# Form of Geometric Sequence of Integers

## Theorem

Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n + 1$ consisting of integers only.

Then the $j$th term of $P$ is given by:

$a_j = k p^{n - j} q^j$

where:

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

### Corollary

Let $p$ and $q$ be integers.

Then the finite sequence $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ defined as:

$a_j = p^j q^{n - j}$

is a geometric sequence whose common ratio is $\dfrac p q$.

## Proof

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

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:

$P = \paren {a, a \dfrac q p, a \dfrac {q^2} {p^2}, \ldots, a \dfrac {q^n} {p^n} }$

All the elements of $P$ are integers, so, in particular:

$a \dfrac {q^n} {p^n} \in \Z$

We have that:

$p \perp q$
$q^n \perp p^n$

and so from Euclid's Lemma:

$p^n \divides a$

Thus:

$a = k p^n$

for some $k \in \Z$, and so:

$P = \paren {k p^n, k q p^{n - 1}, k q^2 p^{n - 2}, \ldots, k q^{n - 1} p, k q^n}$

$\blacksquare$