nyuuzyou/wikis huggingface dataset:
Odd Divisor of Even Number also divides its Half

== Theorem ==
Let $a, b \in \Z$ be integers.
Let $a$ be odd and $b$ be even.
Let:
:$a \divides b$
where $\divides$ denotes divisibility.
Then:
:$a \divides \dfrac b 2$ 
== Proof ==
By definition of an even number:
:$\exists r \in \Z: b = 2 r$
By definition of an odd number:
:$\exists s \in \Z: a = 2 s + 1$
Thus:
:$a \divides 2 \iff a = \pm 1$
in which case from One Divides all Integers:
:$a \divides \dfrac b 2$
We have that:
:$a \divides 2 r$
and so by Euclid's Lemma:
:$a \divides r = \dfrac b 2$
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Odd Function Times Even Function is Odd

== Theorem ==
Let $X \subset \R$ be a symmetric set of real numbers:
:$\forall x \in X: -x \in X$
Let $f: X \to \R$ be an odd function.
Let $g: X \to \R$ be an even function.
Let $f \cdot g$ denote the pointwise product of $f$ and $g$.
Then $\paren {f \cdot g}: X \to \R$ is an odd function.
== Proof ==
Thus, by definition, $\paren {f \cdot g}$ is an odd function.
== Also see ==
* Even Function Times Even Function is Even
* Odd Function Times Odd Function is Even
Category:Even Functions
Category:Odd Functions
Odd Function Times Odd Function is Even

== Theorem ==
Let $S \subset \R$ be a symmetric set of real numbers:
:$\forall x \in S: -x \in X$
Let $f, g: X \to \R$ be two odd functions.
Let $f \cdot g$ denote the pointwise product of $f$ and $g$.
Then $\paren {f \cdot g}: S \to \R$ is an even function.
== Proof ==
Thus, by definition, $\paren {f \cdot g}$ is an even function.
== Also see ==
* Even Function Times Even Function is Even
* Odd Function Times Even Function is Odd
Category:Even Functions
Category:Odd Functions
Odd Function of Zero is Zero

== Theorem ==
Let $f: \R \to \R$ be an odd function.
Let $f$ be defined at the point $x = 0$.
Then:
:$\map f 0 = 0$
== Proof ==
By definition of odd function:
:$\map f {-x} = -\map f x$
and so:
The only real number $a$ for which $a = -a$ is $0$.
Hence the result.
== Sources ==
* : Chapter One: $\S 4$. Even and Odd Functions
Category:Odd Functions
Odd Integer Modulo 4

== Theorem ==
Let $n$ be an odd integer.
Then $n$ can be expressed either as:
:$n = 4 k + 1$
or as:
:$n = 4 k + 3$
== Proof ==
By the Division Theorem, $n$ can be expressed as:
:$n = 4 k + r$
where:
:$k, r \in \Z$
:$0 \le r < 4$
That is, one of the following holds:
Of these:
and:
and so are even by definition.
Then:
and:
and so are odd by definition.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm: Problems $2.1$: $3 \ \text{(a)}$
Category:Odd Integers
Odd Integers do not form Integral Domain

== Theorem ==
Let $S$ denote the set of odd integers.
Then $\struct {S, +, \times}$ is not an integral domain.
== Proof ==
Consider the odd integers $1$ and $3$.
We have that $1 + 3 = 4$.
But $4$ is not odd.
Thus addition on $\struct {S, +, \times}$ is not closed.
Hence $\struct {S, +, \times}$ is not even a ring, let alone an integral domain.
== Sources ==
* : Exercises: Chapter $1$: Exercise $1 \ \text{(ii)}$
Category:Integers
Category:Integral Domains
Odd Integers under Addition do not form Group

