Condition for Existence of Third Number Proportional to Two Numbers

Theorem
Let $a, b, c \in \Z$ be integers.

Let $\left({a, b, c}\right)$ be a geometric progression.

In order for this to be possible, both of these conditions must be true:
 * $(1): \quad a$ and $b$ cannot be coprime
 * $(2): \quad a \mathop \backslash b^2$

where $\backslash$ denotes divisibility.

Proof
Let $P = \left({a, b, c}\right)$ be a geometric progression

Then by definition their common ratio is:
 * $\dfrac b a = \dfrac c b$

From Two Coprime Integers have no Third Integer Proportional it cannot be the case that $a$ and $b$ are coprime.

Thus condition $(1)$ is satisfied.

From Form of Geometric Progression of Integers, $P$ is in the form:
 * $\left({k p^2, k p q, k q^2}\right)$

from which it can be seen that:
 * $k p^2 \mathop \backslash k^2 p^2 q^2$

demonstrating that condition $(2)$ is satisfied.