Numbers between which exists one Mean Proportional are Similar Plane

Theorem
Let $a, b \in \Z$ such that the geometric mean is an integer.

Then $a$ and $b$ are similar plane numbers.

Proof
Let the geometric mean of $a$ and $b$ be an integer $m$.

Then, in the language of, $m$ is a mean proportional of $a$ and $b$.

Thus:
 * $\left({a, m, b}\right)$

is a geometric progression.

From Form of Geometric Progression of Integers:
 * $\exists k, p, q \in \Z: a = k p^2, b = k q^2$

So $a$ and $b$ are plane numbers whose sides are:
 * $k p$ and $p$

and
 * $k q$ and $q$

respectively.

Then:
 * $\dfrac {k p} {k q} = \dfrac p q$

demonstrating that $a$ and $b$ are similar plane numbers by definition.