By going over the tests of previous years in graph theory, I've come across an interesting (in my opinion) question:
$G$ is 3-connected, non-bipartite graph. Prove that $G$ contains at least 4 odd cycles.
I tried the following way: as $G$ is non-bipartite, it has an odd cycle $C$. Now, since $G$ 3-connected, there should be $v \in V(G-C)$ with 3 paths to $C$. From here it should be a game of combining odd/even paths, to get what is needed. But there are too much options.
Is there any other way?