Number Squared making Cube is itself Cube
Jump to navigation
Jump to search
Theorem
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 $6$)
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 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.
$\blacksquare$
Historical Note
This proof is Proposition $6$ of Book $\text{IX}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{IX}$. Propositions