nyuuzyou/wikis huggingface dataset:
Thales' Theorem

== Theorem ==
Let $A$ and $B$ be two points on opposite ends of the diameter of a circle.
Let $C$ be another point on the circle such that $C \ne A, B$.
Then the lines $AC$ and $BC$ are perpendicular to each other.
:
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Historical Note ==
== Sources ==
* : $\text I$: Introductory: The Greeks and Mathematics
* 
* $(3)$
* $(3)$
Category:Circles
Category:Thales' Theorem
Thales' Theorem/Converse

== Theorem ==
Let $O$ be a circle.
Let $AOB$ be a diameter of $O$.
Then $\angle APB$ is a right angle $P$ lies on the circle $O$.
== Proof ==
=== Necessary Condition ===
Draw $OC \parallel PB$.
By Parallelism implies Equal Corresponding Angles:
:$\angle ACO = \angle APB$ and both are right angles.
$\triangle ACO$ and $\triangle APB$ share $\angle OAC$.
$\triangle ACO$ and $\triangle APB$ are both right triangles.
Now draw $OP$.
$OC$ is shared.
But $OA$ is a radius of the circle. 
Therefore $OP$ is also a radius.
So $P$ must lie on the circle.
=== Sufficient Condition ===
:
Given:
:$AO = CO = BO$
By definition of isosceles triangles:
:$\triangle AOC$ and $\triangle BOC$ are isosceles triangles
The result follows.
=== Also see ===
The Sufficient Condition is Thales' Theorem.
== Sources ==
* : Chapter $10$: Conversely ...
Category:Thales' Theorem
Category:Circles
Category:Right Angles
Thales' Theorem/Historical Note

== Theorem ==
This result was known by the ancient Babylonians as early as $2000$ BCE.
The proof of what is now known as Thales' Theorem that actually used is unknown. Whether he was aware of the Interior Angles Theorem and used it in his proof cannot be decided.
The Interior Angles Theorem is traditionally ascribed to , who lived near , and may have met him.
Thus it is possible that learned of the Interior Angles Theorem either directly or indirectly from himself.
== Sources ==
* 
* : Chapter $\text {A}.1$: Thales (ca. B.C.)
Category:Historical Notes
Thales' Theorem/Proof 1

== Theorem ==
== Proof ==
:
Let $O$ be the center of the circle, and define the vectors:
:$\mathbf u = \overrightarrow{OC}$
:$\mathbf v = \overrightarrow{OB}$
:$\mathbf w = \overrightarrow{OA}$
Thus:
:$\overrightarrow{AC} = \mathbf u - \mathbf w$
:$\overrightarrow{BC} = \mathbf u - \mathbf v$
Then:
From Non-Zero Vectors are Orthogonal iff Perpendicular, it follows that $AC$ and $BC$ are perpendicular to each other.
== Historical Note ==
Category:Thales' Theorem
Thales' Theorem/Proof 3

== Theorem ==
== Proof ==
Let $O$ be the center of $ACB$.
We have that $AO$, $BO$ and $CO$ are radii of the same circle.
Thus by definition:
:$AO = BO = CO$
Thus $\triangle AOC$ and $\triangle BOC$ are isosceles triangles.
From External Angle of Triangle equals Sum of other Internal Angles:
:$\angle COB = \angle ACO + \angle CAO$
Hence:
Similarly, from External Angle of Triangle equals Sum of other Internal Angles:
:$\angle COA = \angle BCO + \angle CBO$
Hence:
Therefore:
But $\angle COB + \angle COA$ equals two right angles.
Hence $\angle ACB$ is a right angle.
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.1$: Thales (ca. B.C.)
Category:Thales' Theorem
That which produces Medial Whole with Medial Area is Irrational

