nyuuzyou/wikis huggingface dataset:
Preorder Category is Category

== Theorem ==
Let $\struct {S, \precsim}$ be a preordered set.
Let $\mathbf S$ be its associated preorder category.
Then $\mathbf S$ is a category.
== Proof ==
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
Note that for objects $a, b$ of $\mathbf S$, there is at most one morphism $a \to b$, by definition of $\mathbf S$.
Suppose that $a \to b$ and $b \to c$ are morphisms of $\mathbf S$.
Then we have $a \precsim b$ and $b \precsim c$, and as $\precsim$ is a preordering:
:$a \precsim c$
which is to say that there is a unique morphism $a \to c$.
By uniqueness, this morphism is the composite of $a \to b$ and $b \to c$.
Also, for any $a \in S$, there is a morphism $a \to a$ by virtue of:
:$a \precsim a$
since $\precsim$ is a preordering.
The uniqueness of the morphisms mentioned above implies directly that $a \to a$ classifies as the identity morphism $\operatorname{id}_a$ for $a$.
The uniqueness of morphisms also immediately implies associativity.
Hence $\mathbf S$ is a metacategory.
Since both $S$ and $\precsim$ are sets, $\mathbf S$ is also a category.
Category:Preorder Categories
Preorder Induced by Preorder Category

== Theorem ==
Let $\struct {S, \precsim}$ be a preordered set.
Let $\mathbf S$ be its associated preorder category.
Let $\precsim'$ be the preorder induced by $\mathbf S$ as on Category Induces Preorder.
Then $\precsim'$ is the same as $\precsim$.
== Proof ==
Suppose that for some $a, b \in S$, we have:
:$a \precsim' b$
By Category Induces Preorder, this happens there exists an $\mathbf S$-morphism $f: a \to b$.
By definition of $\mathbf S$, this $f: a \to b$ exists :
:$a \precsim b$
Hence the result.
Category:Preorder Categories
Preordering/Examples/Finite Set Difference on Natural Numbers

== Examples of Preorderings ==
Consider the relation $\RR$ on the powerset of the natural numbers:
:$\forall a, b \in \powerset \N: a \mathrel \RR b \iff a \setminus b \text { is finite}$
where $\setminus$ denotes set difference.
Then $\RR$ is a preordering on $\powerset \N$, but not an ordering on $\powerset \N$.
== Proof ==
We need to show that $\RR$ is both reflexive and transitive, but specifically not antisymmetric.
We have from Set Difference with Self is Empty Set:
:$\forall a \in \powerset \N: a \setminus a = \O$
and so as $\O$ is a finite set:
:$\forall a \in \powerset \N: a \mathrel \RR a$
That is, $\RR$ is reflexive.
Let $a, b, c \in \powerset \N$ such that $a \setminus b$ is finite and $b \setminus c$ is finite.
In order to show that $\RR$ is transitive we need to show that $a \setminus c$ is also finite.
 $a \setminus c$ is not finite.
Then $a$ is not finite.
For $x \in \powerset \N$, let $x'$ denote the complement of $x$ in $\N$.
We have that:
We have from De Morgan's Laws: Set Difference with Intersection that:
:$\paren {a \setminus b} \cup \paren {a \setminus c} = a \setminus \paren {b \cap c}$
and so:
:$\paren {a \setminus b} \cup \paren {a \setminus c} = \paren {a \setminus b} \cup \paren {a \cap b \cap c'}$
Hence if $a \setminus c$ is not finite, then $a \cap b \cap c'$ is not finite.
But then $b \cap c'$ is not finite.
That is, $b \setminus c$ is not finite.
This contradicts our supposition that $b \setminus c$ is finite.
Hence by Proof by Contradiction it follows that $a \setminus c$ is finite.
That is, $a \mathrel \RR c$ is transitive
Let $a = \set {1, 2, 3}$ and $b = \set {2, 3, 4}$.
Then:
:$a \setminus b = \set 1$
:$b \setminus a = \set 4$
Both $a \setminus b$ and $b \setminus a$ are finite, and so:
:$a \mathrel \RR b$ and $b \mathrel \RR a$
but it is not the case that $a = b$.
That is, $\RR$ is not antisymmetric.
We have shown that $\RR$ is both reflexive and transitive, but specifically not antisymmetric.
Hence $\RR$ is a preordering on $\powerset \N$ which is not an ordering on $\powerset \N$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $38$
Category:Examples of Preorderings
Preordering induces Equivalence Relation

