nyuuzyou/wikis huggingface dataset:
Factorization of Continuous Linear Transformation between Topological Vector Spaces

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau_X}$ and $\struct {Y, \tau_X}$ be topological vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $N$ be a vector subspace of $X$ with $N \subseteq \ker T$. 
Let $\struct {X/N, \tau_N}$ be the quotient topological vector space of $X$ modulo $N$.
Let $\pi : X \to X/N$ be the quotient mapping.
Then $T$ is continuous :
:there exists a continuous linear transformation $\Lambda : X/N \to Y$ such that:
::$T x = \map \Lambda {\map \pi x}$
:for each $x \in X$. 
== Proof ==
=== Necessary Condition ===
Suppose that $T$ is continuous.
From Condition for Mapping from Quotient Vector Space to be Well-Defined, there exists a linear transformation $\Lambda : X/N \to Y$ such that $T x = \map \Lambda {\map \pi x}$ for each $x \in X$.
It remains to show that $\Lambda$ is continuous.
Let $V \subseteq Y$.
We want to show that:
:$\Lambda^{-1} \sqbrk V = \pi \sqbrk {T^{-1} \sqbrk V}$
Let $\map \pi x \in \Lambda^{-1} \sqbrk V$.
Then, we have $\map \Lambda {\map \pi x} \in V$.
That is, $T x \in V$.
So, we have $x \in T^{-1} \sqbrk V$. 
Then, we have $\map \pi x \in \pi \sqbrk {T^{-1} \sqbrk V}$. 
So we have that $\Lambda^{-1} \sqbrk V \subseteq \pi \sqbrk {T^{-1} \sqbrk V}$.
Now let $\map \pi x \in \pi \sqbrk {T^{-1} \sqbrk V}$.
Then there exists $z \in T^{-1} \sqbrk V$ such that $\map \pi x = \map \pi z$. 
Then we have, from Quotient Mapping is Linear Transformation, we have $\map \pi {x - z} = {\mathbf 0}_{X/N}$.
So from Kernel of Quotient Mapping, we have $x - z \in N$.
So, we have $x - z \in \ker T$, and so $T x = T z$.
We have $T z \in V$ and $T x = \map \Lambda {\map \pi x}$, so:
:$\map \Lambda {\map \pi x} \in V$
So, we have $\map \pi x \in \Lambda^{-1} \sqbrk V$.
So we obtain that $\pi \sqbrk {T^{-1} \sqbrk V} \subseteq \Lambda^{-1} \sqbrk V$.
Hence we have shown:
:$\Lambda^{-1} \sqbrk V = \pi \sqbrk {T^{-1} \sqbrk V}$
for all $V \subseteq Y$.
Now let $V$ be open in $\struct {Y, \tau_Y}$. 
Since $T$ is continuous, $T^{-1} \sqbrk V$ is open in $\struct {X, \tau_X}$. 
From Quotient Mapping on Quotient Topological Vector Space is Open Mapping, $\pi$ is an open mapping.
Hence, $\pi \sqbrk {T^{-1} \sqbrk V}$ is open in $\struct {X/N, \tau_N}$.
We have shown that $\Lambda^{-1} \sqbrk V = \pi \sqbrk {T^{-1} \sqbrk V}$, so:
:$\Lambda^{-1} \sqbrk V$ is open in $\struct {X/N, \tau_N}$.
Since $V$ was an arbitrary open set in $\struct {Y, \tau_Y}$, we have that:
:$\Lambda : X/N \to Y$ is continuous.
=== Sufficient Condition ===
Suppose that there exists a continuous linear transformation $\Lambda : X/N \to Y$ such that:
:$T x = \map \Lambda {\map \pi x}$
for each $x \in X$. 
From the definition of the quotient topology, $\pi$ is continuous.
Since $\Lambda$ is continuous, we have that $\Lambda \circ \pi = T$ is continuous from Composite of Continuous Mappings is Continuous.
Category:Quotient Topological Vector Spaces
Category:Topological Vector Spaces
Factorization of Limit Ordinals