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line.
Let a straight line $BC$ such that:
:$BC$ is incommensurable in square with $AB$
:$AB^2 + BC^2$ is medial
:twice the rectangle contained by $AB$ and $BC$ is medial
:$AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$
be cut off from $AB$.
Let $DI$ be a rational straight line.
Using :
:Let $DE$ be a parallelogram set out on $DI$ equal to $AB^2 + BC^2$.
Let its breadth be $DG$.
Similarly:
:Let $DH$ be a parallelogram set out on $DI$ equal to $2 \cdot AB \cdot BC$.
From :
:$FE = AC^2$
We have that $AB^2 + BC^2$ is medial and equal to $DE$.
Therefore $DE$ is medial.
We have that $DE$ has been applied to the rational straight line $DI$ producing $DG$ as breadth.
So from :
:$DG$ is rational and incommensurable in length with $DI$.
We have that $2 \cdot AB \cdot BC$ is medial.
But $2 \cdot AB \cdot BC = DH$.
Therefore $DH$ is medial.
We have that $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$.
Therefore $DE$ is incommensurable with $DH$.
But from :
:$DE : DH = DG : DF$
Therefore by :
:$DG$ is incommensurable with $DF$.
But both $GD$ and $DF$ are rational.
Therefore $GD$ and $DF$ are rational straight lines which are commensurable in square only.
Therefore, by definition, $FG$ is an apotome.
From :
:a rectangle contained by a rational and an irrational straight line is irrational.
Hence its "side" is irrational.
But $AC$ is the "side" of $FE$.
Therefore $AC$ is irrational.
Such a straight line is known as that which produces with a medial area a medial whole.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
That which produces Medial Whole with Rational Area is Irrational

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line.
Let a straight line $BC$ such that:
:$BC$ is incommensurable in square with $AB$
:$AB^2 + BC^2$ is medial
:the rectangle contained by $AB$ and $BC$ is rational
be cut off from $AB$.
We have that:
:$AB^2 + BC^2$ is medial
while:
:$2 \cdot AB \cdot BC$ is rational.
Therefore $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$.
From:
:
and:
:
it follows that:
:$2 \cdot AB \cdot BC$ is incommensurable with $AC^2$.
But $2 \cdot AB \cdot BC$ is rational.
Therefore $AC^2$ is irrational.
Therefore $AC$ is irrational.
Such a straight line is known as that which produces with a rational area a medial whole.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
The Bulldozers and the Bee

== Problem ==
Two (presumably driven by a pair of insane enemies) are $20$ miles apart, heading towards each other at $10$ miles per hour, on a collision course.
At the same time, a takes off from the of one bulldozer at $20$ miles per hour, towards the other bulldozer.
As soon as the bee reaches the other bulldozer, it reverses direction instantaneously and heads off at $20$ miles per hour back towards the first bulldozer.
It continues to do this until the bulldozers collide, squashing the bee between them and killing her.
The question is: how far does the bee fly before the collision?
== Solution ==
This is frequently asked as a trick question.
=== The Long Answer ===
=== The Short Answer ===
The bulldozers are travelling at $10$ mph and are $20$ miles apart.
Therefore they travel $10$ miles each and collide after $1$ hour.
The bee is flying at $20$ mph and therefore travels $20$ miles in that time.
== Pointless quibbles ==
Whether a bee can actually fly at $20$ miles per hour is doubtful, let alone sustain that speed for a whole hour. I may be completely wrong. This may be completely reasonable.
Even if she could, she could not reverse direction instantaneously. The laws of physics are completely against it.
== Historical Note ==
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.1$: The bulldozers and the bee
Category:Classic Problems
Category:The Bulldozers and the Bee
The Bulldozers and the Bee/Historical Note

== Historical Note on The Bulldozers and the Bee ==
The problem of The Bulldozers and the Bee has been phrased in several different forms.
One many apocryphal tales concerning is that he was asked this question. He instantly gave the answer.
:"So you've heard this one then? You solved it the quick way?" he was asked.
:"I solved it by summing an infinite geometric progression. There's a quicker way?" was the reply.
, giving the answer to this problem based in a different context, reports that a French professor of mathematics exclaimed:
:"Mon Dieu, quelle série!"
also completely overlooking the simple manner of solution.
The point is that there are (at least) two ways to solve the problem, and they come to the same value.
That is:
:$\ds 20 \times 2 \sum_{n \mathop \ge 1} \paren {\frac 1 3}^n = 20$
== Sources ==
* : Solutions: $46$. -- Baxter's Dog
* : Answers: $74$. Baxter's Dog
Category:Historical Notes
The Bulldozers and the Bee/The Long Answer