== Theorem ==
Let $S$ be the set of odd integers:
:$S = \set {x \in \Z: \exists n \in \Z: x = 2 n + 1}$
Let $\struct {S, +}$ denote the algebraic structure formed by $S$ under the operation of addition.
Then $\struct {S, +}$ is not a group.
== Proof ==
It is to be demonstrated that $\struct {S, +}$ does not satisfy the group axioms.
Let $a$ and $b$ be odd integers.
Then $a = 2 n + 1$ and $b = 2 m + 1$ for some $m, n \in \Z$.
Then:
and it is seen that $a + b$ is even.
That is:
:$a + b \notin S$
Thus $\struct {S, +}$ does not fulfil .
Hence the result.
== Sources ==
* : $\S 1$: Some examples of groups
Category:Odd Integers
Category:Integer Addition
Odd Integers under Addition do not form Subgroup of Integers

== Theorem ==
Let $S$ denote the set of odd integers.
Then $\struct {S, +}$ is not a subgroup of the additive group of integers $\struct {\Z, +}$.
== Proof ==
Consider the odd integers $1$ and $3$.
We have that $1 + 3 = 4$.
But $4$ is not odd.
Thus addition on $\struct {S, +}$ is not closed.
Hence $\struct {S, +}$ is not a group, let alone a subgroup of $\struct {\Z, +}$
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{E iii}$
Category:Integer Addition
Category:Integral Domains
Odd Integers under Multiplication do not form Group

== Theorem ==
Let $S$ be the set of odd integers:
:$S = \set {x \in \Z: \exists n \in \Z: x = 2 n + 1}$
Let $\struct {S, \times}$ denote the algebraic structure formed by $S$ under the operation of multiplication.
Then $\struct {S, \times}$ is not a group.
== Proof ==
It is to be demonstrated that $\struct {S, \times}$ does not satisfy the group axioms.
First it is noted that Integer Multiplication is Closed.
Then from Odd Number multiplied by Odd Number is Odd, $S$ is closed under $\times$.
Thus $\struct {S, \times}$ fulfils .
From Integer Multiplication is Associative, we have that $\times$ is associative on $S$.
Thus $\struct {S, \times}$ fulfils .
Then we have that:
:$\forall x \in S: 1 \times x = x = x \times 1$
and as $1 \in S$ it follows that $1$ is the identity element of $\struct {S, \times}$
Thus $\struct {S, \times}$ fulfils .
Now consider $3 \in S$.
There exists no $x \in S$ such that $3 \times x = 1$.
Thus $x$ has no inverse element in $S$.
Thus $\struct {S, \times}$ does not fulfil .
Thus it has been demonstrated that $\struct {S, \times}$ does not satisfy the group axioms.
Hence the result.
== Sources ==
* : Chapter $1$: Definitions and Examples: Exercise $1 \ \text{(c)}$
Category:Odd Integers
Category:Integer Multiplication
Category:Examples of Groups
Odd Integers whose Smaller Odd Coprimes are Prime

== Theorem ==
Let $n \in \Z_{>0}$ be an odd positive integer such that all smaller odd integers greater than $1$ which are coprime to it are prime.
The complete list of such $n$ is as follows:
:$1, 3, 5, 7, 9, 15, 21, 45, 105$
== Proof ==
First it is demonstrated that $105$ itself satisfies this property.
Let $d \in \Z_{> 1}$ be odd and coprime to $105$.
Then $d$ does not have $3$, $5$ or $7$ as a prime factor.
Thus $d$ must have at least one odd prime as a divisor which is $11$ or greater.
The smallest such composite number is $11^2$.
But $11^2 = 121 > 105$.
Thus $d$ must be prime.
Thus it has been demonstrated that all odd integers greater than $1$ and smaller than $105$ which are coprime to $105$ are prime.
Using an argument similar to the above, we see that for an integer to have this property, 
if it is greater than $p^2$ for some odd prime $p$, then it must be divisible by $p$.
If not, it will be coprime to $p^2$, a composite number.
Let $p_n$ denote the $n$th prime.
Suppose $N$ has this property.
By the argument above, if $p_{n + 1}^2 \ge N > p_n^2$, we must have $p_2 p_3 \cdots p_n \divides N$.
By Absolute Value of Integer is not less than Divisors, we have $p_2 p_3 \cdots p_n \le N$.
Bertrand-Chebyshev Theorem asserts that there is a prime between $p_n$ and $2 p_n$.
Thus we have $2 p_n > p_{n + 1}$.
Hence for $n \ge 5$:
This is a contradiction. 
Hence we must have $N \le p_5^2 = 121$.
From the argument above we also have:
:$3 \divides N$ for $9 < N \le 25$
:$3, 5 \divides N$ for $25 < N \le 49$
:$3, 5, 7 \divides N$ for $49 < N \le 121$
So we end up with the list $N = 1, 3, 5, 7, 9, 15, 21, 45, 105$.
== Sources ==
* 
* : $105$
Category:Coprime Integers
Category:105
Odd Integers whose Smaller Odd Coprimes are Prime/Mistake

