Is $L(K_{2n+1})$ vertex-critical?

Question

Let $n>1$ be an integer. We consider the line graphs of $K_{2n}$ and $K_{2n+1}$.

Since $\chi(L(K_{2n})) = \omega(L(K_{2n})) = 2n-1$ we get that removing a point from $L(K_{2n})$ wouldn't change its chromatic number. (By $\omega(\cdot)$we denote the clique number.)

But since $\chi(L(K_{2n+1})) = 2n+1$ and $\omega(L(K_{2n+1})) = 2n$, the above argument does not apply, and this raises the question: is $L(K_{2n+1})$ vertex-critical for all integers $n$?

Answer 1

It is not vertex-critical for $n>1$, since we may have at most $n$ edges of the same color, thus at most $2n^2<n(2n+1)-1$ edges totally of $2n$ colors.