avewright/proofwiki-math huggingface dataset:
Theorem: Product of Coprime Pairs is Coprime

Let $a, b, c, d$ be integers.
Let: 
:$a \perp c, b \perp c, a \perp d, b \perp d$
where $a \perp c$ denotes that $a$ and $c$ are coprime.
Then:
:$a b \perp c d$
{{:Euclid:Proposition/VII/26}}

Proof:
Let $e = a b, f = c d$.
{{begin-eqn}}
{{eqn | l = a
      | o = \perp
      | r = c
      | c = 
}}
{{eqn | ll= \land
      | l = b
      | o = \perp
      | r = c
      | c = 
}}
{{eqn | n = 1
      | lll= \implies
      | l = a b
      | o = \perp
      | r = c
      | c = {{EuclidPropLink|book = VII|prop = 24|title = Integer Coprime to Factors is Coprime to Whole}}
}}
{{eqn | l = a
      | o = \perp
      | r = d
      | c = 
}}
{{eqn | ll= \land
      | l = b
      | o = \perp
      | r = d
      | c = 
}}
{{eqn | n = 2
      | lll= \implies
      | l = a b
      | o = \perp
      | r = d
      | c = {{EuclidPropLink|book = VII|prop = 24|title = Integer Coprime to Factors is Coprime to Whole}}
}}
{{eqn | l = a b
      | o = \perp
      | r = c
      | c = from $(1)$
}}
{{eqn | ll= \land
      | l = a b
      | o = \perp
      | r = d
      | c = from $(2)$
}}
{{eqn | lll= \implies
      | l = a b
      | o = \perp
      | r = c d
      | c = {{EuclidPropLink|book = VII|prop = 24|title = Integer Coprime to Factors is Coprime to Whole}}
}}
{{end-eqn}}
{{qed}}
{{Euclid Note|26|VII}}
Theorem: Powers of Coprime Numbers are Coprime

Let $a, b$ be coprime integers:
:$a \perp b$
Then:
:$\forall n \in \N_{>0}: a^n \perp b^n$
{{:Euclid:Proposition/VII/27}}

Proof:
Proof by induction:
Let  $a \perp b$.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$a^n \perp b^n$
$\map P 1$ is true, as this just says:
:$a \perp b$
Theorem: Numbers are Coprime iff Sum is Coprime to Both

Let $a, b$ be integers.
Then:
:$a \perp b \iff a \perp \paren {a + b}$
where $a \perp b$ denotes that $a$ and $b$ are coprime.
{{:Euclid:Proposition/VII/28}}

Proof:
=== Necessary Condition ===
Let $a \perp b$.
Suppose $a + b$ is not coprime to $a$.
Then:
:$\exists d \in \Z_{>1}: d \divides a, d \divides \paren {a + b}$
But then:
:$d \divides \paren {\paren {a + b} - a}$
and so:
:$d \divides b$
and so $a$ and $b$ are not coprime.
From this contradiction it follows that $a + b$ is coprime to $a$.
{{qed|lemma}}
Theorem: Composite Number has Prime Factor

Let $a$ be a composite number.
Then there exists a prime number $p$ such that:
:$p \divides a$
where $\divides$ means '''is a divisor of'''.
{{:Euclid:Proposition/VII/31}}

Proof:
By definition of composite number:
:$\exists b \in \Z: b \divides a$
If $b$ is a prime number then the proof is complete.
Otherwise $b$ is composite.
By definition of composite number:
:$\exists c \in \Z: c \divides b$
and so:
:$c \divides a$
Again, if $c$ is a prime number then the proof is complete.
Continuing in this manner, some prime number will be found which will divide the number before it.
This will be a divisor of $a$.
Because if not, then an infinite series of numbers will be divisors of $a$, each of which is less than the other.
This is impossible in numbers.
Hence the result.
{{qed}}
{{Euclid Note|31|VII|}}
Theorem: Natural Number is Prime or has Prime Factor

Let $a$ be a natural number greater than $1$.
Then either:
:$a$ is a prime number
or:
:there exists a prime number $p \ne a$ such that $p \divides a$
where $\divides$ denotes '''is a divisor of'''.
{{:Euclid:Proposition/VII/32}}