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure such that $\RR$ is a preordering.
Let $\sim$ be the relation induced by $\RR$:
:$x \sim y$ $x \precsim y$ and $y \precsim x$.
Then $\sim$ is an equivalence relation.
== Proof ==
To show that $\sim$ is an equivalence relation, we must show that it is reflexive, transitive, and symmetric.
By the definition of preordering, $\precsim$ is transitive and reflexive.
=== Transitive ===
Let $p, q, r \in S$.
Suppose that $p \sim q$ and $q \sim r$.
Then $p \precsim q$, $q \precsim r$, $r \precsim q$, and $q \precsim p$.
Since $\precsim$ is transitive:
:$p \precsim r$ and $r \precsim p$.
Thus by the definition of $\sim$, $p \sim r$.
Since this holds for all $p$, $q$, and $r$, $\sim$ is transitive.
=== Reflexive ===
Let $p \in S$.
Since $\precsim$ is reflexive:
: $p \precsim p$
Thus by the definition of $\sim$:
: $p \sim p$
As this holds for all $p$, $\sim$ is reflexive.
=== Symmetric ===
Let $p, q \in S$ with $p \sim q$.
Then $p \precsim q$ and $q \precsim p$.
Thus $q \sim p$.
Since this holds for all $p$ and $q$, $\sim$ is symmetric.
== Sources ==
* : Chapter $1$: Sets and mappings: Further exercises: $5$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $39 \ \text{(a)}$
Category:Preorderings
Category:Examples of Equivalence Relations
Preordering induces Ordering

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure such that $\RR$ is a preordering.
Let $\sim_\RR$ denote the equivalence on $S$ induced by $\RR$:
:$x \sim_\RR y$ $x \mathrel \RR y$ and $y \mathrel \RR x$
Let $\preccurlyeq_\RR$ be the relation defined on the quotient set $S / {\sim_\RR}$ by:
:$\eqclass x {\sim_\RR} \preccurlyeq_\RR \eqclass y {\sim_\RR} \iff x \mathrel \RR y$
where $\eqclass x {\sim_\RR}$ denotes the equivalence class of $x$ under $\sim_\RR$.
Then $\preccurlyeq_\RR$ is an ordering.
== Proof ==
First we note from Ordering Induced by Preordering is Well-Defined that $\preccurlyeq_\RR$ is a well-defined relation.
Checking in turn each of the criteria for an ordering:
=== Reflexivity ===
Let $A \in S / {\sim_\RR}$.
By the definition of quotient set, $A$ is non-empty.
Thus there exists $a \in A$.
Since $\RR$ is a preordering, it is reflexive.
Hence:
:$a \mathrel \RR a$
So by definition of $\preccurlyeq_\RR$:
:$\eqclass a {\sim_\RR} \preccurlyeq_\RR \eqclass a {\sim_\RR}$
Thus $\preccurlyeq_\RR$ has been shown to be reflexive.
=== Transitivity ===
Let $A, B, C \in S / {\sim_\RR}$ such that $A \preccurlyeq_\RR B$ and $B \preccurlyeq_\RR C$.
Then:
:$\exists a \in A, b_1 \in B: a \mathrel \RR b_1$
:$\exists b_2 \in B, c \in C: b_2 \mathrel \RR c$
By the definition of quotient set:
:$b_1 \sim_\RR b_2$
By definition, $\sim_\RR$ is the equivalence on $S$ induced by $\RR$.
Hence:
:$b_1 \mathrel \RR b_2$
We have:
:$a \mathrel \RR b_1$
:$b_1 \mathrel \RR b_2$
:$b_2 \mathrel \RR c$
:$\RR$ is transitive
Hence from Transitive Chaining:
:$a \precsim c$
Thus by the definition of $\RR$:
:$A \preccurlyeq_\RR C$
We have that $A$, $B$ and $C$ are arbitrary.
It follows that $\preccurlyeq_\RR $ is transitive.
=== Antisymmetry ===
Let $A, B \in S / {\sim_\RR}$ such that $A \preccurlyeq_\RR B$ and $B \preccurlyeq_\RR A$.
Then:
:$\eqclass a {\sim_\RR} \preccurlyeq_\RR \eqclass b {\sim_\RR}$
:$\eqclass b {\sim_\RR} \preccurlyeq_\RR \eqclass a {\sim_\RR}$
That is:
By definition of $S / {\sim_\RR}$:
:$a \mathrel \RR b$
:$b \mathrel \RR a$
But by definition of $\sim_\RR$, that means:
:$a \sim_\RR b$
Hence:
:$a, b \in \eqclass a {\sim_\RR}$
and:
:$a, b \in \eqclass b {\sim_\RR}$
Hence by Fundamental Theorem on Equivalence Relations:
:$\eqclass a {\sim_\RR} = \eqclass b {\sim_\RR}$
So $\preccurlyeq_\RR$ has been shown to be antisymmetric.
$\preccurlyeq_\RR$ has been shown to be reflexive, transitive and antisymmetric.
Hence by definition it is an ordering.
== Sources ==
* : Chapter $1$: Sets and mappings: Further exercises: $5$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $40 \ \text {(b)}$
Category:Preorderings
Category:Orderings
Preordering is not necessarily Ordering

