Find the edge chromatic number of $K_n$ when $n$ is a positive integer. Find the edge chromatic number of $K_n$ when $n$ is a positive integer. Note: $K_n$ is
the complete graph on $n$ vertices。

I draw for $n = 1,2,3,4,5,6$ cases but have no general solution or proof about this question.
Let $n&gt;1$. At any vertex $n-1$ edges meet and therefore the chromatic number is at least $n-1$.  If such a colouring exists then consider one particular colour, red say. Each red edge   pairs up vertices and every vertex would be in such a pair. Such a colouring is therefore impossible if $n$ is odd. 

Conversely, we can  number the vertices and the colours $0,1,...,n-1$. We  can then colour the edge from $i$ to $j$ with colour $i+j \mod n$ and we thus obtain an edge-colouring with $n$ colours. 

We now know that the chromatic index is precisely $n$ when $n$ is odd.

Now consider $n=2k+2$ and delete one vertex, $v$ say. The graph which remains, on $2k+1$ vertices, has a colouring with $2k+1$ colours obtained as in the second paragraph. Then vertex $i$ is incident with edges of all colours apart from colour $2i$ (modulo $2k+1$).  Add vertex $v$ back in and join it to vertex $i$ with colour $2i$ (modulo $2k+1$). These colours are all different since Now $2i\equiv 2j \mod 2k+1$ if and only if  $i\equiv j \mod 2k+1$. 

So the chromatic index is precisely $n-1$ when $n$ is even.
Let $n&gt;1$. At any vertex $n-1$ edges meet and therefore the chromatic number is at least $n-1$.  If such a colouring exists then consider one particular colour, red say. Each red edge   pairs up vertices and every vertex would be in such a pair. Such a colouring is therefore impossible if $n$ is odd. 

Conversely, we can  number the vertices and the colours $0,1,...,n-1$. We  can then colour the edge from $i$ to $j$ with colour $i+j \mod n$ and we thus obtain an edge-colouring with $n$ colours. 

We now know that the chromatic index is precisely $n$ when $n$ is odd.

Now consider $n=2k+2$ and delete one vertex, $v$ say. The graph which remains, on $2k+1$ vertices, has a colouring with $2k+1$ colours obtained as in the second paragraph. Then vertex $i$ is incident with edges of all colours apart from colour $2i$ (modulo $2k+1$).  Add vertex $v$ back in and join it to vertex $i$ with colour $2i$ (modulo $2k+1$). These colours are all different since Now $2i\equiv 2j \mod 2k+1$ if and only if  $i\equiv j \mod 2k+1$. 

So the chromatic index is precisely $n-1$ when $n$ is even.