Proof:
By definition of composite number $a$ is either prime or composite.
Let $a$ be prime.
Then the statement of the result is fulfilled.
Let $a$ be composite.
Then by {{EuclidPropLink|book = VII|prop = 31|title = Composite Number has Prime Factor}}:
:$\exists p: p \divides a$
where $p$ is a prime number.
The result follows by Proof by Cases.
{{qed}}
{{Euclid Note|32|VII|}}
Theorem: LCM of Three Numbers

Let $a, b, c \in \Z: a b c \ne 0$.
The lowest common multiple of $a, b, c$, denoted $\lcm \set {a, b, c}$, can always be found.
{{:Euclid:Proposition/VII/36}}

Proof:
Let $d = \lcm \set {a, b}$.
This exists from {{EuclidPropLink|book = VII|prop = 34|title = Existence of Lowest Common Multiple}}.
Either $c \divides d$ or not, where $\divides$ denotes divisibility.
Suppose $c \divides d$.
But by definition of lowest common multiple, $a \divides d$ and $b \divides d$ also.
Suppose $a, b, c$ are divisors of $e$ where $e < d$.
Then $a, b$ are divisors of $e$.
That is, $e$ is a common divisor of $a$ and $b$ which is lower than $d$.
But by {{EuclidPropLink|book = VII|prop = 35|title = LCM Divides Common Multiple}}:
:$d \divides e$
which is impossible.
It follows that there can be no such $e$.
Therefore $d = \lcm \set {a, b, c}$.
Now suppose $c \nmid d$.
Let $e = \lcm \set {c, d}$.
This exists from {{EuclidPropLink|book = VII|prop = 34|title = Existence of Lowest Common Multiple}}.
Since $a$ and $b$ are both divisors of $d$, it follows that:
:$a \divides e$
:$b \divides e$
But we have that $c \divides e$ as well.
Suppose $a, b, c$ are divisors of $f$ where $f < e$.
Then $a, b$ are divisors of $f$.
So by {{EuclidPropLink|book = VII|prop = 35|title = LCM Divides Common Multiple}}:
:$d = \lcm \set {a, b} \divides f$
But also $c \divides f$.
Therefore $c$ and $d$ are divisors of $f$.
By {{EuclidPropLink|book = VII|prop = 35|title = LCM Divides Common Multiple}}:
:$e = \lcm \set {c, d} \divides f$
But this is impossible as by hypothesis $f < e$.
Therefore $a, b, c$ are divisors of no number smaller than $e$.
Therefore $e = \lcm \set {a, b, c}$.
{{qed}}
Theorem: Negative of Real Function that Increases Without Bound

Let $f: \R \to \R$ be a real function.
Then:
:$(1): \quad \ds \lim_{x \mathop \to +\infty} \map f x = +\infty \implies \lim_{x \mathop \to +\infty} -\map f x = -\infty$
:$(2): \quad \ds \lim_{x \mathop \to -\infty} \map f x = +\infty \implies \lim_{x \mathop \to -\infty} -\map f x = -\infty$

Proof:
Suppose $\ds \lim_{x \mathop \to +\infty} \map f x = +\infty$.
Then by the definition of infinite limits at infinity:
:$\forall M > 0: \exists N > 0: x > N \implies \map f x > M$
But $M > 0 \iff -M < 0$.
Likewise $\map f x > M \iff -\map f x < -M$.
Putting $M' = -M$:
:$\forall M' < 0: \exists N > 0: x > N \implies -\map f x < M'$
The result then follows from the definition of negative infinite limit at infinity.
The proof for $\ds \lim_{x \mathop \to -\infty} \map f x = +\infty$ is analagous.
{{qed}}
Theorem: Integer Divided by Divisor is Integer

Let $a, b \in \N$.
Then:
:$b \divides a \implies \dfrac 1 b \times a \in \N$
where $\divides$ denotes divisibilty.
{{:Euclid:Proposition/VII/37}}