== Theorem ==
Let $S$ be a set.
Let $\RR$ be a preordering on $S$.
Then it is not necessarily the case that $\RR$ is also an ordering on $S$.
== Proof ==
Consider the relation $\RR$ on the powerset of the natural numbers:
:$\forall a, b \in \powerset \N: a \mathrel \RR b \iff a \setminus b \text { is finite}$
where $\setminus$ denotes set difference.
It is demonstrated in Preordering Example: Finite Set Difference on Natural Numbers that;
:$\RR$ is a preordering on $\powerset \N$
but:
:$\RR$ is not an ordering on $\powerset \N$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $38$
Category:Preorderings
Category:Orderings
Preordering of Products under Operation Compatible with Preordering

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\precsim$ be a preordering on $S$.
Then $\precsim$ is compatible with $\circ$ :
:$\forall x_1, x_2, y_1, y_2 \in S: x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ y_1} \precsim \paren {x_2 \circ y_2}$
== Proof ==
By definition, $\precsim$ is compatible with $\circ$ :
=== Sufficient Condition ===
Let $\precsim$ be compatible with $\circ$.
Then for all $x_1, x_2, y_1, y_2 \in S$:
As $\precsim$ is a preordering it is by definition transitive.
Thus it follows that:
:$x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ y_1} \precsim \paren {x_2 \circ y_2}$
=== Necessary Condition ===
Let $\precsim$ fulfil the property that:
:$\forall x_1, x_2, y_1, y_2 \in S: x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ x_2} \precsim \paren {y_1 \circ y_2}$
As $\precsim$ is a preordering it is by definition reflexive.
That is:
:$\forall z \in S: z \precsim z$
Make the substitutions:
It follows that:
:$\forall x, y, z \in S: x \precsim y \implies \paren {x \circ z} \precsim \paren {y \circ z}$
Similarly, make the substitutions:
It follows that:
:$\forall x, y, z \in S: x \precsim y \implies \paren {z \circ x} \precsim \paren {z \circ y}$
So for all $x, y, z \in S$, both conditions are fulfilled:
for $\precsim$ to be compatible with $\circ$.
The result follows.
Category:Preorder Theory
Category:Compatible Relations
Pressure in Ideal Gas

== Theorem ==
Let $B$ be a body of an ideal gas $\GG$.
Then the pressure in $B$ at a height $h$ is given by:
:$p_h = p_0 \map \exp {-\dfrac {\rho_0 g h} {p_0} }$
where:
:$p_0$ is the pressure at height $h = 0$
:$\rho_0$ is the (mass) density of $\GG$ at height $h = 0$
:$g$ is the acceleration due to gravity
== Proof ==
== Sources ==
* 
* 
Category:Pressure
Pressure in Static Body of Liquid

