Complete Graph is Hamiltonian for Order Greater than 2

Theorem
Let $n \in \Z$ be an integer such that $n > 2$.

Let $K_n$ denote the complete graph of order $n$.

Then $K_n$ is Hamiltonian.

Proof
First we note that when $n = 2$ there is one edge in $K_n$.

So if you start at one vertex $u$ and travel along that edge to the other vertex $v$, you cannot return to $u$ except by using that same edge.

Consequently $K_2$ is not Hamiltonian.

Let $n = 4$.

Let us take two vertices $u, v$ of $K_n$:

From Complete Graph is Regular, the degrees of $u$ and $v$ are given by:
 * $\map \deg u = \map \deg v = n - 1$

Let us remove the edge $u v$ which joins them.

Let the resulting graph be denoted $G$.

We now have that:
 * $\map \deg u = \map \deg v = n - 2$

From Ore's Theorem, $G$ is Hamiltonian if for each pair of non-adjacent vertices $u, v \in V$:
 * $\deg u + \deg v \ge n$

In $G$, we have that:

Hence:

Thus $G$ is Hamiltonian.

A Hamiltonian graph with another edge added is still Hamiltonian.

Thus restoring the edge $u v$ to $G$ to turn it back into $K_n$ means that $K_n$ is Hamiltonian.

When $n = 3$ we cannot use this theorem, as you then find for the resulting $G$:
 * $\deg u + \deg v = 2$

which is less than $3$.

Instead we inspect the complete graph $K_3$ and see it is is the cycle graph $C_3$


 * K3.png

The result follows from Cycle Graph is Hamiltonian.