Proof:
Let $b \divides a$.
By definition of divisibilty:
:$\exists c \in \N: c \times b = a$
Then also:
:$c \times 1 = c$
So by {{EuclidPropLink|book = VII|prop = 15|title = Alternate Ratios of Multiples}}:
:$1 : b = c : a$
Hence the result.
{{qed}}
{{Euclid Note|37|VII|}}
Theorem: Divisor is Reciprocal of Divisor of Integer

Let $a, b, c \in \Z_{>0}$.
Then:
:$b = \dfrac 1 c \times a \implies c \divides a$
where $\divides$ denotes divisibilty.
{{:Euclid:Proposition/VII/38}}

Proof:
Let $a$ have an aliquot part $b$.
Let $c$ be an integer called by the same name as the aliquot part $b$.
Then:
:$1 = \dfrac 1 c \times c$
and so by {{EuclidPropLink|book = VII|prop = 15|title = Alternate Ratios of Multiples}}:
:$ 1 : c = b : a$
Hence the result.
{{qed}}
{{Euclid Note|38|VII|}}
Theorem: Negative of Real Function that Decreases Without Bound

Let $f: \R \to \R$ be a real function.
Then:
:$(1): \quad \ds \lim_{x \mathop \to +\infty} \map f x = -\infty \implies \lim_{x \mathop \to +\infty} -\map f x = +\infty$
:$(2): \quad \ds \lim_{x \mathop \to -\infty} \map f x = -\infty \implies \lim_{x \mathop \to -\infty} -\map f x = +\infty$

Proof:
Suppose $\ds \lim_{x \mathop \to +\infty} \map f x = -\infty$.
Then by the definition of negative infinite limit at infinity:
:$\forall M < 0: \exists N > 0: x > N \implies \map f x < M$
But:
:$M < 0 \iff -M > 0$
Likewise:
:$\map f x < M \iff -\map f x > -M$
Putting $M' = -M$:
:$\forall M' > 0: \exists N > 0: x > N \implies -\map f x > M'$
The result then follows from the definition of infinite limit at infinity.
The proof for $\ds \lim_{x \mathop \to -\infty} \map f x = -\infty$ is analogous.
{{qed}}
Theorem: Construction of Geometric Sequence in Lowest Terms

It is possible to find a geometric sequence of integers $G_n$ of length $n + 1$ with a given common ratio such that $G_n$ is in its lowest terms.
{{:Euclid:Proposition/VIII/2}}

Proof:
Let $r = \dfrac a b$ be the given common ratio.
Let the required geometric sequence have a length of $4$.
Let $a^2 = c$.
Let $a b = d$.
Let $b^2 = e$.
Let:
:$a c = f$
:$a d = g$
:$a e = h$
and let:
:$b e = k$
As:
:$a^2 = c$
:$a b = d$
it follows from {{EuclidPropLink|book = VII|prop = 17|title = Multiples of Ratios of Numbers}} that:
:$\dfrac a b = \dfrac c d$
As:
:$a b = d$
:$b^2 = e$
it follows from {{EuclidPropLink|book = VII|prop = 18|title = Ratios of Multiples of Numbers}} that:
:$\dfrac a b = \dfrac d e$
As:
:$a c = f$
:$a d = g$
it follows from {{EuclidPropLink|book = VII|prop = 17|title = Multiples of Ratios of Numbers}} that:
:$\dfrac c d = \dfrac f g$
As:
:$a d = g$
:$a e = h$
it follows from {{EuclidPropLink|book = VII|prop = 17|title = Multiples of Ratios of Numbers}} that:
:$\dfrac d e = \dfrac g h$
As:
:$a e = h$
:$b e = k$
it follows from {{EuclidPropLink|book = VII|prop = 18|title = Ratios of Multiples of Numbers}} that:
:$\dfrac a b = \dfrac h k$
Putting the above together:
:$c, d, e$ are in geometric sequence with common ratio $\dfrac a b$
:$f, g, h, k$ are in geometric sequence with common ratio $\dfrac a b$
We have that $a$ and $b$ are the smallest numbers with the same ratio.
So by {{EuclidPropLink|book = VII|prop = 22|title = Numbers forming Fraction in Lowest Terms are Coprime}}:
:$a \perp b$
where $\perp$ denotes coprimality.
We also have that:
:$a^2 = c, b^2 = e$
and:
:$a c = e, b e = k$
so by {{EuclidPropLink|book = VII|prop = 27|title = Powers of Coprime Numbers are Coprime}}:
:$c \perp e$
:$f \perp k$
But from {{EuclidPropLink|book = VIII|prop = 1|title = Geometric Sequence with Coprime Extremes is in Lowest Terms}}, these are the least of those with the same common ratio.
Hence the result.
{{qed}}
Theorem: Construction of Geometric Sequence in Lowest Terms

