The Clebsch graph is a $5$-regular graph on 16 vertices, defined as follows. Take the vertices and edges of the $4$-cube, and then add edges between antipodal pairs of vertices.
Apparently, the edges of the complete graph $K_{16}$ can be partitioned into three copies of the Clebsch graph, and this is an important example in Ramsey theory.
Is there an easy way to see this partition?
Note: I found the following explanation: an equivalent definition of the Clebsch graph is that its vertices correspond to elements of the finite field $GF(16)$ and two vertices are connected by an edge whenever their difference is a perfect cube, but I don't quite see how that helps.
Also, I found a picture of the partition here, but this still doesn't shed light on the structure to me.
I prefer the hypercube definition to the finite-field definition, but any help will be much appreciated.
Let $G_1$ be our graph.
Now let $x$ be a generator of $GF(16)$ and let $f_x$ be the map $p\to xp$. Define a second graph $G_2$ on the elements of $GF(16)$ by: two vertices $p$ and $q$ are connected by an edge whenever $x(p-q)$ is a perfect cube. Then clearly $f_x$ defines an isomorphism between $G_1$ and $G_2$.
Do the same again with the map $p\to x^2p$ to get a graph $G_3$.
You now have 3 copies of the Clebsch graph.
Since for every pair $p,q$ exactly one of $p-q,x(p-q),x^2(p-q)$ is a perfect cube this proves that every edge is in exactly one of the copies.