== Problem ==
== The Long Answer ==
Let $d$ be the total distance the bee travels.
Let $D_1$ be the initial separation of the bulldozers in miles.
Let $d_n$ be the distance the bee travels on each leg of her journey.
Let $d'_n$ be the distance that one of the bulldozers travels during the time the bee travels $d_n$.
Let $D_n$ be the distance the bulldozers are apart at the start of each leg of the journey.
The bee travels twice as fast as each of the bulldozers. So on each leg, $d_n = 2 d'_n$.
Consider the $m$th leg of the journey.
The bee travels $d_m$, and the bulldozers travel $\dfrac {d_m} 2$. These two together equal $D_m$.
Therefore $d_m = \dfrac {2 D_m} 3$, while $d'_m = \dfrac {D_m} 3$.
At the start of leg $m + 1$, both bulldozers have covered the distance $\dfrac {D_m} 3$. So at the start of the second leg, the bulldozers are $D_{m + 1} = D_m - \dfrac {2 D_m} 3 = \dfrac {D_m} 3$.
This gives us a recurrence formula:
$d_{n + 1} = \begin{cases}
\dfrac {2 D_1} 3 & : n = 1 \\
\dfrac {d_n} 3 & : n > 1 
\end{cases}$
It can be seen that the answer can be calculated by Sum of Geometric Sequence, and comes out as $20$ miles.
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.13$: Example
Category:The Bulldozers and the Bee
Theorem of Even Perfect Numbers

== Theorem ==
Let $a \in \N$ be an even perfect number.
Then $a$ is in the form:
:$2^{n - 1} \paren {2^n - 1}$
where $2^n - 1$ is prime.
Similarly, if $2^n - 1$ is prime, then $2^{n - 1} \paren {2^n - 1}$ is perfect.
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
== Also see ==
* Definition:Mersenne Prime
* Are All Perfect Numbers Even?
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* 
* 
* 
* 
* 
* 
* 
Category:Theorem of Even Perfect Numbers
Category:Perfect Numbers
Category:Mersenne Numbers
Category:Mersenne Primes
Category:Number Theory
Category:Named Theorems
Theorem of Even Perfect Numbers/Historical Note

== Theorem ==
The first part of this proof of the Theorem of Even Perfect Numbers was documented by in : .
The second part was achieved by .
Hence, the hunt for even perfect numbers reduces to the hunt for prime numbers of the form $2^n - 1$.
From Primes of form Power Less One, we see that for $2^n - 1$ to be prime, $n$ itself must be prime.
== Sources ==
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
Category:Historical Notes
Theorem of Even Perfect Numbers/Necessary Condition

== Theorem ==
Let $a \in \N$ be an even perfect number.
Then $a$ is in the form:
:$2^{n - 1} \paren {2^n - 1}$
where $2^n - 1$ is prime.
== Proof ==
Let $a \in \N$ be an even perfect number.
We can extract the highest power of $2$ out of $a$ that we can, and write $a$ in the form:
:$a = m 2^{n - 1}$
where $n \ge 2$ and $m$ is odd.
Since $a$ is perfect and therefore $\map {\sigma_1} a = 2 a$:
 
So:
:$\map {\sigma_1} m = \dfrac {m 2^n} {2^n - 1}$
But $\map {\sigma_1} m$ is an integer and so $2^n - 1$ divides $m 2^n$.
From Consecutive Integers are Coprime, $2^n$ and $2^n - 1$ are coprime.
So from Euclid's Lemma $2^n - 1$ divides $m$.
Thus $\dfrac m {2^n - 1}$ divides $m$.
Since $2^n - 1 \ge 3$ it follows that:
:$\dfrac m {2^n - 1} < m$
Now we can express $\map {\sigma_1} m$ as:
:$\map {\sigma_1} m = \dfrac {m 2^n} {2^n - 1} = m + \dfrac m {2^n - 1}$
This means that the sum of all the divisors of $m$ is equal to $m$ itself plus one other divisor of $m$.
Hence $m$ must have exactly two divisors, so it must be prime by definition.
This means that the other divisor of $m$, apart from $m$ itself, must be $1$.
That is:
:$\dfrac m {2^n - 1} = 1$
Hence the result.
== Historical Note ==
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Conjecture $2$
* : $28$
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : $28$
Category:Theorem of Even Perfect Numbers
Theorem of Even Perfect Numbers/Necessary Condition/Historical Note

== Historical Note on Theorem of Even Perfect Numbers ==
 stated in $1638$ that he had a proof that every even perfect number is of the form $2^{p - 1} \paren {2^p - 1}$, but failed to actually produce it.
The first actual published proof was made by .
== Sources ==
* : Preface
* : $28$
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : $28$
Category:Historical Notes
Theorem of Even Perfect Numbers/Sufficient Condition

== Theorem ==
Let $n \in \N$ be such that $2^n - 1$ is prime.
Then $2^{n - 1} \paren {2^n - 1}$ is perfect.
== Proof ==
Suppose $2^n - 1$ is prime.
Let $a = 2^{n - 1} \paren {2^n - 1}$.
Then $n \ge 2$ which means $2^{n - 1}$ is even and hence so is $a = 2^{n - 1} \paren {2^n - 1}$.
Note that $2^n - 1$ is odd.
Since all divisors (except $1$) of $2^{n - 1}$ are even it follows that $2^{n - 1}$ and $2^n - 1$ are coprime.
Let $\map {\sigma_1} n$ be the divisor sum of $n$, that is, the sum of all divisors of $n$ (including $n$).
From Divisor Sum Function is Multiplicative, it follows that $\map {\sigma_1} a = \map {\sigma_1} {2^{n - 1} } \map {\sigma_1} {2^n - 1}$.
But as $2^n - 1$ is prime, $\map {\sigma_1} {2^n - 1} = 2^n$ from Divisor Sum of Prime Number.
Then we have that $\map {\sigma_1} {2^{n - 1} } = 2^n - 1$ from Divisor Sum of Power of Prime.
Hence it follows that $\map {\sigma_1} a = \paren {2^n - 1} 2^n = 2 a$.
Hence from the definition of perfect number it follows that $2^{n - 1} \paren {2^n - 1}$ is perfect.
== Sources ==
* : Preface
* : Book $\text{IX}$. Propositions
* : $\text {3-5}$ The Use of Computers in Number Theory: Conjecture $1$
* : $28$
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : $28$
* 
* 
Category:Theorem of Even Perfect Numbers
Theorema Egregium

== Theorem ==
The Gaussian curvature of a surface does not change if one bends the surface without stretching it.
That is, Gaussian curvature can be determined entirely by measuring angles, distances and their rates on the surface itself, without further reference to the particular way in which the surface is embedded in the ambient $3$-dimensional Euclidean space.
Thus the Gaussian curvature is an intrinsic invariant of a surface.
== Proof ==
== Historical Note ==
== Linguistic Note ==
Category:Differential Geometry
Category:Named Theorems
Theorema Egregium/Linguistic Note

== Theorem ==
Theorema Egregium is Latin for remarkable theorem.
Note that the word egregium, literally meaning outside the flock, or possibly outside the realm, is the root of the English word egregious.
At one time, egregious meant outstandingly good, but the meaning has drifted over time, and now it means appallingly bad.
Category:Linguistic Notes
Theoretical Justification for Cycle Notation

== Theorem ==
Let $\N_k$ be used to denote the initial segment of natural numbers:
:$\N_k = \closedint 1 k = \set {1, 2, 3, \ldots, k}$
Let $\rho: \N_n \to \N_n$ be a permutation of $n$ letters.
Let $i \in \N_n$.
Let $k$ be the smallest (strictly) positive integer for which $\map {\rho^k} i$ is in the set:
:$\set {i, \map \rho i, \map {\rho^2} i, \ldots, \map {\rho^{k - 1} } i}$
Then:
:$\map {\rho^k} i = i$
== Proof ==
 $\map {\rho^k} i = \map {\rho^r} i$ for some $r > 0$.
As $\rho$ has an inverse in $S_n$:
:$\map {\rho^{k - r} } i = i$
This contradicts the definition of $k$, because $k - r < k$
Thus:
:$r = 0$
The result follows.
== Sources ==
* : Chapter $9$: Permutations: Proposition $9.1$
Category:Permutations
Theory of Algebraically Closed Fields of Characteristic p is Complete

