Geometric Sequence of Integers with Integer Common Ratio
Theorem
Let $P = \sequence {a_j}_{1 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n$ consisting entirely of integers.
Let $r$ be the common ratio of $P$.
Then $r$ is an integer if and only if:
- $\forall i, j \in \set {1, 2, \ldots, n}, i \le j: a_i \divides a_j$
That is, terms of $P$ divide later terms of $P$ if and only if $r$ is an integer.
Proof
Necessary Condition
Let $r$ be an integer.
By definition of geometric sequence, 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.
$\Box$
Sufficient Condition
From Common Ratio in Integer Geometric Sequence 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 Sequence of Integers, the terms of $P$ are in the form:
- $(1): \quad a_j = k q^{j - 1} p^{n - j}$
Let $a_i \divides a_j$.
From Powers of Coprime Numbers are Coprime:
- $(2): \quad q^{j - i} \perp p^{j - i}$
Then:
\(\ds a_i\) | \(\divides\) | \(\ds a_j\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds k q^{i - 1} p^{n - i}\) | \(\divides\) | \(\ds k q^{j - 1} p^{n - j}\) | from $(1)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists m \in \Z: \, \) | \(\ds m \paren {k q^{j - 1} p^{n - i} } q^{j - i}\) | \(=\) | \(\ds \paren {k q^{j - 1} p^{n - i} } p^{j - i}\) | Definition of Divisor of Integer | |||||||||
\(\ds \leadsto \ \ \) | \(\ds m q^{j - i}\) | \(=\) | \(\ds p^{j - i}\) | dividing both sides by $k q^{j - 1} p^{n - i}$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds q^{j - i}\) | \(\divides\) | \(\ds p^{j - i}\) | Definition of Divisor of Integer |
But from $(2)$:
- $q^{j - i} \perp p^{j - i}$
Thus $q^{j - i} \divides p^{j - i}$ can happen only when $q^{j - 1} = q = 1$.
That is, when $r$ is an integer.
$\blacksquare$