== Source Work ==
:
:The Dictionary
::$105$
== Mistake ==
:$105$ is the largest integer such that every odd integer less than it and prime to it is a prime number.
While all integers are coprime to $1$ (and $105$ in particular), remember that $1$ is not classified as a prime number.
The canonical exposition of this result is for $1$ to be specifically mentioned.
== Sources ==
* : $105$
Category:Mistakes/Curious and Interesting Numbers
Odd Multiple Angle Formula for Sine

== Theorem ==
:$\ds \frac {\map \sin {2 n + 1} \theta} {\sin \theta} = \paren {2 n + 1} \prod_{k \mathop = 1}^n \paren {1 - \frac {\sin^2 \theta} {\map {\sin^2} {\frac {k \pi} {2 n + 1} } } }$
for $\sin \theta \ne 0$.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $161 \ \text{(b)}$
Category:Sine Function
Odd Number Coprime to Number is also Coprime to its Double

== Theorem ==
Let $a, b \in \Z$ be integers.
Let $a$ be odd.
Let:
:$a \perp b$
where $\perp$ denotes coprimality.
Then:
:$a \perp 2 b$
== Proof ==
By definition of odd number:
:$a \perp 2$
The result follows from Integer Coprime to Factors is Coprime to Whole.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Odd Number Theorem

== Theorem ==
:$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1} = n^2$
That is, the sum of the first $n$ odd numbers is the $n$th square number.
=== Corollary ===
== Visual Demonstration ==
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $1^2 = 1$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds k^2 = \sum_{j \mathop = 1}^k \paren {2 j - 1}$
Then we need to show:
:$\ds \paren {k + 1}^2 = \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$
== Also presented as ==
This result can also be seen as:
:$\ds \sum_{j \mathop = 0}^{n - 1} \paren {2 j + 1} = n^2$
== Historical Note ==
== Sources ==
* : $\S 19$: Arithmetic Series: $19.3$
* : $\text {1-1}$ Principle of Mathematical Induction: Exercise $5$
* : Appendix $\text{A}.6$: Mathematical Induction
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $1 \ \text {(b)}$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $10$
* : $25$
* : $\S 1.2.1$: Mathematical Induction
* : $25$
* : $\S 21$: Series of Constants: Arithmetic Series: $21.3.$
Category:Square Numbers
Category:Proofs by Induction
Category:Sums of Sequences
Category:Named Theorems
Category:Odd Number Theorem
Odd Number minus Even Number is Odd

== Theorem ==
== Proof ==
Let $a$ be odd and $b$ be even.
Then by definition of odd number:
:$\exists c \in \Z: a = 2 c + 1$
and by definition of even number:
:$\exists d \in \Z: b = 2 d$
So:
Hence the result by definition of odd number.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Odd Number minus Odd Number is Even

