James Van Lint and RM Wilson's book A Course in Combinatorics poses the following problem:
Let the edges of $K_7$ be colored red and blue. Show that there are at least four subgraphs $K_3$ >with all three edges the same color, i.e. four monochromatic triangles.
$K_n$ means a complete graph on $n$ vertices, i.e. a graph with $n$ vertices where every pair of vertices is connected by unique undirected edge.
In a previous example, the textbook proves that $K_6$ has at least one monochromatic triangle.
The book states that the number of monochromatic triangles in a $2-$colored $K_n$ is at least ${n \choose 3}−\lfloor \frac n2 \lfloor {n - 1 \choose 2}^2 \rfloor \rfloor$. However, the book instructs that we don't use this result.
Below I have formulated my best attempt at a solution. I'm wondering if this works. I'm also wondering if, assuming this works, there's a way for me to clean this up and make it more readable.
Suppose there is only one monochromatic triangle. We know that every $K_6$ must have at least one monochromatic triangle. For ease of reference, I call the vertices in this monochromatic triangle $A$, $B$, and $C$; I call the other three vertices $D$, $E$, and $F$. Suppose, without loss of generality, that the triangle formed by $A$, $B$, and $C$ is red. At most one of the edges connecting $D$ to $A$, $B$, and $C$ can be red; otherwise we have must have at least one more red monochromatic triangle. We can reason similarly for $E$ and $F$. Now take $D$ and $E$, and notice that, by the Pigeonhole Principle, there must be at least one vertex of $A$, $B$, and $C$ such that $D$ and $E$ are both connecting by a blue edge to that vertex. If $D$ and $E$ are connected by a blue edge, then we have a monochromatic triangle, so they must be connected by a red edge. We can reason similarly for the edge connecting $D$ and $F$ and the edge connecting $E$ and $F$. But then $D$, $E$, and $F$ must form a red triangle. So we have a contradiction and there must be more than one monochromatic triangle in $K_6$.
Now I turn to the proof involving $K_7$.
For ease of reference, I refer to the seven vertices in $K_7$ by the letters $A$, $B$, $C$, $D$, $E$, $F$, and $G$. The subgraph of $K_7$ made up of $A$, $B$, $C$, $D$, $E$, and $F$ must have two monochromatic triangles. Without loss of generality, suppose $A$ is a vertex in one of the monochromatic triangles of the subgraph with vertices $A$, $B$, $C$, $D$, $E$, and $F$. Then the subgraph made up of up $B$, $C$, $D$, $E$, $F$, and $G$ has a monochromatic triangle that wasn't in the subgraph made up of $A$, $B$, $C$, $D$, $E$, and $F$ or the subgraph made up of $A$, $B$, $C$, $D$, $E$, and $F$ had at least three monochromatic triangles at least two of which are in the subgraph consisting of $B$, $C$, $D$, $E$, and $F$. Either way, this means that $K_7$ has at least three monochromatic triangles. If $K_7$ has three monochromatic triangles, then by Pigeonhole Principle at least one vertex must be in two monochromatic triangles, which means the subgraph of $K_7$ consisting the six vertices other than this vertex must have at least two monochromatic triangles. So $K_7$ must have at least four monochromatic triangles.