TITLE: Chromatic index of a complete graph
QUESTION [7 upvotes]: Looking to show that $\forall n \in \mathbb{N}$
$$\chi^{'}(K_{2n+2})=\chi^{'}(K_{2n+1})=2n+1$$
I'm trying to construct a colouring of the edges of $K_{2n+1}$ that leaves colour $i$ missing at vertex $i$ so I can move to $K_{2n+2}$ without increasing the index (for my induction) but the details of that initial colouring are hard to work out.

REPLY [3 votes]: Just another solution: Label the vertices $1,2,\ldots, 2n+1$ and for each $i$, colour edge $(i+k,i-k)$ using colour $i$ for every $k=1,2,\ldots,n$ (operations modulo $2n+1$). This is a proper edge colouring using $2n+1$ colours with the property that no edge coloured $i$ is incident to vertex $i$. 
Example for $2n+1=7$: