Geometric Sequences in Proportion have Same Number of Elements

Theorem
Let $P = \left\langle{a_j}\right\rangle_{0 \mathop \le j \mathop \le n}$ be a geometric progression of integers of length $n$.

Let $r$ be the common ratio of $P$.

Let $Q = \left\langle{b_j}\right\rangle_{0 \mathop \le j \mathop \le m}$ be a geometric progression of integers of length $m$.

Let $r$ be the common ratio of $Q$.

Let $b_0$ and $b_m$ be such that $\dfrac {b_0} {b_m} = \dfrac {a_0} {a_n}$.

Then $m = n$.

Proof
Let $S = \left\langle{c_j}\right\rangle_{0 \mathop \le j \mathop \le n}$ be a geometric progression of integers of length $n$ such that:
 * $r$ is the common ratio of $S$
 * $S$ is in its lowest terms.

From Geometric Progression in Lowest Terms has Coprime Extremes, $c_0$ is coprime to $c_n$.

Then:
 * $\dfrac {c_0} {c_n} = \dfrac {a_0} {a_n} = \dfrac {b_0} {b_m}$

But:
 * $c_n = r^n c_0$

and so:
 * $b_m = r^n b_0$

The result follows.