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Commentary on Proposition IX.8¶

If $1, a, a_2, a_3, \dots$ be a geometrical progression, then $a_2, a_4, a_6, \dots$ are squares; $a_3, a_6, a_9, \dots$ are cubes; $a_6, a_{12}, \dots$ are both squares and cubes.

Since

$$\begin{aligned} 1 : a &= a : a_2, \\ a_2 &= a^2. \end{aligned}$$

And, since $a_2, a_3, a_4$ are in geometrical progression and $a_2\ (= a^2)$ is a square,

$$a_4\text{ is a square. \hfill [VIII. 22]}$$

Similarly $a_6, a_8, \dots$ are squares.

Next,

$$\begin{aligned} 1 : a &= a_2 : a_3 \\ &= a^2 : a_3, \end{aligned}$$

whence $a_3 = a^3$, a cube number.

And, since $a_3, a_4, a_5, a_6$ are in geometrical progression, and $a_3$ is a cube,

$$a_6\text{ is a cube. \hfill [VIII. 23]}$$

[Folio 0398 | p. 392]

Similarly $a_9, a_{12}, \dots$ are cubes.

Clearly then $a_6, a_{12}, a_{18}, \dots$ are both squares and cubes. The whole result is of course obvious if the geometrical progression is written, with our notation, as

$$1, a, a^2, a^3, a^4, \dots a^n.$$