Problem. Prove that for $n$ vertices, where $n$ is odd, it is possible to colour all edges of the complete graph formed by those vertices in $n$ colours so that for any choice of three distinct colours from those $n$, there is a triangle (edges connecting three vertices) coloured with that triplet of colours.
Partial solution. Arrange the vertices in a regular $n$-sided polygon and colour its sides, one colour per side. Each diagonal in this polygon is parallel to a side: colour it with the same colour as the corresponding parallel side. Now, all edges of the graph are coloured. There is exactly the same number of triangles in this graph as there is choices of three colours out of $n$. Hence, the statement holds if there are no triangles with more than one edge coloured by the same colour–and there aren't any, because that would ask for the triangle to have two parallel sides.
Question. This solution does not observe the case of duplicate triangles, i.e. what if there is a triangle parallel to another triangle, hence having the same colouring and "wasting" a triangle in the process? How do we eliminate that possibility?
Best Answer
Let $n=2k+1$. We can assume the vertex set is $V=\{0,1,\dots,n-1\}$ and use the same set for the set of colours. Colour the edge $\{x,y\}$ with the colour $x+y$ where addition is modulo $n$. (This is the same as your colouring.)
Given three colours $a,b,c$ we have to solve the system of equations modulo $n$ $$x+y=a$$ $$x+z=b$$ $$y+z=c$$ The solution is $$x=\frac{a+b-c}2=(k+1)(a+b-c)$$ $$y=\frac{a+c-b}2=(k+1)(a+c-b)$$ $$z=\frac{b+c-a}2=(k+1)(b+c-a)$$