== Theorem ==
Let $x$ be a limit ordinal.
Then:
:$x = \paren {\omega \times y}$ for some $y \in \On$
where $\omega$ is the minimally inductive set.
== Proof ==
By the Division Theorem for Ordinals:
:$x = \paren {\omega \times y} + z$
for some unique $y$ and $z \in \omega$.
 $z \ne 0$.
Because $z \in \omega$, $z$ is not a limit ordinal.
Therefore, by the definition of limit ordinal:
:$z = w^+$
for some $w \in \omega$.
But this means that:
This means that $x$ is the successor of some ordinal.
Hence $x$ cannot be a limit ordinal.
But this contradicts the assumption that $x$ is a limit ordinal.
It follows that $z = 0$.
Therefore:
Category:Ordinal Arithmetic
Factorization of Natural Numbers within 4 n + 1 not Unique

== Theorem ==
Let:
:$S = \set {4 n + 1: n \in \N} = \set {1, 5, 9, 13, 17, \ldots}$
be the set of natural numbers of the form $4 n + 1$.
Then not all elements of $S$ have a complete factorization by other elements of $S$ which is unique.
== Proof ==
Proof by Counterexample:
Consider the number:
:$m = 693 = 3^2 \times 7 \times 11$
Thus:
:$m = 9 \times 77 = 21 \times 33$
We have that:
The divisors of these numbers are as follows:
Thus $693$ has two different complete factorizations into elements of $S$.
Hence the result.
== Sources ==
* : Chapter $7$: Patterns in Numbers: Euclid
Category:Number Theory
Factorization of Open Linear Transformation between Topological Vector Spaces

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau_X}$ and $\struct {Y, \tau_X}$ be topological vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $N$ be a vector subspace of $X$ with $N \subseteq \ker T$. 
Let $\struct {X/N, \tau_N}$ be the quotient topological vector space of $X$ modulo $N$.
Let $\pi : X \to X/N$ be the quotient mapping.
Then $T$ is open :
:there exists a open linear transformation $\Lambda : X/N \to Y$ such that:
::$T x = \map \Lambda {\map \pi x}$
:for each $x \in X$. 
== Proof ==
=== Necessary Condition ===
Suppose that $T$ is open.
From Condition for Mapping from Quotient Vector Space to be Well-Defined, there exists a linear transformation $\Lambda : X/N \to Y$ such that $T x = \map \Lambda {\map \pi x}$ for each $x \in X$.
It remains to show that $\Lambda$ is open.
Let $E \subseteq X/N$. 
We want to show that:
:$\Lambda \sqbrk E = T \sqbrk {\pi^{-1} \sqbrk E}$
Let $y \in \Lambda \sqbrk E$.
So there exists $\map \pi x \in E$ such that $\Lambda {\map \pi x} = y$. 
That is, $T x = y$. 
Since $\map \pi x \in E$, we have $x \in \pi^{-1} \sqbrk E$.
So we have $y \in T \sqbrk {\pi^{-1} \sqbrk E}$. 
Hence we have $\Lambda \sqbrk E \subseteq T \sqbrk {\pi^{-1} \sqbrk E}$.
Now let $z \in T \sqbrk {\pi^{-1} \sqbrk E}$.
Then there exists $x \in \pi^{-1} \sqbrk E$ such that $T x = z$. 
Then we have $z = \map \Lambda {\map \pi x}$.
Since $x \in \pi^{-1} \sqbrk E$, we have $\map \pi x \in E$.
So, since $z = \map \Lambda {\map \pi x}$, we have $z \in \Lambda \sqbrk E$.
So we have $T \sqbrk {\pi^{-1} \sqbrk E} \subseteq \Lambda \sqbrk E$.
So we conclude that $\Lambda \sqbrk E = T \sqbrk {\pi^{-1} \sqbrk E}$.
Now let $E$ be an open set in $\struct {X/N, \tau_N}$. 
By the definition of the quotient topology, $\pi$ is continuous.
So $\pi^{-1} \sqbrk E$ is open in $\struct {X, \tau_X}$. 
Since $T$ is open, we have that $T \sqbrk {\pi^{-1} \sqbrk E}$ is open in $\struct {Y, \tau_Y}$.
That is, $\Lambda \sqbrk E$ is open in $\struct {Y, \tau_Y}$.
Since $E$ was an arbitrary open set in $\struct {X/N, \tau_N}$, we have that $\Lambda$ is open.
=== Sufficient Condition ===
Suppose that there exists a open linear transformation $\Lambda : X/N \to Y$ such that:
:$T x = \map \Lambda {\map \pi x}$
for each $x \in X$.
From Quotient Mapping on Quotient Topological Vector Space is Open Mapping, $\pi$ is open.
Since $\Lambda$ is open, we obtain that $\Lambda \circ \pi = T$ is open by Composition of Open Mappings is Open Mapping.
Category:Open Mappings
Category:Topological Vector Spaces
Category:Quotient Topological Vector Spaces
Factors in Absolutely Convergent Product Converge to One

== Theorem ==
Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ be absolutely convergent.
Then:
:$a_n \to 0$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Factors in Convergent Product Converge to One
* Absolutely Convergent Product is Convergent (which relies on this result)
Category:Convergence
Category:Infinite Products
Category:Factors in Absolutely Convergent Product Converge to One
Factors in Absolutely Convergent Product Converge to One/Proof 1

== Theorem ==
== Proof ==
We have that $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent.
By the definition of absolutely convergent product, $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} }$ is convergent.
From the above, we see that $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} } > \sum_{i \mathop = 1}^\infty \size {a_i}$
And since $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} }$ is convergent, then $\ds \sum_{i \mathop = 1}^\infty \size {a_i}$ is convergent
Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent.
By Terms in Convergent Series Converge to Zero:
:$a_n \to 0$
Category:Factors in Absolutely Convergent Product Converge to One
Factors in Absolutely Convergent Product Converge to One/Proof 2

== Theorem ==
== Proof ==
We have that $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent.
Let $b_n = \paren {1 + a_n}$
Then $\ds \prod_{n \mathop = 1}^\infty b_n$ is absolutely convergent.
From Absolutely Convergent Product is Convergent, $\ds \prod_{n \mathop = 1}^\infty b_n$ is convergent.
By Factors in Convergent Product Converge to One:
:$b_n \to 1$
Thus:
:$a_n \to 0$
Category:Factors in Absolutely Convergent Product Converge to One
Factors in Convergent Product Converge to One

== Theorem ==
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ be convergent.
Then $a_n \to 1$.
== Proof ==
By definition of convergent product, there exists $n_0 \in \N$ such that:
:$a_n \ne 0$ for $n \ge n_0$
:the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty a_n$ has a nonzero limit.
Let $p_n$ denote the $n$th partial product of $\ds \prod_{n \mathop = n_0}^\infty a_n$.
For $n > n_0$:
:$a_n = \dfrac {p_n} {p_{n - 1} }$
By the Combination Theorem for Sequences:
:$a_n \to 1$
== Also see ==
* Factors in Absolutely Convergent Product Converge to One
Category:Convergence
Category:Infinite Products
Factors in Uniformly Convergent Product Converge Uniformly to One

== Theorem ==
Let $X$ be a set.
Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let $\sequence {f_n}$ be a sequence of bounded mappings $f_n: X \to \mathbb K$.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge uniformly on $X$.
Then $f_n$ converges uniformly to $1$.
== Proof ==
Follows directly from Uniformly Convergent Product Satisfies Uniform Cauchy Criterion.
== Also see ==
* Tail of Uniformly Convergent Product Converges Uniformly to One
Category:Uniform Convergence
Category:Infinite Products
Factors of Binomial Coefficient