It is possible to find a geometric sequence of integers $G_n$ of length $n + 1$ with a given common ratio such that $G_n$ is in its lowest terms.
{{:Euclid:Proposition/VIII/2}}

Proof:
Let the required length of the geometric sequence $P$ be $n$.
Let $r$ be the given common ratio.
From Common Ratio in Integer Geometric Sequence is Rational, $r$ is a rational number.
Let $r = \dfrac p q$ be in canonical form.
Thus, by definition:
:$p \perp q$
Let $a$ be the first term of $P$.
Then the sequence $P$ is:
:$P = \tuple {a, a \dfrac p q, a \dfrac {p^2} {q^2}, \ldots, a \dfrac {p^n} {q^n} }$
All the elements of $P$ are natural numbers, so, in particular:
:$a \dfrac {p^n} {q^n} \in \N$
From Powers of Coprime Numbers are Coprime:
:$p^n \perp q^n$
and so from Euclid's Lemma:
:$q^n \divides a$
Thus:
:$a = k q^n$
for some $k \in \N$, and so:
:$P = \tuple {k q^n, k p q^{n - 1}, k p^2 q^{n - 2}, \ldots, k p^{n - 1} q, k p^n}$
From Geometric Sequence with Coprime Extremes is in Lowest Terms:
:$k q^n \perp k p^n$
from which it follows that $k = 1$.
It follows that the required geometric sequence is:
:$P = \tuple {q^n, p q^{n - 1}, p^2 q^{n - 2}, \ldots, p^{n - 1} q, p^n}$
{{qed}}
Theorem: Geometric Sequence in Lowest Terms has Coprime Extremes

A geometric sequence of integers in lowest terms has extremes which are coprime.
{{:Euclid:Proposition/VIII/3}}

Proof:
Let $a_0, a_1, a_2, \ldots, a_n$ be natural numbers.
Let $\sequence {a_k}_{0 \mathop \le k \mathop \le n}$ be a geometric sequence with common ratio $r$.
Let $a_0, a_1, \ldots, a_n$ be the smallest such natural numbers.
From {{EuclidPropLink|book = VII|prop = 33|title = Least Ratio of Numbers}}, let $d_0, d_1$ be the smallest natural numbers such that $d_1 = r d_0$.
From {{EuclidPropLink|book = VIII|prop = 2|title = Construction of Geometric Sequence in Lowest Terms}} one can build a sequence of $3, 4, \ldots, n$ terms with the same property.
Let the geometric sequence so constructed with $n$ terms be $\sequence {b_k}_{0 \mathop \le k \mathop \le n}$.
From {{EuclidPropLink|book = VII|prop = 22|title = Numbers forming Fraction in Lowest Terms are Coprime}}, $d_0$ and $d_1$ are coprime.
From {{EuclidPropLink|book = VIII|prop = 2|qual = Porism|title = Construction of Geometric Sequence in Lowest Terms}}, each of the extreme terms of the intermediate geometric sequences is a power of a natural number.
From {{EuclidPropLink|book = VII|prop = 27|title = Powers of Coprime Numbers are Coprime}}, each of those extreme terms is coprime.
We have that $a_0, \ldots, a_n$ are the smallest such natural numbers such that $\sequence {a_k}_{0 \mathop \le k \mathop \le n}$ is a geometric sequence with common ratio $r$.
We have also constructed $b_0, \ldots, b_n$ to have the same property.
Thus:
: $a_0 = b_0, a_1 = b_1, \ldots, a_n = b_n$
and the result follows.
{{qed}}
Theorem: Geometric Sequence in Lowest Terms has Coprime Extremes