== Theorem ==
== Proof ==
Let $a$ and $b$ be odd numbers.
Then by definition of odd number:
:$\exists c \in \Z: a = 2 c + 1$
:$\exists d \in \Z: b = 2 d + 1$
So:
Hence the result by definition of even number.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Odd Number multiplied by Even Number is Even

== Theorem ==
== Proof ==
Let $a$ be odd and $b$ be even.
Then by definition of odd number:
:$\exists c \in \Z: a = 2 c + 1$
and by definition of even number:
:$\exists d \in \Z: b = 2 d$
So:
Hence the result by definition of even number.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Odd Number multiplied by Odd Number is Odd

== Theorem ==
== Proof ==
Let $a$ be odd and $b$ be even.
Then by definition of odd number:
:$\exists c \in \Z: a = 2 c + 1$
:$\exists d \in \Z: b = 2 d + 1$
So:
Hence the result by definition of odd number.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Category:Odd Integers
Odd Numbers Not Expressible as Sum of 4 Distinct Non-Zero Coprime Squares

== Theorem ==
The largest odd positive integer that cannot be expressed as the sum of exactly $4$ distinct non-zero square numbers with greatest common divisor $1$ is $157$.
The full sequence of such odd positive integers which cannot be so expressed is:
:$1, \ldots, 37, 41, 43, 45, 47, 49, 53, 55, 59, 61, 67, 69, 73, 77, 83, 89, 97, 101, 103, 115, 157$
where the sequence contains all odd integers between $1$ and $37$.
This sequence appears not to be documented on the .
== Proof ==
The statement of the result was taken from the paper cited below by and others:
:The largest odd integer not expressible as a sum of $4$ distinct non-zero squares with greatest common divisor $1$ is $157$.
By a brute force exercise, we assemble all the sets of $4$ distinct non-zero integers and calculate the sum of their squares such that those sums are less than $157$.
It is noted in passing that all such sets have a greatest common divisor of $1$.
Thus we have:
It remains to be demonstrated that the above sequence is indeed complete.
The original article by states:
: $3$. Genau dann is $n \in N$ nicht Summe von vier verschiedenen positiven (im Falle $n \not \equiv 0 \bmod 8$ auch teilerfremden) Quadraten, wenn entwieder
::$n = 4^h a \quad$ mit $h \ge 0$ und $a \in \set {1, 2, \ldots, 19, 22, 23, 25, 26, 27, 31, 33, 34, 37, 38, 42, 43, 47, 55, 58, 67, 73, 82, 97, 103}$
:oder
::$n \in \set {21, 29, 35, 41, 45, 49, 53, 59, 61, 69, 77, 83, 89, 101, 115, 157}$
That is, in English:
: $3$. $n \in N$ is not the sum of four different positive squares (in the case where $n \not \equiv 0 \bmod 8$ also non-prime) :
::$n = 4^h a \quad$ where $h \ge 0$ and $a \in \set {1, 2, \ldots, 19, 22, 23, 25, 26, 27, 31, 33, 34, 37, 38, 42, 43, 47, 55, 58, 67, 73, 82, 97, 103}$
:or
::$n \in \set {21, 29, 35, 41, 45, 49, 53, 59, 61, 69, 77, 83, 89, 101, 115, 157}$
== Sources ==
* 
* 
* : $157$
Category:Sums of Squares
Category:157
Odd Numbers not Sum of Prime and Power