== Theorem ==
For all $r \in \R, k \in \Z$:
:$k \dbinom r k = r \dbinom {r - 1} {k - 1}$
where $\dbinom r k$ is a binomial coefficient.
Hence:
:$\dbinom r k = \dfrac r k \dbinom {r - 1} {k - 1}$ (if $k \ne 0$)
and:
:$\dfrac 1 r \dbinom r k = \dfrac 1 k \dbinom {r - 1} {k - 1}$ (if $k \ne 0$ and $r \ne 0$)
=== Corollary 1 ===
=== Corollary 2 ===
=== Complex Numbers ===
== Proof ==
If $k = 0$ then $\dbinom r k = r \dbinom {r - 1} {k - 1} = 0$ by definition.
Otherwise:
If $k \ne 0$, we can divide both sides of:
:$k \dbinom r k = r \dbinom {r - 1} {k - 1}$
by $k$ to obtain:
:$\dbinom r k = \dfrac r k \dbinom {r - 1} {k - 1}$
If $k \ne 0$ and $r \ne 0$, we can divide both sides of:
:$\dbinom r k = \dfrac r k \dbinom {r - 1} {k - 1}$
by $r$ to obtain:
:$\dfrac 1 r \dbinom r k = \dfrac 1 k \dbinom {r - 1} {k - 1}$
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $3 \ \text{(c)}$
* : $\S 1.2.6$: Binomial Coefficients: $\text{C}$
Category:Binomial Coefficients
Factors of Binomial Coefficient/Complex Numbers

== Theorem ==
For all $z, w \in \C$ such that it is not the case that $z$ is a negative integer and $w$ an integer:
:$\dbinom z w = \dfrac z w \dbinom {z - 1} {w - 1}$
where $\dbinom z w$ is a binomial coefficient.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $42$ (Solution)
Category:Binomial Coefficients
Factors of Binomial Coefficient/Corollary 1

== Theorem ==
For all $r \in \R, k \in \Z$:
:$\paren {r - k} \dbinom r k = r \dbinom {r - 1} k$
from which:
:$\dbinom r k = \dfrac r {r - k} \dbinom {r - 1} k$ (if $r \ne k$)
== Proof ==
Then:
:$\dbinom r k = \dfrac r {r - k} \dbinom {r - 1} k$
follows from the 
:$\paren {r - k} \dbinom r k = r \dbinom {r - 1} k$
by dividing both sides by $r - k$.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $\text{C}$
Category:Binomial Coefficients
Factors of Composition Series for Prime Power Group

== Theorem ==
Let $G$ be a group such that $\order G = p^n$ where $p$ is a prime number.
Then $G$ has a composition series in which each factor group is cyclic of order $p$.
== Proof ==
From Composition Series of Group of Prime Power Order‎, $G$ has a sequence of subgroups:
:$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$
such that $\order {G_k} = p^k$, $G_k \lhd G_{k + 1}$ and $G_{k + 1} / G_k$ is cyclic and of order $p$.
From Prime Group is Simple it follows that $G_{k + 1} / G_k$ is a simple group for all $k$.
From Prime Group is Cyclic it follows that $G_{k + 1} / G_k$ is cyclic.
The result follows from Condition for Composition Series.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 74$
Category:Composition Series
Category:Cyclic Groups
Factors of Difference of Two Even Powers

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$\ds x^{2 n} - y^{2 n} = \paren {x - y} \paren {x + y} \prod_{k \mathop = 1}^{n - 1} \paren {x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}$
== Proof ==
From Factorisation of $z^n - a$:
:$\ds z^{2 n} - y^{2 n} = \prod_{k \mathop = 0}^{2 n - 1} \paren {x - \alpha^k y}$
where $\alpha$ is a primitive complex $2 n$th roots of unity, for example:
From Complex Roots of Unity occur in Conjugate Pairs:
:$U_{2 n} = \set {1, \tuple {\alpha, \alpha^{2 n - 1} }, \tuple {\alpha^2, \alpha^{2 n - 2} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k} }, \ldots, \tuple {\alpha^{n - 1}, \alpha^{n + 1} }, -1}$
where $U_{2 n}$ denotes the complex $2 n$th roots of unity:
:$U_{2 n} = \set {z \in \C: z^{2 n} = 1}$
The case $k = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $x - y$.
The case $k = n$ is taken care of by setting $\alpha^k = -1$, from whence we have the factor $x + y$.
Taking the product of each of the remaining factors of $x^{2 n} - y^{2 n}$ in pairs:
Hence the result.
== Also see ==
* Factors of Difference of Two Odd Powers
== Sources ==
* : $\S 2$: Special Products and Factors: $2.22$
* : $\S 2$: Special Products and Factors: $2.22.$
Category:Difference of Two Powers
Factors of Difference of Two Odd Powers

