I was attempting the following problem:
Let $G$ be a connected simple graph. (a) If $G$ is eulerian with an even number of vertices, then it has a spanning subgraph $G'$ such that every node $i$ has degree $\deg i/2$ in $G'$. (b) If $G$ is Eulerian with an odd number of nodes, then it has a spanning subgraph $G'$ such that every node except one $i$ is incident with $\deg i/2$ edges of $G'$. (c) if $G$ is not eulerian, then it has a spanning subgraph $G'$ such that every node $i$ is incident with either $\lfloor\deg i/2\rfloor$ or $\lceil\deg i/2\rceil$ edges of $G'$.
The exercise was in a section about SDPs, but I'm not sure those are even necessary. We can partition the edges of $G$ into disjoint cycles, for part (a) I tried showing that if we remove a cycle with an odd number of edges then $G$ remains connected and we can apply induction.
The hard part is using the connectedness condition of $G$--without it, the claim becomes false, taking the disjoint union of two triangles. I think the second part follows immediately from the first by noticing that you could add one vertex of degree 2 to reduce to the first case, and then the spanning subtree will use precisely 1 edge out of this vertex. The spanning subtree on the original graph thus destroys the condition for only one vertex.
