First Element of Geometric Sequence not dividing Second/Proof 2

Theorem

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

Let $a_0$ not be a divisor of $a_1$.

Then:

$\forall j, k \in \set {0, 1, \ldots, n}, j \ne k: a_j \nmid a_k$

That is, if the initial term of $P$ does not divide the second, no term of $P$ divides any other term of $P$.

In the words of Euclid:

If there be as many numbers as we please in continued proportion, and the first do not measure the second, neither will any other measure any other.

(The Elements: Book $\text{VIII}$: Proposition $6$)

Proof

From Form of Geometric Sequence of Integers, the terms of $P$ are in the form:

$\paren 1: \quad: a_j = k q^j p^{n - j}$

where $p \perp q$.

Let $a_0 \nmid a_1$.

That is:

\(\ds a_0\)

\(\nmid\)

\(\ds a_1\)

\(\ds \leadsto \ \ \)

\(\ds k q^n\)

\(\nmid\)

\(\ds k p q^{n - 1}\)

from $\paren 1$

\(\ds \leadsto \ \ \)

\(\ds \paren {k q^{n - 1} } q\)

\(\nmid\)

\(\ds \paren {k q^{n - 1} } p\)

rearranging

\(\ds \leadsto \ \ \)

\(\ds q\)

\(\nmid\)

\(\ds p\)

Contrapositive of Multiple of Divisor Divides Multiple

Aiming for a contradiction, suppose

$\exists i, j \in \set {0, 1, \ldots, n}, i \ne j: a_i \divides a_j$

Without loss of generality let $i < j$.

\(\ds a_i\)

\(\divides\)

\(\ds a_j\)

\(\ds \leadsto \ \ \)

\(\ds k q^i p^{n - i}\)

\(\divides\)

\(\ds k q^j p^{n - j}\)

from $\paren 1$

\(\ds \leadsto \ \ \)

\(\ds m \paren {k q^j p^{n - i} } q^{j - i}\)

\(=\)

\(\ds \paren {k q^j p^{n - i} } p^{j - i}\)

for some $m \in \Z$

\(\ds \leadsto \ \ \)

\(\ds m q^{j - i}\)

\(=\)

\(\ds p^{j - i}\)

Contrapositive of Multiple of Divisor Divides Multiple

\(\ds \leadsto \ \ \)

\(\ds q^{j - i}\)

\(\divides\)

\(\ds p^{j - i}\)

But we have that $p \perp q$.

From Powers of Coprime Numbers are Coprime:

$q^{j - i} \perp p^{j - i}$

This can happen only when $q = 1$.

This is incompatible with $q \nmid p$.

From this contradiction, the result follows.

$\blacksquare$

Historical Note

This proof is Proposition $6$ of Book $\text{VIII}$ of Euclid's The Elements.
It is the converse of Proposition $7$: First Element of Geometric Sequence that divides Last also divides Second.