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
== Proof ==
From Factorisation of $z^n - a$:
:$x^{2 n + 1} - y^{2 n + 1} = \ds \prod_{k \mathop = 0}^{2 n} \paren {x - \alpha^k y}$
where $\alpha$ is a primitive complex $2 n + 1$th roots of unity, for example:
From Complex Roots of Unity occur in Conjugate Pairs:
:$U_{2 n + 1} = \set {1, \tuple {\alpha, \alpha^{2 n} }, \tuple {\alpha^2, \alpha^{2 n - 1} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k + 1} }, \ldots, \tuple {\alpha^n, \alpha^{n + 1} } }$
where $U_{2 n + 1}$ denotes the complex $2 n + 1$th roots of unity:
:$U_{2 n + 1} = \set {z \in \C: z^{2 n + 1} = 1}$
The case $k = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $x - y$.
Taking the product of each of the remaining factors of $x^{2 n + 1} - y^{2 n + 1}$ in pairs:
Hence the result.
== Also see ==
* Factors of Difference of Two Even Powers
== Sources ==
* : $\S 2$: Special Products and Factors: $2.20$
* : $\S 2$: Special Products and Factors: $2.20.$
Category:Difference of Two Powers
Factors of Group Direct Product are not Subgroups

== Theorem ==
Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups.
Let $\struct {G \times H, \circ}$ be the group direct product of $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Then neither $\struct {G, \circ_1}$ nor $\struct {H, \circ_2}$ is a subgroup of $\struct {G \times H, \circ}$.
== Proof ==
A subgroup is by definition a subset which is a group.
But neither $G$ nor $H$ are actually subsets of their cartesian product $G \times H$.
Hence the result.
Category:Group Direct Products
Category:Subgroups
Factors of Mersenne Number M67

== Theorem ==
The Mersenne number $M_{67}$ has the factors:
:$193 \, 707 \, 721$
:$761 \, 838 \, 257 \, 287$
== Proof ==
First we calculate $M_{67}$, which is $2^{67} - 1$:
:$2^2 = 2 \times 2 = 4$
as follows:
 256
 x 256
------
 1 536
12 800
51 200
------
65 536
------
as follows:
 65 536
 x 65 536
 -------
 393 216
 1 966 080
 32 768 000
 327 680 000
3 932 160 000
-------------
4 294 967 296
-------------
as follows:
 4 294 967 296
 x 4 294 967 296
 --------------
 25 769 803 776
 386 547 056 640
 858 993 459 200
 30 064 771 072 000
 257 698 037 760 000
 3 865 470 566 400 000
 17 179 869 184 000 000
 386 547 056 640 000 000
 858 993 459 200 000 000
17 179 869 184 000 000 000
--------------------------
18 446 744 073 709 551 616
--------------------------
 1 233 444 664 532 221 1
as follows:
 18 446 744 073 709 551 616
x 8
---------------------------
147 573 952 589 676 412 928
---------------------------
 63 355 33 525 74 414 14
Subtracting $1$ to get $M_{67}$:
Now to calculate $193 \, 707 \, 721 \times 761 \, 838 \, 257 \, 287$:
 193 707 721
 x 761 838 257 287
--------------------------
 1 355 954 047
 15 496 617 680
 38 741 544 200
 1 355 954 047 000
 9 685 386 050 000
 38 741 544 200 000
 1 549 661 768 000 000
 5 811 231 630 000 000
 154 966 176 800 000 000
 193 707 721 000 000 000
 11 622 463 260 000 000 000
135 595 404 700 000 000 000
---------------------------
147 573 952 589 676 412 927
---------------------------
 1 223 254 435 432 22 1
Thus:
as we were to demonstrate.
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.12$: Mersenne ()
Category:Mersenne Numbers
Category:Classic Problems
Factors of Mersenne Number M67/Historical Note

