If First of Three Numbers in Geometric Sequence is Square then Third is Square

Theorem

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

Let $a$ be a square number.

Then $c$ is also a square number.

In the words of Euclid:

If three numbers be in continued proportion, and the first be square, the third will also be square.

(The Elements: Book $\text{VIII}$: Proposition $22$)

Proof

From Form of Geometric Sequence of Integers:

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

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

If $a = k p^2$ is a square number it follows that $k$ is a square number: $k = r^2$, say.

So:

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

and so $c = r^2 q^2 = \paren {r q}^2$.

$\blacksquare$

Historical Note

This proof is Proposition $22$ 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