TITLE: Existence of k-regular graph
QUESTION [6 upvotes]: In a few examples i noted that the existence of $k$-regular graph on n vertices is :

True , for k or n even.
False , for k and n odd . But we can find a graph with $n-1$ vertices with
degree k and one vertex with degree  $k-1$. There doesn't exists a k-regular graph for 
k and n odd  because $k=\deg(G) = 2*|E(G)| / |V(G)|$
$|E(G)| = k*n/2$, and $|E(G)|= m$ is not a natural number if $n$ and $k$ is odd.

Any proof idea ??

REPLY [2 votes]: A lot easier way: the sum of the degrees is $2|E|$. Therefore the sum of the degrees must be an even number. Since an odd times an odd is always an odd, and the sum of the degrees of an $\boldsymbol k$-regular graph is $k\cdot n$, $n$ and $k$ cannot both be odd.