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 $K_3$?

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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.

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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).

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On p. 43 of the survey "Small Ramsey Numbers" by Stanisław Radziszowski

https://www.combinatorics.org/ojs/index.php/eljc/article/view/DS1

it is stated that

$$51\le R_4(3)=R(3,3,3,3)\le62.$$

The reference for the lower bound is F. R. K. Chung, On the Ramsey numbers $N(3,3,\dots,3;2)$, Discrete Math. 5 (1973), 317-321. The reference for the upper bound is S. Fettes, R. L. Kramer, and S. P. Radziszowski, An upper bound of $62$ on the classical Ramsey number $R(3,3,3,3)$, Ars Combinatoria 72 (2004), 41-63.