertex named $v$ then it have $n-1$ adjacent edges with other $n-1$ $\color{blue}{\text{vertices}}$. Now partition those $n-1$ edges into $4$ pieces with two or more edges in $\color{green}{\text{one partition}}$. If there is $\color{red}{\text{any edge}}$ among these $\color{blue}{\text{vertices}}$ (which is obviously exist because our graph is $K_n$) on that $\color{green}{\text{partition}}$, then there is a triangle mean $K_3$ graph. Put this $\color{red}{\text{edge}}$ in that $\color{green}{\text{partition}}$. Then one of the pieces contains a $K_3$. We can put rest edges randomly into those partition. I guess for $n\geq6$ do the work as the number of adjacent edges are more then $4$ which guarantee one partition contain two adjacent edges.
But the answer is $n\geq R(6,6)$, where n is a Ramsey number $n=R(r,b)$, without any detail.
I was just lost how they get it from? It would be nice if anyone show me the right way.
This is equivalent to calculating the multichromatic Ramsey number $R(3,3,3,3)$. This article here discusses it (although they use different notation). $R(6,6)$ is not equal to $R(3,3,3,3)$ as we know $R(6,6)\in [ 102,165]$ and $R(3,3,3,3)\in[50,65]$ (look at the article I linked).