== Theorem ==
The sequence of odd numbers which cannot be expressed as the sum of a perfect power and a prime number begins:
:$1, 5, 1549, 1 \, 771 \, 561, \ldots$
It is not known if there are any more terms.
== Proof ==
The cases $1$ and $5$ are trivial.
Now we show that $1549 - a^b$ is never prime for $a \ge 1$ and $b \ge 2$.
It suffices to show the result for prime values of $b$.
We first prove:
For $(1)$:
Suppose $a$ is odd.
Then so is $a^b$.
Therefore $1549 - a^b$ is even.
Hence $1549 - a^b$ is prime only if it is equal to $2$.
But $1547$ is not a perfect power.
For $(2)$:
Suppose $b = 2$ and $a$ is not divisible by $3$.
By Corollary to Square Modulo $3$:
:$3 \divides \paren {a^2 - 1}$
Hence:
:$3 \divides \paren {1548 - a^2 + 1} = 1549 - a^b$
So $1549 - a^b$ is prime only if it is equal to $3$.
But $1546$ is not a perfect square.
For $(3)$:
Suppose $a^b \equiv 4 \pmod {10}$.
Then $1549 - a^b \equiv 5 \pmod {10}$
So $1549 - a^b$ is prime only if it is equal to $5$.
But $1544$ is not a perfect power.
Hence we only need to check the cases:
:For $b = 2: \: 6^2, 12^2, 18^2, 24^2, 30^2, 36^2$ since $42^2 > 1549$
:For $b \ne 2: \: 2^3, 2^5, 2^7, 4^3, 4^5, 6^3, 8^3, 10^3$ since $2^{11}, 4^7, 6^5 > 1549$
of which $12^2 \equiv 18^2 \equiv 4^3 \equiv 4^5 \equiv 4 \pmod {10}$, so they are rejected.
We have:
and none of the above are prime.
For $11^6$, note the factorizations from Difference of Two Powers:
:$11^6 - m^2 = \paren {11^3 - m} \paren {11^3 + m}$
:$11^6 - n^3 = \paren {11^2 - n} \paren {1 + 11^2 n + 11^4 n^2}$
therefore if $11^6 - a^b$ is a prime, $b$ cannot be divisible by $2$ or $3$, unless $11^3 - m$ or $11^2 - n$ is equal to $1$.
For $m = 11^3 - 1$ and $n = 11^2 - 1$:
:$11^6 - m^2 = 2661 = 3 \times 887$
:$11^6 - n^2 = 1757161 = 7 \times 173 \times 1451$
and both are not primes.
Hence none of the above cases hold.
Since $\log_2 \paren {11^6} \approx 20.8$, we must have $b \le 19$.
By the conditions above, $b$ can only be one of $5, 7, 11, 13, 17, 19$.
Moreover $a$ cannot be odd unless $11^6 - a^b = 2$.
However $11^6 - 2 = 1771559$ is prime, and thus not a perfect power.
The cases we need to check are:
:$a^b = 2^5, 6^5, 10^5, 12^5, 14^5, 2^7, 6^7, 2^{11}, 2^{13}, 2^{17}, 2^{19}$
since $18^5, 10^7, 6^{11} > 11^6$, and $4, 8, 16$ are square/cube.
We have:
and none of the above are prime.
== Sources ==
* : $1549$
* : $1549$
Category:Powers
Category:Prime Numbers
Odd Order Complete Graph is Eulerian

== Theorem ==
Let $K_n$ be the complete graph of $n$ vertices.
Then $K_n$ is Eulerian $n$ is odd.
If $n$ is even, then $K_n$ is traversable iff $n = 2$.
== Proof ==
From the definition, the complete graph $K_n$ is $n-1$-regular.
That is, every vertex of $K_n$ is of degree $n-1$.
Suppose $n$ is odd. Then $n-1$ is even, and so $K_n$ is Eulerian.
Suppose $n$ is even. Then $n-1$ is odd.
Hence for $n \ge 4$, $K_n$ has more than $2$ odd vertices and so can not be traversable, let alone Eulerian.
If $n = 2$, then $K_n$ consists solely of two odd vertices (of degree $1$).
Hence, by Characteristics of Traversable Graph (or trivially, by inspection), $K_2$ has an Eulerian trail, and so is traversable (although not Eulerian).
== Historical Note ==
Category:Eulerian Graphs
Category:Complete Graphs
Odd Order Complete Graph is Eulerian/Historical Note

