In space there are given $p_n=\lfloor{en!}\rfloor+1$ points. Each pair of points are connected by a line, and each line is coloured by one of the $n$ colours. Prove that there is at least one triangle of the same coloured sides.
I am not getting how to counter the $e$ here. Please help.
I however made one guess though.. Is $p_n = R(\underbrace{3,3,\cdots,3}_{n \ times})$ ?. I guessed so just because It holds true for $n=2,3$.
NB: $R$ is the Ramsey number.
Let $f(n)$ be the maximum size of a complete graph that can have its edges colored with $n$ colors without triangles.
Notice that $f(n+1)\leq (n+1)(f(n))+1$. To see this pick an arbitrary vertex $v$ and consider the connections to the remaining $f(n+1)-1$ vertices. If we had $f(n+1)-1>(n+1)f(n)$ then by the pigeon-hole principle there would be a set with $f(n)+1$ vertices that are all connected to $v$ by the same color, this is a contradiction, because those $f(n)+1$ vertices would have to have all of their connections with only the remaining $n$ colors, and a triangle would be formed.
We have $f(2)=5\leq \lfloor e2!\rfloor $.
We now prove by induction that $f(n)\leq\lfloor en!\rfloor$
We just have to prove that $(n+1)\lfloor en!\rfloor+1\leq \lfloor e(n+1)!\rfloor$.
this is equivalent to proving that $(n+1)\lfloor en!\rfloor<\lfloor e(n+1)!\rfloor=\lfloor en!(n+1)\rfloor$
All we have to do is show that the fractional part of $en!$ is at least $\frac{1}{n+1}$ but the fractional part of $en!$ is $\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}\dots$