Number does not divide Number iff Cube does not divide Cube

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Theorem

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

Then:

$a \nmid b \iff a^3 \nmid b^3$

where $a \nmid b$ denotes that $a$ is not a divisor of $b$.


In the words of Euclid:

If a cube number do not measure a cube number, neither will the side measure the side; and, if the side do not measure the side, neither will the cube measure the cube.

(The Elements: Book $\text{VIII}$: Proposition $17$)


Proof

Let $a \nmid b$.

Aiming for a contradiction, suppose:

$a^3 \mathrel \backslash b^3$

where $\backslash$ denotes divisibility.

Then by Number divides Number iff Cube divides Cube:

$a \mathrel \backslash b$

From Proof by Contradiction it follows that $a^2 \mathrel \backslash b^2$ is false.

Thus $a^3 \nmid b^3$.

$\Box$


Let $a^3 \nmid b^3$.

Aiming for a contradiction, suppose

$a \mathrel \backslash b$

Then by Number divides Number iff Cube divides Cube:

$a^3 \mathrel \backslash b^3$

From Proof by Contradiction it follows that $a \mathrel \backslash b$ is false.

Thus $a \nmid b$.

$\blacksquare$


Historical Note

This theorem is Proposition $17$ of Book $\text{VIII}$ of Euclid's The Elements.


Sources