Last Element of Geometric Sequence with Coprime Extremes has no Integer Proportional as First to Second

Theorem
Let $G_n = \left\langle{a_j}\right\rangle_{0 \mathop \le j \mathop \le n}$ be a geometric progression of integers such that $a_0 \ne 1$.

Let $a_0 \perp a_n$, where $\perp$ denotes coprimality.

Then there does not exist an integer $b$ such that:
 * $\dfrac {a_0} {a_1} = \dfrac {a_n} b$

Proof
Suppose $\dfrac {a_0} {a_1} = \dfrac {a_n} b$.

Then:
 * $\dfrac {a_0} {a_n} = \dfrac {a_1} b$

By Ratios of Fractions in Lowest Terms:
 * $a_0 \mathop \backslash a_1$

where $\backslash$ denotes divisibility.

From Divisibility of Elements in Geometric Progression of Integers:
 * $a_0 \mathop \backslash a_n$

But $a_0 \perp a_n$.

From this contradiction it follows that there can be no such $b$.