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Commentary on Proposition IX.15¶

If a, b, c be a geometrical progression in the least terms which have a given common ratio, (b + c), (c + a), (a + b) are respectively prime to a, b, c.

Let α : β be the common ratio in its lowest terms, so that the geometrical progression is

$$\alpha^2,\quad \alpha\beta,\quad \beta^2. \qquad \text{[VIII. 2]}$$

Now, α, β being prime to one another, α + β is prime to both α and β. [VII. 28]

Therefore (α + β), α are both prime to β.

Hence (α + β) α is prime to β, [VII. 24]

and therefore to $\beta^2$ ; [VII. 25]

i.e. $\alpha^2 + \alpha\beta$ is prime to $\beta^2$, or $a + b$ is prime to $c$.

Similarly, $\alpha\beta + \beta^2$ is prime to $\alpha^2$, or $b + c$ is prime to $a$.

Lastly, $\alpha + \beta$ being prime to both $\alpha$ and $\beta$, $(\alpha + \beta)^2$ is prime to $\alpha\beta$, [VII. 24, 25]

or $\alpha^2 + \beta^2 + 2\alpha\beta$ is prime to $\alpha\beta$ : whence $\alpha^2 + \beta^2$ is prime to $\alpha\beta$.

The latter inference, made in two steps, may be proved by reductio ad absurdum as Commandinus proves it.

If $\alpha^2 + \beta^2$ is not prime to $\alpha\beta$, let $x$ measure them ; therefore $x$ measures $\alpha^2 + \beta^2 + 2\alpha\beta$ as well as $\alpha\beta$ ; hence $\alpha^2 + \beta^2 + 2\alpha\beta$ and $\alpha\beta$ are not prime to one another, which is contrary to the hypothesis.