Form of Geometric Sequence of Integers from One

Theorem

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

Let $a_0 = 1$.

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

$a_j = a^j$

where:

the common ratio of $Q_n$ is $a$

$a = a_1$.

Thus:

$Q_n = \tuple {1, a, a^2, \ldots, a^n}$

Proof

From Form of Geometric Sequence of Integers, the $j$th term of $Q_n$ is given by:

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

where:

the common ratio of $Q_n$ expressed in canonical form is $\dfrac q p$

$k$ is an integer.

As $a_0 = 1$ it follows that:

$1 = k p^{n - j}$

from which it follows that:

$k = 1$

$p = 1$

and the common ratio of $Q_n$ is $q$.

Thus:

$a_1 = q$

Setting $a = a_1$ yields the result as stated.

$\blacksquare$