== Historical Note on Factors of Mersenne Number $M_{67}$ ==
While had demonstrated in $1876$ that $M_{67}$ is composite, he had not established what its divisors are.
The Factors of Mersenne Number $M_{67}$ were demonstrated by in a famously dramatic presentation On The Factorization of Large Numbers to a meeting of the American Mathematical Society in October $1903$. 
When called to give his lecture, he walked to the blackboard, and worked out the calculation, longhand, of $2^{67}$. Then he carefully subtracted $1$.
Moving to another area of the board, he then multiplied out $193, 707, 721 \times 761, 838, 257, 287$.
The numbers matched.
 returned to his seat to thunderous applause, having delivered the only lecture in history in which not a word was spoken.
When asked how long it had taken him to find these factors, he reportedly replied:
:Three years of Sundays.
It is noted that had originally listed $M_{67}$ as one of the integers of the form $2^p - 1$ to be prime.
== Sources ==
* : Preface
* 
* : $2^{67} - 1$
* : Chapter $\text {A}.12$: Mersenne ()
* : $2^{67} - 1$
Category:Historical Notes
Factors of Polynomial with Integer Coefficients have Integer Coefficients

== Theorem ==
Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map h X \in \Q \sqbrk X$ have coefficients all of which are integers.
Let it be possible to express $\map h X$ as:
:$\map h X = \map f X \, \map g X$
where $\map f X, \map g X \in \Q \sqbrk X$.
Then it is also possible to express $\map h X$ as:
:$\map h X = \map {f'} X \, \map {g'} X$
where:
:$\map {f'} X, \map {g'} X \in \Q \sqbrk X$
:the coefficients of $\map {f'} X$ and $\map {g'} X$ are all integers
:$\map {f'} X = a \map f X$ and $\map {g'} X = b \map f X$, for $a, b \in \Q$.
== Proof ==
Let $\cont h$ denote the content of $\map h X$.
From Polynomial has Integer Coefficients iff Content is Integer:
:$\cont h \in \Z$
Let $\map h X = \map f X \, \map g X$ as suggested.
Then from Rational Polynomial is Content Times Primitive Polynomial:
From the above, $\map {f'} X$ and $\map {g'} X$ are primitive.
Hence by definition:
:$\cont {f'} = \cont {g'} = 1$
From Polynomial has Integer Coefficients iff Content is Integer, both $\map {f'} X$ and $\map {g'} X$ have coefficients which are all integers.
We also have by definition of content that $\cont f$ and $\cont g$ are rational numbers.
The result follows.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Theorem $63$
Category:Polynomial Theory
Factors of Repunits

== Theorem ==
The prime factors of the first few repunits are as follows:
:
$R_n$	Factors of $n$ 	Also see
$2$ 	$11$ 	
$3$ 	$3 \times 37$ 	
$4$ 	$11 \times 101$ 	
$5$ 	$41 \times 271$ 	
$6$ 	$3 \times 7 \times 11 \times 13 \times 37$ 	
$7$ 	$239 \times 4649$ 	
$8$ 	$11 \times 73 \times 101 \times 137$ 	
$9$ 	$3^2 \times 37 \times 333 \, 667$ 	
$10$ 	$11 \times 41 \times 271 \times 9091$ 	
$11$ 	$21 \, 649 \times 513 \, 239$ 	
$12$ 	$3 \times 7 \times 11 \times 13 \times 37 \times 101 \times 9901$	
$13$ 	$53 \times 79 \times 265 \, 371 \, 653$ 	
== Sources ==
* : Tables: $7$ The Factors of the Repunits from $11$ to $R40$
* : Tables: $7$ The Factors of the Repunits from $11$ to $R40$
Category:Repunits
Factors of Solvable Group are Prime

