Graph theory: Prove $k$-regular graph $\#V$ = odd, $\chi'(G)> k$

Solution 1:

Let $G=\langle V,E\rangle$ be a $k$-regular graph with $n=2m+1$ vertices; as you say, clearly $k=2\ell$ for some $\ell$, so $G$ has $$\frac{kn}2=\frac{2\ell(2m+1)}2=\ell(2m+1)$$ edges. Suppose that $c:E\to\{1,\dots,k\}$ is a coloring of the edges of $G$. $$\frac{\ell(2m+1)}k=m+\frac12\;,$$ so there is some color that is used on at least $m+1$ edges. $G$ has only $2m+1$ vertices, so two of these edges must share a vertex, and $c$ therefore cannot be a proper coloring.

Solution 2:

Another way to prove this fact is to notice that in any proper edge coloring, every set of edges that share a color must form a matching. But for any given color, the matching touches an even number of vertices, so there must be one vertex missing that color. Since that vertex has $k$ edges, all of a different color, together there must be at least $k + 1$ colors.