Geometric Sequence of Integers with Integer Common Ratio

Theorem
Let $P = \left\langle{a_j}\right\rangle_{1 \mathop \le j \mathop \le n}$ be a geometric progression of length $n$ consisting entirely of integers.

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

Then $r$ is an integer iff:
 * $\forall i, j \in \left\{{1, 2, \ldots, n}\right\}, i \le j: a_i \mathop \backslash a_j$

That is, terms of $P$ divide later terms of $P$ iff $r$ is an integer.

Necessary Condition
Let $r$ be an integer.

By definition of geometric progression, the terms of $P$ are in the form:
 * $a_k = b r^{k - 1}$

where $b$ and $k$ are integers.

It follows from Integer Multiplication is Closed that $a_k$ is an integer.

Sufficient Condition
From Common Ratio in Integer Geometric Progression is Rational, $r$ is a rational number.

Let $r$ be expressed in canonical form as:
 * $r = \dfrac p q$

where, by definition of canonical form, $p \perp q$, that is, $p$ is coprime to $q$.

From Form of Geometric Progression of Integers, the terms of $P$ are in the form:
 * $(1): \quad a_j = k q^{j - 1} p^{n - j}$

Let $a_i \mathop \backslash a_j$.

From Powers of Coprime Numbers are Coprime:
 * $(2): \quad q^{j - i} \perp p^{j - i}$

Then:

But from $(2)$:
 * $q^{j - i} \perp p^{j - i}$

Thus $q^{j - i} \mathop \backslash p^{j - i}$ can happen only when $q^{j - 1} = q = 1$.

That is, when $r$ is an integer.