Form of Geometric Sequence of Integers

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

Then the $j$th term of $P$ is given by:
 * $a_j = k q^{j - 1} p^{n - j}$

where:
 * the common ratio of $P$ expressed in canonical form is $\dfrac q p$
 * $k$ is an integer.

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

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

Let $r = \dfrac q p$ be in canonical form.

Thus, by definition:
 * $p \perp q$

Let $a$ be the first term of $P$.

Then the sequence $P$ is:
 * $P = \left({a, a \dfrac q p, a \dfrac {q^2} {p^2}, \ldots, a \dfrac {q^{n - 1} } {p^{n - 1} } }\right)$

All the elements of $P$ are integers, so, in particular:
 * $a \dfrac {q^{n - 1} } {p^{n - 1} } \in \Z$

We have that:
 * $p \perp q$

From Powers of Coprime Numbers are Coprime:
 * $q^{n - 1} \perp p^{n - 1}$

and so from Euclid's Lemma:
 * $p^{n - 1} \mathop \backslash a$

Thus:
 * $a = k p^{n - 1}$

for some $k \in \Z$, and so:
 * $P = \left({k p^{n - 1}, k q p^{n - 2}, k q^2 p^{n - 3}, \ldots, k q^{n - 2} p, k q^{n - 1} }\right)$