A geometric sequence of integers in lowest terms has extremes which are coprime.
{{:Euclid:Proposition/VIII/3}}

Proof:
Let $P$ be a geometric sequence of natural numbers of length $n$.
Let the common ratio of $P$ be expressed in canonical form as $\dfrac p q$.
From Construction of Geometric Sequence in Lowest Terms:
:$P = \paren {q^n, p q^{n - 1}, p^2 q^{n - 2}, \ldots, p^{n - 1} q, p^n}$
By definition of canonical form:
:$p \perp q$
It follows from Powers of Coprime Numbers are Coprime that:
:$p^n \perp q^n$
Hence the result.
{{qed}}
Theorem: General Associativity Theorem/Formulation 3

Let $\struct {S, \circ}$ be a semigroup.
Let $a_i$ denote elements of $S$.
Let $\circ$ be associative.
Let $n \in \Z$ be a positive integer such that $n \ge 3$.
Then all possible parenthesizations of the expression:
:$a_1 \circ a_2 \circ \cdots \circ a_n$
are equivalent.

Proof:
Let $\circ$ be associative.
It will be shown that any parenthesization of $a_1 \circ a_2 \circ \dots \circ a_n$ is equal to the '''left-associated expression''':
:$\paren {\paren {\paren {\cdots \paren {a_1 \circ a_2} \circ a_3} \circ \cdots} \circ a_n}$
The proof proceeds by induction on $n$.
Theorem: First Element of Geometric Sequence not dividing Second

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 $P$.
{{:Euclid:Proposition/VIII/6}}

Proof:
Let $P_a = \tuple {a_0, a_1, \ldots, a_n}$ be a geometric sequence of natural numbers such that $a_0 \nmid a_1$.
{{AimForCont}} $a_0 \divides a_k$ for some $k: 2 \le k \le n$.
Let $b_0, b_1, \ldots, b_k$ be the least natural numbers which have the same common ratio as $a_0, a_1, \ldots, a_k$.
These can be found by means of {{EuclidPropLink|book = VII|prop = 33|title = Least Ratio of Numbers}}.
From {{EuclidPropLink|book = VII|prop = 14|title = Proportion of Numbers is Transitive}}
:$a_0 : a_k = b_0 : b_k$
Also:
:$a_0 : a_1 = b_0 : b_1$
and so as $a_0 \nmid a_1$ it follows by {{EuclidDefLink|VII|20|Proportional}}:
:$b_0 \nmid b_1$
From One Divides all Integers it follows that:
:$b_0 \ne 1$
From {{EuclidPropLink|book = VIII|prop = 3|title = Construction of Sequence of Numbers with Given Ratios}}:
:$b_0 \perp b_k$
But as:
:$a_0 : a_k = b_0 : b_k$
it follows that:
:$a_0 \nmid a_k$
Now suppose $a_j \divides a_k$ such that $0 < j < k$.
Let $b_j, \ldots, b_k$ be the least natural numbers which have the same common ratio as $a_j, \ldots, a_k$.
These can be found by means of {{EuclidPropLink|book = VII|prop = 33|title = Least Ratio of Numbers}}.
From {{EuclidPropLink|book = VII|prop = 14|title = Proportion of Numbers is Transitive}}:
:$a_j : a_k = b_j : b_k$
The other cases follow similarly.
{{qed}}
Theorem: First Element of Geometric Sequence not dividing Second

