Can a connected Eulerian graph have an even number of vertices and an odd number of edges?

Question

I know that the requirements for a Eulerian graph are that all vertex degrees are even and that it is connected. But I am not sure how that would work with the amount of vertices and edges.

So, if this is possible, how would we draw a graph like that? If it isn't possible, how can we prove that?

Thank you!

Answer 1

Edit: Sorry, my previous answer was incorrect.

In fact, there is such a graph. Consider a hexagon with a triangle inside:

Answer 2

There are infinitely many such graphs, but the smallest have 6 vertices. No graph of order 2 is Eulerian, and the only connected Eulerian graph of order 4 is the 4-cycle with (even) size 4.

The only possible degrees in a connected Eulerian graph of order 6 are 2 and 4. Any such graph with an even number of vertices of degree 4 has even size, so our graphs must have 1, 3, or 5 vertices of degree 4. Up to isomorphism, there is exactly one graph of each type.

With one vertex of degree 4 and size 7:

This generalizes to any even cycle and odd cycle joined at a vertex.

With three vertices of degree 4 and size 9:

This generalizes to the square of any cycle on $4k+2$ vertices.

With five vertices of degree 4 and size 11:

This generalizes to any complete graph on $4k+1$ vertices with a subdivided edge.

I expect there are many, many more examples of such graphs.