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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.

Rivers McForge
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Shahrzadeh A
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    and I had another problem too :D prove if we coloring k6 with two color,then we have 4-cycle with same color. – Shahrzadeh A Jul 16 '20 at 10:53
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    In a complete graph, everything's equivalent, so only the total number of vertices with each color matters. Just divide into cases based on the number of same-colored vertices, and apply the Pigeonhole Principle. – aschepler Jul 16 '20 at 10:57
  • Ummm.But this problem is about coloring edges, not coloring vertex – Shahrzadeh A Jul 16 '20 at 12:35
  • Aha, that does make it much more interesting. I'm not one of the downvoters, but it may help if you share some of your thoughts so far instead of just stating the problem. You could certainly outline your proof for one cycle in $K_6$, and say a little about why it's difficult to apply ideas from that proof to the harder problem, or what's still missing after trying those ideas. – aschepler Jul 16 '20 at 18:30
  • You know, The good thing in K6 is every vertex has 5 edges.But K10 doesn't have same thing. – Shahrzadeh A Jul 16 '20 at 18:46

1 Answers1

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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.

enter image description here

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.

Alex Ravsky
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    Actually if the edges of $K_7$ are colored with two colors, there will be at least $4$ monochromatic triangles. The number of $2$-colored angles is at most $3\cdot3+3\cdot3+3\cdot3+3\cdot3+3\cdot3+3\cdot3+2\cdot4=62$, so the number of nonmonochromatic triangles is at most $\frac{62}2=31$, so the number of monochromatic triangles is at least $\binom73-31=4$. See also A014557. – bof Jul 19 '20 at 11:05
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    If the edges of $K_{10}$ are colored with two colors, there will be three edge-disjoint monochromatic triangles, so there will be two edge-disjoint triangles of the same color. I wonder if that's what the problem was supposed to be. – bof Jul 19 '20 at 11:17