An undirected graph possesses an Eulerian circuit if and only if it is…

Concept — Euler’s circuit theorem. An Eulerian circuit is a closed trail that traverses every edge of a graph exactly once and returns to its starting vertex. In an undirected graph such a circuit exists precisely when the graph is connected across its non-isolated vertices and every vertex has even degree. The governing idea is a parity argument: a trail passing through a vertex uses one incident edge to arrive and a different incident edge to leave, so the edges at a vertex are always consumed two at a time.

Application to this question.

1. Fix any vertex v and follow the circuit. Every time the circuit enters v it must also leave v, and those two moves use two different edges incident to v.

2. The starting vertex is no exception, because the circuit is closed: its very first departure pairs with its very final arrival.

3. An Eulerian circuit uses every edge exactly once, and each edge at v belongs to exactly one arrival–departure pair, so deg(v) = 2 × (number of visits the circuit makes to v).

4. Hence deg(v) is even. The vertex v was arbitrary, so every vertex of a graph with an Eulerian circuit has even degree — that settles the “only if” direction.

5. The “if” direction holds too (Hierholzer’s construction): in a connected graph whose degrees are all even, start anywhere and walk without repeating edges. Even degree guarantees an unused exit edge whenever you enter a vertex, so the walk can only get stuck back at the start, closing a circuit; any edges still unused form further all-even subgraphs whose cycles are spliced into that circuit until every edge is covered.

6. Connectedness plus even degree at every vertex is therefore both necessary and sufficient, so the missing condition is that the vertices are all of even degree.

Cross-check against the other stated conditions.

• Vertex count is a separate property from vertex degree: the 4-vertex path has an even number of vertices yet two vertices of degree 1, while the 5-cycle C5 has an odd number of vertices and every degree 2, and only the latter carries an Eulerian circuit.

• Odd degrees block the pairing directly: at a vertex of odd degree one incident edge can never be matched with an arrival-and-departure partner, so no closed trail can use every edge exactly once and still return to where it began. The handshaking lemma adds only that the number of odd-degree vertices is always even — it never rescues such a graph.

• The Königsberg bridge problem is exactly the “of any degree” case Euler settled in 1736: connectedness alone is not enough.

Result. An undirected graph possesses an Eulerian circuit if and only if it is connected and all of its vertices are of even degree.