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 $P$.
{{:Euclid:Proposition/VIII/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:
{{begin-eqn}}
{{eqn | l = a_0
      | o = \nmid
      | r = a_1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = k q^n
      | o = \nmid
      | r = k p q^{n - 1}
      | c = from $\paren 1$
}}
{{eqn | ll= \leadsto
      | l = \paren {k q^{n - 1} } q
      | o = \nmid
      | r = \paren {k q^{n - 1} } p
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = q
      | o = \nmid
      | r = p
      | c = Contrapositive of Multiple of Divisor Divides Multiple
}}
{{end-eqn}}
{{AimForCont}}
: $\exists i, j \in \set {0, 1, \ldots, n}, i \ne j: a_i \divides a_j$
{{WLOG}} let $i < j$.
{{begin-eqn}}
{{eqn | l = a_i
      | o = \divides
      | r = a_j
      | c = 
}}
{{eqn | ll= \leadsto
      | l = k q^i p^{n - i}
      | o = \divides
      | r = k q^j p^{n - j}
      | c = from $\paren 1$
}}
{{eqn | ll= \leadsto
      | l = m \paren {k q^j p^{n - i} } q^{j - i}
      | r = \paren {k q^j p^{n - i} } p^{j - i}
      | c = for some $m \in \Z$
}}
{{eqn | ll= \leadsto
      | l = m q^{j - i}
      | r = p^{j - i}
      | c = Contrapositive of Multiple of Divisor Divides Multiple
}}
{{eqn | ll= \leadsto
      | l = q^{j - i}
      | o = \divides
      | r = p^{j - i}
      | c = 
}}
{{end-eqn}}
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.
{{qed}}
Theorem: First Element of Geometric Sequence that divides Last also divides Second

Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers of length $n$.
Let $a_0$ be a divisor of $a_n$.
Then $a_0$ is a divisor of $a_2$.
{{:Euclid:Proposition/VIII/7}}

Proof:
By hypothesis, let $a_0$ be a divisor of $a_n$.
{{AimForCont}} $a_0$ is not a divisor of $a_2$.
From First Element of Geometric Sequence not dividing Second it would follow that $a_0$ does not divide $a_n$.
From this contradiction follows the result.
{{qed}}
{{Euclid Note|7|VIII}}
Theorem: Geometric Sequences in Proportion have Same Number of Elements

Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers of length $n$.
Let $r$ be the common ratio of $P$.
Let $Q = \sequence {b_j}_{0 \mathop \le j \mathop \le m}$ be a geometric sequence of integers of length $m$.
Let $r$ be the common ratio of $Q$.
Let $b_0$ and $b_m$ be such that $\dfrac {b_0} {b_m} = \dfrac {a_0} {a_n}$.
Then $m = n$.
{{:Euclid:Proposition/VIII/8}}

Proof:
Let $S = \sequence {c_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers of length $n$ such that:
:$r$ is the common ratio of $S$
:$S$ is in its lowest terms.
From Geometric Sequence in Lowest Terms has Coprime Extremes, $c_0$ is coprime to $c_n$.
Then:
:$\dfrac {c_0} {c_n} = \dfrac {a_0} {a_n} = \dfrac {b_0} {b_m}$
But:
:$c_n = r^n c_0$
and so:
:$b_m = r^n b_0$
The result follows.
{{qed}}
{{Euclid Note|8|VIII}}
Theorem: Elements of Geometric Sequence between Coprime Numbers

Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers of length $n + 1$.
Let $a_0$ be coprime to $a_n$.
Then there exist geometric sequences of integers $Q_1$ and $Q_2$ of length $n + 1$ such that:
:the initial term of both $Q_1$ and $Q_2$ is $1$
:the final term of $Q_1$ is $a_0$
:the final term of $Q_2$ is $a_n$.
{{:Euclid:Proposition/VIII/9}}

Proof:
Let the common ratio of $P$ be $r$.
By Form of Geometric Sequence of Integers with Coprime Extremes, the $j$th term of $P$ is given by:
:$a_j = q^j p^{n - j}$
such that:
:$a_0 = p^n$
:$a_n = q^n$
For $j \in \set {0, 1, \ldots, n}$, let $r_j = q^j$.
Let the finite sequence $Q_2 = \sequence {t_j}_{0 \mathop \le j \mathop \le n}$ be defined as:
:$\forall j \in \set {0, 1, \ldots, n}: t_j = p^{n - j}$
Then $Q_1$ is a geometric sequence of integers of length $n$ such that:
:$t_0 = 1$
:$t_n = p^n$
Let the finite sequence $Q_1 = \sequence {s_j}_{0 \mathop \le j \mathop \le n}$ be defined as:
:$\forall j \in \set {0, 1, \ldots, n}: s_j = q^j$
Then $Q_1$ is a geometric sequence of integers of length $n$ such that:
:$s_0 = 1$
:$s_n = q^n$
Hence the result.
{{qed}}
{{Euclid Note|9|VIII|{{EuclidNoteConverse|title = Product of Geometric Sequences from One|book = VIII|prop = 10}}}}
Theorem: Product of Geometric Sequences from One

