Geometric Sequence of Integers with Integer Common Ratio

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$