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