Given a connected multigraph $G$, consider the number of forests in it – subsets of the edge set of $G$ that form no cycle, empty set included. This is a specific value $T_G(2,1)$ of the Tutte polynomial; I played around with some graphs using this relation and saw the following pattern:
The number of forests is odd iff the graph is Eulerian.
For example, any simple tree with $m\ge1$ edges is not Eulerian and all its edge subsets are forests, hence $2^m$ possible forests, an even number. The octahedral graph has $1083$ forests and the $6$-regular Shrikhande graph $187888382977$. The dipole graph with $m$ edges has $m+1$ forests.
If the above observation is always true, how can I prove it? It seems that I would need to use the deletion-contraction recurrence satisfied by the count of forests (and spanning trees, and all other graph invariants encompassed by the Tutte polynomial), but all I get from this is that removing an edge from Eulerian graph always gives a non-Eulerian graph, and contracting always gives an Eulerian graph.
But isn't this just what you need? Let $f(G)$ denote the number of forests of $G$ and let us prove the statement for arbitrary (not necessarily connected) multigraphs by induction on the number $m+n$ of vertices plus edges. If $m=0$, then the graph is Eulerian and has $1$ forest, so the statement holds. Now if $G$ is Eulerian and $e$ is an edge of $G$, $f(G) = f(G-e) + f(G / e)$ is a sum of an odd and an even number, so it is odd. If $G$ is non-Eulerian then either $G$ is a single edge, in which case the number of forests is $2$, as needed, or you can find an edge of $G$ such that neither $G-e$ nor $G/e$ is Eulerian: if there is a vertex with even degree, take any edge incident to it, and otherwise just take any edge. In this case $f(G)$ is a sum of two odd numbers, so it is even.