Prove that if we edge-color $K_{10}$ with two colors, we have at least two monochromatic triangles with the same color

Question

Prove that if we color the edges of $K_{10}$ with two colors, we have two $3$-cycles in it with the same edge color. You know! I had a problem similar to this, and I solved it: Prove that if we color $K_6$ with two colors, then we have a $3$-cycle with the same color.

Answer 1

if we coloring the edges of $K_{10}$ with two color , we have two 3-cycle in it with same edge color

The claim holds even for a graph $G=K_8$. Indeed, since $G$ contains a copy of $K_6$, it has a monochromatic triangle $T_1$. Let $v_1$ be an arbitrary vertex of $T_1$. Since $G$ with the vertex $v_1$ removed contains a copy of $K_6$, it has a monochromatic triangle $T_2$. Let $v_2$ be an arbitrary vertex of $T_2$. Since $G$ with the vertices $v_1$ and $v_2$ removed contains a copy of $K_6$, it has a monochromatic triangle $T_3$. Then two of the triangles $T_1$, $T_2$, and $T_3$ have the same color.

if we coloring k6 with two color,then we have 4-cycle with same color.

There is well-known question (see, this my answer): what is the maximum number $E(n)$ of edges an $n$-vertex graph can have so that the resulting graph contains no 4-cycle? According to Proposition 1 from the linked answer, $E(n)\le\frac{n+n\sqrt{4n-3}}4$, which follows $E(6)\le 8$. But, similarly to the case of $n=10$, we can refine this bound using the fact that when the sum of $l_i$’s is fixed the minimum of $\sum {l_i \choose 2}$ is attained when $l_i$’s differs by at most $1$ (see the proof of Proposition 1). So if $\sum l_i=16$ then the minimum of $\sum {l_i \choose 2}$ is attained when four of $l_i$’s equal $3$ and two of $l_i$’s equal $2$. Then $\sum {l_i \choose 2}=4\cdot {3 \choose 2}+2\cdot {2 \choose 2}=14$. Since $K_6$ has ${6 \choose 2}=15$ edges, we see that at most one edge is missed to a complete graph. Then we have a 4-cycle, a contradiction. Thus $\sum l_i\le 14$ and $E(6)\le 7$. The following Joseph DeVincentis’ example from Erich Friedman’s page shows that this bound is exact.

Thus if we color edges of $K_6$ in two colors then at least $8$ of edges are monochromatic. Since $E(6)<8$, there exists a 4-cycle of this color.