== Theorem ==
Let $G$ be a solvable group.
Let $\HH$ be a composition series of $G$.
Then all factor groups of $\HH$ must be prime.
== Proof ==
By definition, a composition series is a normal series whose factor groups are all simple.
A solvable group, by definition, is one which has a composition series whose factor groups are all cyclic.
From Cyclic Group is Simple iff Prime, it follows that all the factor groups of a composition series of a solvable group must all be prime.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 75$. Solvable Groups
Category:Solvable Groups
Category:Prime Groups
Factors of Sum of Two Even Powers

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$x^{2 n} + y^{2 n} = \ds \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {\paren {2 k - 1} \pi} {2 n} + y^2}$
== Proof ==
From Factorisation of $z^n + a$:
:$z^{2 n} + y^{2 n} = \ds \prod_{k \mathop = 0}^{2 n - 1} \paren {x - \alpha_k y}$
where $\alpha_k$ are the complex $2n$th roots of negative unity:
Then we have that:
:$U_{2 n} = \set {\tuple {\alpha_0, \alpha_{2 n - 1} }, \tuple {\alpha_1, \alpha_{2 n - 2} }, \ldots, \tuple {\alpha_k, \alpha_{2 n - k - 1} }, \ldots, \tuple {\alpha_{n - 1}, \alpha_n } }$
where $U_{2 n}$ denotes the complex $2n$th roots of negative unity:
:$U_{2 n} = \set {z \in \C: z^{2 n} = -1}$
Taking the product, $p_k$, of the factors of $x^{2 n} + y^{2 n}$ in pairs:
However
Consider the permutation:
:$\sigma = \begin{pmatrix}
 1 & 2 & \cdots & k & \cdots & n - 1 & n \\
 n & n - 1 & \cdots & n - k & \cdots & 2 & 1
\end{pmatrix}$
From Permutation of Indices of Product:
:$\ds \prod_{\map R k} p_k = \prod_{\map R {\map \sigma k} } p_{\map \sigma k}$
Hence:
== Sources ==
* : $\S 2$: Special Products and Factors: $2.23$
* : $\S 2$: Special Products and Factors: $2.23.$
Category:Sum of Two Powers
Factors of Sums of Powers of 100,000

== Theorem ==
All integers $n$ of the form:
:$n = \ds \sum_{k \mathop = 0}^m 10^{5 k}$ for $m \in \Z_{> 0}$
are composite.
=== General Result ===
== Proof ==
Notice that:
:$\ds \sum_{k \mathop = 0}^m 10^{5 k} \times R_5 = R_{5 \paren {m + 1} }$
where $R_i$ is the $i$th repunit.
Suppose $p \divides m + 1$, where $p$ is a prime that is not $5$.
By Divisors of Repunit with Composite Index:
:$R_p \divides R_{5 \paren {m + 1} }$
By Prime not Divisor implies Coprime, $p$ and $5$ are coprime.
By Condition for Repunits to be Coprime, $R_p$ and $R_5$ are coprime.
By Euclid's Lemma:
:$R_p \divides \dfrac {R_{5 \paren {m + 1} } } {R_5} = n$
Suppose $25 \divides m + 1$.
By Divisors of Repunit with Composite Index:
:$R_5 \divides R_{25}$
and:
:$R_{25} \divides R_{5 \paren {m + 1} }$
So we have:
:$\dfrac {R_{25}} {R_5} \divides \dfrac {R_{5 \paren {m + 1} } } {R_5} = n$
The final case is $m + 1 = 5$.
This is the case $n = 100 \, 001 \, 000 \, 010 \, 000 \, 100 \, 001$.
We have:
:$n = 21 \, 401 \times 25 \, 601 \times 182 \, 521 \, 213 \, 001$
Thus all cases are covered.
== Historical Note ==
== Sources ==
* : $100,001$
Category:Composite Numbers
Category:Factors of Sums of Powers of 100,000