Proof of Hamilton Cycle in a Complete Bipartite Graph

A complete Bipartite graph $K_{m,n}$ has a Hamilton cycle if and only if $m=n$.

I want to know if the following proof technique is correct. My proof will consider using proof by contradiction.

Assume that $m\not = n$. Let $H$ be the (Hamilton) cycle that goes through every vertex in $K_{m,n}$. $ H = v_0e_0v_1e_1...v_ie_i$. Since $K$ is bipartite, the cycle must alternate between the vertices on each side. Since $m\not = n$ there exists a $v_a = v_b, a < b$ inside cycle H. This leads to a contradiction since a cycle cannot have repeating vertices. Hence, a complete Bipartite graph $K_{m,n}$ has a Hamilton cycle if and only if $m= n$.

Is this correct?

Solution 1:

A complete bipartite graph $K_{m,n}$ is Hamiltonian if and only if $m = n$ , for all $m, n \geq 2$.

Proof: Suppose that a complete bipartite graph $K_{m, n}$ is Hamiltonian. Then, it must have a Hamiltonian cycle which visits the two partite sets alternately. Therefore, there can be no such cycle unless the two partite sets have the same number of vertices. If $m = n = 1$, it is clear that $K_{m, n}$ contains no Hamiltonian cycle.

Thus, we get $m = n$, for all $m, n \geq 2$.

Conversely, it is easy to see a Hamiltonian cycle $x_1, y_1, x_2, y_2, x_3, y_3, \dots, x_n, y_n, x_1$ for such graphs.