== Theorem ==
Let $p$ be either $0$ or a prime number.
Let $ACF_p$ be the theory of algebraically closed fields of characteristic $p$ in the language $\LL_r = \set {0, 1, +, -, \cdot}$ for rings, where:
:$0, 1$ are constants
and:
:$+, -, \cdot$ are binary functions. 
Then $ACF_p$ is complete.
== Proof ==
By the Łoś-Vaught Test, it suffices to show that $ACF_p$ is satisfiable, has no finite models, and is $\kappa$-categorical for some uncountable $\kappa$.
=== Satisfiability ===
$\C$ is an algebraically closed field of characteristic $0$.
If $p$ is a prime, then the algebraic closure of $\Z / \Z_p$ is an algebraically closed field of characteristic $p$.
Thus $ACF_p$ is satisfiable.
=== No Finite Models ===
From Algebraically Closed Field is Infinite:
$ACF_p$ has no finite models
=== $\kappa$-Categorical ===
From:
:Field of Uncountable Cardinality $\kappa$ has Transcendence Degree $\kappa$
and:
:Algebraically Closed Fields are Isomorphic iff they have the same Characteristic and Transcendence Degree
it follows that:
:$ACF_p$ is $\kappa$-categorical for all uncountable $\kappa$
Hence:
:$ACF_p$ is $\kappa$-categorical for some uncountable $\kappa$
as was to be proved.
Category:Model Theory for Predicate Logic
Theory of Set of Formulas is Theory

== Theorem ==
Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\FF$ be a set of $\LL$-formulas.
Let $\map T \FF$ be the $\LL$-theory of $\FF$.
Then $\map T \FF$ is a theory.
== Proof ==
Let $\phi$ be a $\LL$-formula.
Suppose that $\map T \FF \models_{\mathscr M} \phi$.
By definition of $\map T \FF$:
:$\FF \models_{\mathscr M} \psi$
for every $\psi \in \map T \FF$.
Hence by definition of semantic consequence:
:$\FF \models_{\mathscr M} \map T \FF$
By Semantic Consequence is Transitive, it follows that:
:$\FF \models_{\mathscr M} \phi$
Finally, by definition of $\map T \FF$:
:$\phi \in \map T \FF$.
Since $\phi$ was arbitrary, the result follows.
== Sources ==
* : $\S 2.10$: Exercise $2.17$
Category:Formal Semantics
Theory of Structure is Complete

== Theorem ==
Let $\AA$ be a structure for a signature for predicate logic $\LL$.
Let $\map {\operatorname{Th}} \AA$ be the theory of $\AA$.
Then $\map {\operatorname{Th}} \AA$ is complete.
== Proof ==
By definition of $\map {\operatorname{Th}} \AA$ be the theory of $\AA$:
:$\AA \models \map {\operatorname{Th}} \AA$
so that $\map {\operatorname{Th}} \AA$ is consistent.
Now let $\mathbf A$ be an $\LL$-sentence.
Let $\map {\operatorname{val}_\AA} {\mathbf A}$ be the value of $\mathbf A$ in $\AA$.
Then either $\map {\operatorname{val}_\AA} {\mathbf A} = T$ or $\map {\operatorname{val}_\AA} {\mathbf A} = F$.
By definition of value, the latter implies:
:$\map {\operatorname{val}_\AA} {\neg \mathbf A} = T$
Therefore, by definition of the models relation $\models$:
:$\AA \models \mathbf A$ or $\AA \models \mathbf A$
That is, by definition of the theory $\map {\operatorname{Th}} \AA$:
:$\mathbf A \in \map {\operatorname{Th}} \AA$ or $\neg \mathbf A \in \map {\operatorname{Th}} \AA$
By Element is Semantic Consequence of Set, it follows that:
:$\map {\operatorname{Th}} \AA \models \mathbf A$ or $\map {\operatorname{Th}} \AA \models \neg \mathbf A$
Since $\mathbf A$ was arbitrary, it follows that $\map {\operatorname{Th}} \AA$ is complete.
== Sources ==
* : $\text{II}.8$ Further Semantic Notions: Lemma $\text{II}.8.22$
Category:Model Theory for Predicate Logic
There Exists No Universal Set

== Theorem ==
There exists no set which is an absolutely universal set.
That is:
:$\map \neg {\exists \, \UU: \forall T: T \in \UU}$
where $T$ is any arbitrary object at all.
That is, a set that contains everything cannot exist.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* 
Category:There Exists No Universal Set
Category:Universal Set
Category:Naive Set Theory
There Exists No Universal Set/Proof 1

