When is a linegraph Eulerian?

Question

So I know that for a simple Eulerian graph, its linegraph is also Eulerian. (Since then, each degree of the nodes of the linegraph are also even). I'm asked to give (sufficing) conditions for a graph $G$ to ensure that its linegraph $L(G)$ is Eulerian.

Answer 1

This is just a humble suggestion.

$$L(G) \; \text{is Eulerian} \iff \text {Each vertex in $L(G)$ has even degree} $$

This will be true if and only if every edge in $G$ is adjacent to an even number of other edges. Considering vertices incident to each edge in $G$, this condition will be satisfied if for each pair of adjacent vertices $ u, v $ in $V(G), \;$ $\deg(u) + \deg(v) = 2m; \;\; m \in \Bbb N$. This is because the neighbours of each vertex $u$ and $v$ contribute to the degree of $uv$ in terms of $L(G)$. Hope this helps..