== Theorem ==
Let $B$ be a body of liquid $\LL$ which is stationary.
Then the pressure in $B$ at a depth $h$ is given by:
:$p_h = \rho g h$
where:
:$\rho$ is the (mass) density of $\LL$
:$g$ is the acceleration due to gravity.
== Proof ==
== Sources ==
* 
* 
Category:Pressure
Pressure of Gas for Density and Temperature

== Theorem ==
The pressure, density and temperature of a body of gas are linked by the equation:
:$p = k \rho \paren {1 + \alpha t}$
where:
:$p$ denotes the pressure of the gas
:$\rho$ denotes the density of the gas
:$t$ denotes the temperature of the gas
:$k$ and $\alpha$ are constants.
== Proof ==
== Sources ==
* : Chapter $\text I$: Functions and their Graphs: $1$. Constants and Variables
Category:Pneumatics
Prim's Algorithm

== Algorithm ==
The purpose of this algorithm is to produce a minimum spanning tree $T$ for any given weighted graph $G$.
:Step 1: Choose any vertex of $G$, and add it to $T$.
:Step 2: Add an edge of minimum weight $e$ to join a vertex in $T$ to one not in $T$.
:Step 3: If $T$ is a spanning tree $G$, stop. Otherwise, go to Step 2.
== Proof ==
The above constitutes an algorithm, for the following reasons:
=== Finiteness ===
For each iteration through the algorithm, step 2 is executed, which increases the number of edges in $T$ by $1$.
As a tree with $n$ nodes has $n - 1$ edges, the algorithm will terminate after $n - 1$ iterations.
=== Definiteness ===
:Step 1: Trivially definite.
:Step 2: As the edges connecting $T$ to the remaining vertices can be arranged in order of weight, there is a definite edge (or set of edges) with minimum weight.
:Step 3: It is straightforward to determine whether all the vertices are connected.
=== Inputs ===
The input to this algorithm is the weighted graph $G$.
=== Outputs ===
The output from this algorithm is the minimum spanning tree $T$.
=== Effective ===
Each step of the algorithm is basic enough to be done exactly and in a finite length of time.
== Also known as ==
== Also see ==
* Prim's Algorithm produces Minimum Spanning Tree
* Kruskal's Algorithm
== Historical Note ==
== Sources ==
* 
* 
* 
* 
* 
Category:Prim's Algorithm
Category:Graph Theory
Category:Greedy Algorithms
Prim's Algorithm/Also known as

== Prim's Algorithm: Also known as == 
It is clear that Prim's Algorithm is a greedy algorithm: at each stage the minimum possible weight is chosen, without any analysis as to whether there may be a combination of larger weights which may produce a smaller-weight spanning tree.
For this reason, it is sometimes called Prim's Greedy Algorithm.
In this case, the greedy algorithm does produce the minimum spanning tree.
Category:Prim's Algorithm
Prim's Algorithm/Historical Note

== Historical Note on Prim's Algorithm ==
Prim's Algorithm was initially developed in $1930$ by Czech mathematician .
 independently discovered it in $1957$, and in $1959$ it was rediscovered once more by .
For these reasons it is also known as the DJP Algorithm, the Jarník Algorithm, or the Prim-Jarník Algorithm.
== Sources ==
* 
* 
Category:Historical Notes
Prim's Algorithm produces Minimum Spanning Tree

