Number Squared making Cube is itself Cube

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

Let $a^2$ be a cube number.

Then $a$ is a cube number.

In the words of Euclid:

If a number by multiplying itself make a cube number, it will itself also be cube.

(The Elements: Book $\text{IX}$: Proposition $6$)


$a^2$ 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 sequence of integers for some $m_1, m_2 \in \Z$.

By Geometric Sequences in Proportion have Same Number of Elements:

$a^2, m_3, m_4, a$

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

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


Historical Note

This proof is Proposition $6$ of Book $\text{IX}$ of Euclid's The Elements.