== Historical Note on Odd Order Complete Graph is Eulerian ==
This was noted in $1809$ by , who was unaware of Euler's more general result.
The remarkable point is that he gave an ingenious method for finding such a Eulerian circuit, which is a far from trivial exercise even for modestly sized complete graphs, for example those for $n < 100$.
Category:Historical Notes
Odd Order Derivative of Even Function Vanishes at Zero

== Theorem ==
Let $X$ be a symmetric subset of $\R$ containing $0$. 
Let $n$ be a positive integer. 
Let $f:X \to \R$ be an even function. 
Let $f$ be at least $\paren{2 n + 1}$-times differentiable.
Then:
:$\map {f^{\paren {2 n + 1} } } 0 = 0$
 
== Proof == 
From the definition of an even function, for all $x \in X$ we have: 
:$\map f x = \map f {-x}$ 
Differentiating $2 n + 1$ times, we have, by the Chain Rule for Derivatives: 
:$\map {f^{\paren {2 n + 1} } } x = \paren {-1}^{2 n + 1} \map {f^{\paren {2 n + 1} } } {-x} = -\map {f^{\paren {2 n + 1} } } {-x}$
Setting $x = 0$ gives: 
:$\map {f^{\paren {2 n + 1} } } 0 = -\map {f^{\paren {2 n + 1} } } {-0}$
That is: 
:$2 \map {f^{\paren {2 n + 1 } } } 0 = 0$
The result follows on division by $2$. 
Category:Differential Calculus
Category:Even Functions
Odd Order Group Element is Square

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $x \in G$.
Let the order $\order x$ be odd.
Then:
:$\exists y \in G: y^2 = x$
=== Corollary ===
== Proof ==
Let $\order x$ be odd.
Then:
:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
from the definition of the order of an element.
Conversely, suppose that 
:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
Then $\order x$ is a divisor of $2 n - 1$ from Element to Power of Multiple of Order is Identity.
Hence $\order x$ is odd.
So $\order x$ is odd :
:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
Then:
Hence the result.
== Warning ==
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $15$
Category:Odd Order Group Element is Square
Category:Order of Group Elements
Odd Order Group Element is Square/Corollary

== Corollary to Odd Order Group Element is Square ==
Let $\struct {G, \circ}$ be a finite group.
Then:
:$\forall x \in G: \exists y \in G: y^2 = x$
 $\order G$ is odd.
== Proof ==
Suppose $\order G$ is odd.
Then from Order of Element Divides Order of Finite Group, all elements of $G$ are of odd order.
Hence:
:$\forall x \in G: \exists y \in G: y^2 = x$
from Odd Order Group Element is Square.
Now suppose that:
:$\forall x \in G: \exists y \in G: y^2 = x$
From Odd Order Group Element is Square it follows that all elements of $G$ are of odd order.
 $\order G$ is even.
From Cauchy's Lemma (Group Theory) it follows that there exist elements in $G$ of even order.
From that contradiction we conclude that $\order G$ is odd.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.12 \ \text{(b)}$
Category:Odd Order Group Element is Square
Odd Order Group Element is Square/Warning

== Warning on Odd Order Group Element is Square ==
== False Statement ==
It is completely false to say:
:$\exists y \in G: y^2 = x$
:
:the order $\order x$ is odd
An order $2$ element in $C_4$ refutes the converse.
This mistake can arise by supposing that this:
:$\exists y \in G: y^2 = x$
implies:
:$\exists n \in \N: \paren {x^n}^2 = x$
The second step can only be used if every $x$ can be expressed in the terms of $y^2$.
Category:Odd Order Group Element is Square
Odd Perfect Number has at least Three Distinct Prime Factors

