Color edges of $K_6$, there is a cycle of length 4

Question

If we color the edges of $K_6$ (complete graph on $6$ vertices) red or blue, how to prove that there is a cycle of length $4$ with monochromatic edges ?

Answer 1

I will start with an uncolored $K_6$, and color the graph step by step, each time doing what I have to do to avoid creating a four-cycle, and show that I can't avoid making a four-cycle at some point. I would advise you to draw on a piece of paper as you follow along, although I will show a picture halfway.

First, we name the vertices $1, 2, 3, 4, 5$ and $6$. Without loss of generality, we may assume $1$ is connected to each of $2, 3$ and $4$ with a red edge (if there are three or more blue edges connected to $1$, swap the coloring, and if now $1$ isn't connected to each of $2, 3$ and $4$ with a red edge, rename the other vertices so that it is).

From $5$ to $\{2, 3, 4\}$, we can't have two red edges, because that would make a four-cycle with $1$, so at least two of them are blue, say $35$ and $45$ (note that we don't know what $25$ is yet, so we leave that for now). Now we look at connecting $6$ to $\{2,3, 4\}$. For the same reason as for $5$, we can't have two red, which means we need at least two blue. Those two blue cannot be to $3$ and $4$, because that would make a four-cycle with $5$. That means that we must have $26$ blue, along with either $36$ and $46$. By symmetry it doesn't matter, so say $36$.

Lastly, because we don't what to make a blue four-cycle, we must have $25$ and $46$ both red (otherwise the cycles $2536$ or $3546$ would be blue). This is what the graph looks like so far:

At this point we turn our attention to the triangle $234$. Because they're all connected by red to $1$, we cannot have two red edges in this triangle, so we must have two blue. If one of the blues is $24$, then either $2453$ or $2436$ are blue four-cycles, so $24$ is red, while $23$ and $34$ are both blue.

This gets us into trouble with coloring the edge $65$ because if we make it red, then $5246$ is a red four-cycle, while if we make it blue, $6235$ is a blue four-cycle. Thus it is impossible to two-color $K_6$ without making a mono-colored $C_4$.

Answer 2

By Ramsey's theorem we know that at least two triangles in $K_6$ are monochromatic.
Let we assume that both of them are blue: in such a case, we may also assume that they do not share an edge (since otherwise they would trivially give a blue $C_4$). If their vertices are $A,B,C$ and $C,D,E$, then the edges $AE,AD$ and $BE$ have to be red to avoid a blue $C_4$. For the same reason, $BD$ has to be blue and at most one edge through $F$ is blue. In such a case, however, either $FAEB$ or $FEAD$ is a red $C_4$. If $ABC$ and $DEF$ are blue triangles and at least two edges from $\{A,B,C\}$ to $\{D,E,F\}$ are blue, there is a blue $C_4$. If no edges from $\{A,B,C\}$ to $\{D,E,F\}$ are blue, there is a red $C_4$. So we may assume that $BD$ is blue, and in that case $AFCD$ is a red $C_4$. I leave to you to finish the proof: you just need to understand what happens if we have two monochromatic triangles with opposite colors. If they share a vertex, say they are $ABC$ and $CDE$, you may consider the constraints on the edges through $F$ and on the edges joining $\{A,B,C\}$ and $\{C,D,E\}$. If they do not share a vertex, say they are $ABC$ and $DEF$, it is enough to consider the constraints on the $9$ edges joining $\{A,B,C\}$ and $\{D,E,F\}$.