If First of Four Numbers in Geometric Sequence is Cube then Fourth is Cube

Theorem

Let $P = \tuple {a, b, c, d}$ be a geometric sequence of integers.

Let $a$ be a cube number.

Then $d$ is also a cube number.

In the words of Euclid:

If four numbers be in continued proportion, and the first be cube, the fourth will also be cube.

(The Elements: Book $\text{VIII}$: Proposition $23$)

Proof

From Form of Geometric Sequence of Integers:

$P = \tuple {k p^3, k p^2 q, k p q^2, k q^3}$

for some $k, p, q \in \Z$.

If $a = k p^3$ is a cube number it follows that $k$ is a cube number: $k = r^3$, say.

So:

$P = \tuple {r^3 p^3, r^3 p^2 q, r^3 p q^2, r^3 q^3}$

and so $d = r^3 q^3 = \paren {r q}^3$.

$\blacksquare$

Historical Note

This proof is Proposition $23$ of Book $\text{VIII}$ 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 $\text{VIII}$. Propositions