Proof: By Euclid
(related to Lemma: Lem. 10.028.1: Finding Two Squares With Sum Also Square)
- Let the two numbers $AB$ and $BC$ be laid down.
- And let them be either (both) even or (both) odd.
- And since, if an even (number) is subtracted from an even (number) , or if an odd (number is subtracted) from an odd (number) , then the remainder is even [Prop. 9.24], [Prop. 9.26], the remainder $AC$ is thus even.
- Let $AC$ have been cut in half at $D$.
- And let $AB$ and $BC$ also be either similar plane (numbers) , or square (numbers) - which are themselves also similar plane (numbers) .
- Thus, the (number created) from (multiplying) $AB$ and $BC$, plus the square on $CD$, is equal to the square on $BD$ [Prop. 2.6].
- And the (number created) from (multiplying) $AB$ and $BC$ is square - inasmuch as it was shown that if two similar plane (numbers) make some (number) by multiplying one another then the (number so) created is square [Prop. 9.1].
- Thus, two square numbers have been found - (namely,) the (number created) from (multiplying) $AB$ and $BC$, and the (square) on $CD$ - which, (when) added (together), make the square on $BD$.
- And (it is) clear that two square (numbers) have again been found - (namely,) the (square) on $BD$, and the (square) on $CD$ - such that their difference - (namely,) the (rectangle) contained by $AB$ and $BC$ - is square whenever $AB$ and $BC$ are similar plane (numbers) .
- But, when they are not similar plane numbers, two square (numbers) have been found - (namely,) the (square) on $BD$, and the (square) on $DC$ - between which the difference - (namely,) the (rectangle) contained by $AB$ and $BC$ - is not square.
- (Which is) the very thing it was required to show.
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
