Size of Regular Graph in terms of Degree and Order

Theorem
Let $G = \struct {V, E}$ be an $r$-regular graph of order $p$.

Let $q$ denote the size of $G$.

Then:
 * $q = \dfrac {p r} 2$

when such an $r$-regular graph exists.

If an $r$-regular graph of order $p$ does exist, then $p r$ is an even integer.

Proof
Let $G$ be of order $p$, size $q$ and $r$-regular.

Then by definition $G$ has:
 * $p$ vertices each of which is of degree $r$.
 * $q$ edges.

Thus:

If $p r$ is odd, that means $p$ is odd and $r$ is odd.

That means there is an odd number of odd vertices.

From the corollary to the Handshake Lemma, this is not possible

Thus for $G$ to exist, it is necessary for $p r$ to be even.

The result follows.