Number Squared making Cube is itself Cube

Theorem
Let $a \in \Z$ be an integer.

Let $a^2$ be a cube number.

Then $a$ is a cube number.

Proof
$a^2$ and $a^3$ are both cube numbers.

From Between two Cubes exist two Mean Proportionals the sequence:
 * $a^3, m_1, m_2, a^2$

is a geometric progression of integers for some $m_1, m_2 \in \Z$.

By Geometric Progressions in Proportion have Same Number of Elements:
 * $a^2, m_3, m_4, a$

is a geometric progression of integers for some $m_3, m_4 \in \Z$.

From If First of Four Numbers in Geometric Progression is Cube then Fourth is Cube it follows that $a$ is a cube number.