Commentary on Proposition IX.9¶
Back to Proposition IX.9.
If $1, a^2, a_2, a_3, a_4, \dots$ are in geometrical progression, $a_2, a_3, a_4, \dots$ are all squares ; and, if $1, a^3, a_2, a_3, a_4, \dots$ are in geometrical progression, $a_2, a_3, \dots$ are all cubes.
By IX. 8, $a_2, a_4, a_6, \dots$ are all squares. And, $a^2, a_2, a_3$ being in geometrical progression, and $a^2$ being a square,
$$a_3\text{ is a square. \hfill [VIII. 22]}$$
For the same reason $a_5, a_7, \dots$ are all squares.
By IX. 8, $a_3, a_6, a_9, \dots$ are all cubes.
Now
$$1 : a^3 = a^3 : a_2.$$
Therefore
$$a_2 = a^3 \cdot a^3\text{, which is a cube, by IX. 3.}$$
And, $a^3, a_2, a_3, a_4$ being in geometrical progression, and $a^3$ being cube,
$$a_4\text{ is cube. \hfill [VIII. 23]}$$
Similarly we prove that $a_5$ is cube, and so on.
The results are of course obvious in our notation, the series being
$$\begin{aligned} (1)&\quad 1, a^2, a^4, a^6, \dots a^{2n}, \\ (2)&\quad 1, a^3, a^6, a^9, \dots a^{3n}. \end{aligned}$$