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Let $G$ be a $k$-regular bipartite graph, and let $G$ contain a bridge. Prove that $k=1$.

I have tried to remove the bridge and find a contradiction that an component should not be bipartite, but I failed.

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  • $\begingroup$ Hint: consider edge counts $\endgroup$ – Bob Krueger Nov 26 '17 at 15:56
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Suppose that the removal of the bridge in a $k$-regular graph results in a component of a graph. This component is either a single vertex, in which case the problem is solved or it must be a bipartite graph with vertex partition sets say $V$ and $W$. WLOG, assume that one of the vertices of the bridge was a vertex in $V$. Then because the sum of the degrees of vertices in each partition sets $V$ and $W$ must be equal in a bipartite graph, so $$|W|k=(|V|-1)k+(k-1)\Longrightarrow (|V|-|W|)k=1.$$ Since all quantities are integers, so $k=1$.

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  • $\begingroup$ you are welcome! $\endgroup$ – PJK Nov 27 '17 at 2:03

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