Square of Cube Number is Cube

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Let $a \in \N$ be a natural number.

Let $a$ be a cube number.

Then $a^2$ is also a cube number.

In the words of Euclid:

If a cube number by multiplying itself make some number the product will be cube.

(The Elements: Book $\text{IX}$: Proposition $3$)

Proof 1

By the definition of cube number:

$\exists k \in \N: k^3 = a$


\(\ds a^2\) \(=\) \(\ds \paren {k^3}^2\)
\(\ds \) \(=\) \(\ds k^6\)
\(\ds \) \(=\) \(\ds \paren {k^2}^3\)


$\exists r = k^2 \in \N: a = r^3$

Hence the result by definition of cube number.


Proof 2

From Cube Number multiplied by Cube Number is Cube, if $a$ and $b$ are cube numbers then $a b$ is a cube number.

The result follows by setting $b = a$.


Historical Note

This proof is Proposition $3$ of Book $\text{IX}$ of Euclid's The Elements.
The proof as given here is not that given by Euclid, as the latter is unwieldy and of limited use.