Let $Q_1 = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ and $Q_2 = \sequence {b_j}_{0 \mathop \le j \mathop \le n}$ be geometric sequences of integers of length $n + 1$.
Let $a_0 = b_0 = 1$.
Then the sequence $P = \sequence {p_j}_{0 \mathop \le j \mathop \le n}$ defined as:
:$\forall j \in \set {0, \ldots, n}: p_j = b^j a^{n - j}$
is a geometric sequence.
{{:Euclid:Proposition/VIII/10}}

Proof:
By Form of Geometric Sequence of Integers with Coprime Extremes, the $j$th term of $Q_1$ is given by:
:$a_j = a^j$
and of $Q_1$ is given by:
:$b_j = b^j$
Let the terms of $P$ be defined as:
:$\forall j \in \left\{{0, 1, \ldots, n}\right\}: p_j = b^j a^{n - j}$
Then from Form of Geometric Sequence of Integers it follows that $P$ is a geometric sequence.
{{qed}}
{{Euclid Note|10|VIII|{{EuclidNoteConverse|title = Elements of Geometric Sequence between Coprime Numbers|book = VIII|prop = 9}}}}
Theorem: Powers of Elements of Geometric Sequence are in Geometric Sequence

Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers.
Then the sequence $Q = \sequence {b_j}_{0 \mathop \le j \mathop \le n}$ defined as:
:$\forall j \in \set {0, 1, \ldots, n}: b_j = a_j^m$
where $m \in \Z_{>0}$, is a geometric sequence.
{{:Euclid:Proposition/VIII/13}}

Proof:
From Form of Geometric Sequence of Integers, the $j$th term of $P$ is given by:
:$a_j = k q^j p^{n - j}$
Thus the $j$th term of $Q$ is given by:
:$b_j = k^m \paren {q^m}^j \paren {p^m}^{n - j}$
From Form of Geometric Sequence of Integers, this is a geometric sequence.
{{qed}}
{{Euclid Note|13|VIII}}
Theorem: Number divides Number iff Square divides Square

Let $a, b \in \Z$.
Then:
:$a^2 \divides b^2 \iff a \divides b$
where $\divides$ denotes integer divisibility.
{{:Euclid:Proposition/VIII/14}}

Proof:
From Between two Squares exists one Mean Proportional:
:$\tuple {a^2, ab, b^2}$
is a geometric sequence.
Let $a, b \in \Z$ such that $a^2 \divides b^2$.
Then from First Element of Geometric Sequence that divides Last also divides Second:
:$a^2 \divides a b$
Thus:
{{begin-eqn}}
{{eqn | l = a^2
      | o = \divides
      | r = a b
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists k \in \Z
      | l = k a^2
      | r = a b
      | c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = k a
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a
      | o = \divides
      | r = b
      | c = {{Defof|Divisor of Integer}}
}}
{{end-eqn}}
{{qed|lemma}}
Let $a \divides b$.
Then:
{{begin-eqn}}
{{eqn | l = a
      | o = \divides
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists k \in \Z
      | l = k a
      | r = b
      | c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = k a^2
      | r = a b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = k a b
      | r = b^2
      | c = {{Defof|Geometric Sequence of Integers}}
}}
{{eqn | ll= \leadsto
      | l = k^2 a^2
      | r = b^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^2
      | o = \divides
      | r = b^2
      | c = {{Defof|Divisor of Integer}}
}}
{{end-eqn}}
{{qed}}
{{Euclid Note|14|VIII}}