== Theorem ==
Let $n \in \N$ be an odd perfect number.
Then $n$ has at least $3$ distinct prime factors.
== Proof ==
 the contrary: that $n$ is a perfect number with at most $2$ distinct prime factors.
By Perfect Number has at least Two Distinct Prime Factors, $n$ must have exactly $2$ distinct prime factors.
Hence let $n = p^a q^b$ where $p$ and $q$ are distinct primes and where $a, b \in \N_{>0}$.
As $n$ is odd, both $p$ and $q$ are also odd.
By definition of perfect number:
:$\map {\sigma_1} n = 2 n$
where $\map {\sigma_1} n$ denotes the divisor sum of $n$.
Hence:
Hence the result by Proof by Contradiction.
== Also see ==
* Theorem of Even Perfect Numbers
Category:Perfect Numbers
Odd Perfect Number is Product of Square and Prime/Historical Note

== Historical Note on Odd Perfect Number is Product of Square and Prime ==
The suggestion that an Odd Perfect Number is Product of Square and Prime was made by in $1638$.
It was superseded by 's demonstration Form of Odd Perfect Number which gave more detailed information on its structure.
== Sources ==
* : Preface
* : $28$
* : $28$
Category:Historical Notes
Odd Permutation/Examples/321

== Example of Odd Permutation ==
:$\tuple {3, 2, 1}$ is an odd permutation of $\tuple {1, 2, 3}$.
== Proof ==
The permutation $\tuple {1 3}$ transforms $\tuple {1, 2, 3}$ to $\tuple {3, 2, 1}$.
That is $1$ transposition, which is an odd integer.
Hence the result.
== Sources ==
* 
* 
Category:Examples of Odd Permutations
Odd Power Function is Strictly Increasing/General Result

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be a totally ordered ring.
Let $n$ be an odd positive integer.
Let $f: R \to R$ be the mapping defined by:
:$\map f x = \map {\circ^n} x$
Then $f$ is strictly increasing on $R$.
== Proof ==
Let $x, y \in R$ such that $0 < x < y$.
By Power Function is Strictly Increasing on Positive Elements:
:$\map f x < \map f y$
Suppose that $x < y < 0$.
By Properties of Ordered Ring:
:$0 < -y < -x$
By Power Function is Strictly Increasing on Positive Elements (applied to $-y$ and $-x$):
:$0 < \map f {-y} < \map f {-x}$
By Power of Ring Negative:
:$\map f {-x} = -\map f x$
:$\map f {-y} = -\map f y$
Thus:
:$0 < -\map f y < -\map f x$
By Properties of Ordered Ring:
:$\map f x < \map f y$
By Sign of Odd Power:
:$\map f x < 0 = \map f 0$ when $x < 0$
:$\map f 0 = 0 < \map f x$ when $0 < x$
Thus we have shown that $f$ is strictly increasing on the positive elements and the negative elements, and across zero.
Category:Totally Ordered Rings
Category:Odd Power Function is Strictly Increasing
Odd Power Function is Strictly Increasing/Real Numbers

== Theorem ==
Let $n \in \Z_{> 0}$ be an odd positive integer.
Let $f_n: \R \to \R$ be the real function defined as:
:$\map {f_n} x = x^n$
Then $f_n$ is strictly increasing.
== Proof ==
From the Power Rule for Derivatives:
:$\map {D_x} {x^n} = n x^{n - 1}$
As $n$ is odd, $n - 1$ is even.
Thus by Even Power is Non-Negative:
:$\map {D_x} {x^n} \ge 0$
for all $x$.
From Derivative of Monotone Function, it follows that $f_n$ is increasing over the whole of $\R$.
The only place where $\map {D_x} {x^n} = 0$ is at $x = 0$.
Everywhere else, $f_n$ is strictly increasing.
By Sign of Odd Power:
:$\map {f_n} x < 0 = \map {f_n} 0$ when $x < 0$ and
:$\map {f_n} 0 = 0 < \map {f_n} x$ when $0 < x$.
Thus $f_n$ is strictly increasing on $\R$.
Category:Odd Power Function is Strictly Increasing
Category:Strictly Increasing Real Functions