== Theorem ==
== Proof ==
 such a $\UU$ exists.
Using the Axiom of Specification, we can create the set:
:$R = \set {x \in \UU: x \notin x}$
But from Russell's Paradox, this set cannot exist.
Thus:
:$R \notin \UU$
and so $\UU$ cannot contain everything.
== Sources ==
* : $\S 2$: The Axiom of Specification
Category:There Exists No Universal Set
There Exists No Universal Set/Proof 2

== Theorem ==
== Proof ==
 such a $\UU$ exists.
By No Injection from Power Set to Set, $\powerset \UU$ has no injection to $\UU$.
Let $f : \powerset \UU \to \UU$ be the identity mapping.
By Identity Mapping is Injection, it is an injection, which leads to a contradiction.
Category:There Exists No Universal Set
There Exists No Universal Set/Proof 3

== Theorem ==
== Proof ==
Let $\SS$ be the set of all sets.
Then $\SS$ must be an element of itself:
:$\SS \owns \SS$
Thus we have an infinite descending sequence of membership:
:$\SS \owns \SS \owns \SS \owns \cdots$
But by No Infinitely Descending Membership Chains, no such sequence exists, a contradiction.
Category:There Exists No Universal Set
There Exists No Universal Set/Proof 4

== Theorem ==
== Proof ==
 such a $\UU$ exists.
Using the Axiom of Specification, we can create the set of all ordinals:
:$\set {x \in \UU: x \text{ is an ordinal} }$
But from Burali-Forti Paradox, this set cannot exist, which is a contradiction.
== Sources ==
Category:There Exists No Universal Set
There are Infinitely Many Carmichael Numbers/Historical Note

== Historical Note on There are Infinitely Many Carmichael Numbers ==
When first identified in $1909$ the existence of what are now called Carmichael numbers, he expressed his belief that there were infinitely many.
This was the general (although unproven) belief in the mathematical community, up until $1992$, when , and finally proved it.
Some sources give this date as $1994$, but that was just the date when the article they wrote on the subject was published.
However, this information took some time to be widely disseminated, and in $1997$, was still reporting in his that:
:It is widely believed, but not proved, that there are an infinite number of Carmichael numbers, but they are rare.
== Sources ==
* 
* : $561$
* 
* : $561$
* 
* 
Category:Historical Notes
There are no 120 consecutive numbers which all have exactly 120 divisors

Prove that the term $n=120$ in is $0$:
$24k$ with $k$ coprime to $6$ has exactly $120$ divisors
than
$k$ has exactly $15$ divisors
than
$k$ is a square number
than
$k$ cannot be $== 5, 7, 11 mod 12$ (since $5, 7, 11$ are not quadratic residues $mod 12$)
And we have a theorem:
A number $== 120, 168, 264 mod 288$ cannot have exactly $120$ divisors (since such numbers can be written as $24*k$ with $k$ coprime to $6$ and $k == 5, 7, 11 mod 12$)
So we have:
There are $120$ consecutive integers with exactly $120$ divisors
than
the start number must be $== 0, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287 mod 288$
than
the start number must be $== 0, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31 mod 32$
than
there are $4$ consecutive multiples of $32$ among these $120$ integers
than
one of these $4$ numbers must be == $64 mod 128$
than
the number of divisors of this number must be divisible by $7$ and cannot be $120$
And we have a contradiction, thus the proof is done.
There are no Odd Unitary Perfect Numbers

== Theorem ==
No unitary perfect numbers exist which are odd.
== Proof ==
Let $n$ be an odd number with prime decomposition $n = p_1^{a_1} \cdots p_k^{a_k}$.
By Sum of Unitary Divisors of Integer, the sum of its unitary divisors is $\ds \prod_{1 \mathop \le i \mathop \le k} \paren {1 + p_i^{a_i} }$.
To be a unitary perfect number, this must be equal to $2 n$.
Since each $p$ is odd, each $1 + p_i^{a_i}$ is even.
Hence $\ds \prod_{1 \mathop \le i \mathop \le k} \paren {1 + p_i^{a_i} }$ is divisible by $2^k$.
Since $n$ is odd, $2 n$ is divisible by $2$ but not $4$.
Thus $k = 1$.
So $n$ is a prime power.
By Sum of Unitary Divisors of Power of Prime, the sum of its unitary divisors is $1 + n$.
This cannot be equal to $2 n$, since $n > 1$.
Therefore there are no unitary perfect numbers which are odd.
== Sources ==
* 
* : $87,360$
Category:Unitary Perfect Numbers