Let $G$ be a $k$-regular graph with $m$ edges and $k$ odd. Prove that $k\mid m$.
We can see this statement is true by example, but how can we prove it?
$k$-regular graph
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graph-theory
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1Please edit your Question so that the body is a self-contained statement of the problem you want help with, particularly not relying entirely on the title for representation. While occasionally the title might suffice to state a problem, splitting the content between title and body is a burden to your Readers, and omits the context that helps willing Readers supply you with prompt and cogent Answers. – 2017-01-14
1 Answers
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HINT: Use the handshaking lemma: if $V$ is the vertex set of $G$, then $\sum_{v\in V}\deg v=2m$. Say there are $n$ vertices; what is $\sum_{v\in V}\deg v$ in terms of $n$ and $k$?
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0$m=nk/2$. But, there are no $k -$ regular graph of order $n$ where both $n$ and $k$ are odd. Does that mean that we can't have $k - $ regular graph with even number of vertices $n$ such that $k|m$? – 2017-01-14
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0@user_99: You don’t need to worry about that. Just use the fact that $nk=2m$, so that $k\mid 2m$. Then use the fact that $k$ is odd. – 2017-01-14