== Theorem ==
Prim's Algorithm always produces a minimum spanning tree.
== Proof ==
Suppose that Prim's Algorithm produces a tree $T$.
Let there exist another spanning tree $S$ with a smaller total weight.
Let $e$ be an edge of smallest weight which lies in $T$ but not $S$.
If we add $e$ to $S$, we obtain a cycle, from Equivalent Definitions for Finite Tree.
This cycle contains an edge $e'$ which is in $S$ but not $T$, otherwise $T$ would not be a tree.
So, we replace $e'$ in $S$ with $e$ from $T$, and obtain a new spanning tree $S'$.
From the method of construction of $T$, it follows that the weight of $e$ can not exceed that of $e'$.
So the total weight of $S'$ does not exceed the total weight of $T$.
Also, $S'$ has one more edge in common with $T$ than $S$ has.
We repeat the above procedure, and repeatedly change edges of $S$ for edges of $T$, and each time the weight of the intermediate graph does not exceed that of $T$.
Thus the weight of $T$ does not exceed that of $S$, contradicting the definition of $S$.
Hence $T$ must be a minimum spanning tree.
Category:Prim's Algorithm
Category:Tree Theory
Prime-Counting Function/Approximations

== Approximations to Prime-Counting Function ==
A table of some of the values of the prime-counting ($\pi$) function compared with $\dfrac x {\ln x}$ and the Eulerian logarithmic integral $\ds \map \Li x = \int_2^x \frac {\d t} {\map \ln t}$:
:
$n$ 	$\map \pi n$	$\dfrac x {\ln x}$	$\map \Li x$
$1 \, 000$ 	$168$ 	$145$ 	$178$
$10 \, 000$ 	$1 \, 229$ 	$1 \, 068$ 	$1 \, 246$
$100 \, 000$ 	$9 \, 596$ 	$8 \, 686$ 	$9 \, 630$
$1 \, 000 \, 000$ 	$78 \, 498$ 	$72 \, 382$ 	$78 \, 628$
$10 \, 000 \, 000$	$664 \, 579$	$620 \, 421$ 	$664 \, 918$
== Sources ==
* : Chapter $\text {B}.16$: The Sequence of Primes
Category:Prime-Counting Function
Prime-Counting Function/Examples

== Definition ==
The values of the prime-counting ($\pi$) function for the first few integers are as follows:
:
$n$	$\map \pi n$
$1$	$0$
$2$	$1$
$3$	$2$
$4$	$2$
$5$	$3$
$6$	$3$
$7$	$4$
$8$	$4$
=== Example: $10$ ===
=== Example: $16$ ===
=== Example: $1000$ ===
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Example $2.1.1$
Category:Prime-Counting Function
Prime-Counting Function/Examples/16

== Definition ==
The value of the prime-counting ($\pi$) function for $16$ is $6$.
== Proof ==
The prime numbers less than $16$ are:
:$2, 3, 5, 7, 11, 13$
Hence:
:$\map \pi {16} = 6$
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Ponderable $2.1.1$
Category:Prime-Counting Function
Prime-Counting Function in terms of Eulerian Logarithmic Integral

== Theorem ==
Let $\map \pi x$ denote the prime-counting function of a number $x$.
Let $\map \Li x$ denote the Eulerian logarithmic integral of $x$:
:$\map \Li x := \ds \int_2^x \dfrac {\d t} {\ln t}$
Then:
:$\map \pi x = \map \Li x + \map \OO {x \map \exp {-c \sqrt {\ln x} } }$
where:
:$\OO$ is the big-O notation
:$c$ is some constant.
=== Riemann Hypothesis Holds ===
== Proof ==
== Sources ==
* : $0,5$
Category:Analytic Number Theory
Prime-Generating Quadratic of form 2 x squared minus 1000 x minus 2609

== Theorem ==
The quadratic function:
:$2 x^2 - 1000 x - 2609$
has $602$ prime values among its first $1000$ values.
Some of those prime numbers are negative.
== Proof ==
== Sources ==
* : $41$
Category:Polynomial Expressions for Primes
Prime-Generating Quadratic of form x squared - 79 x + 1601

== Theorem ==
The quadratic function:
:$x^2 - 79 x + 1601$
gives prime values for integer $x$ such that $0 \le x \le 79$.
The primes generated are repeated once each.
== Proof ==
Let $x = z + 40$.
Then:
Thus it can be seen that this is an application of Euler Lucky Number $41$.
== Sources ==
* : $41$
* : $41$
Category:Polynomial Expressions for Primes
Category:Euler Lucky Numbers
Prime-Generating Quadratics of form 2 a squared plus p

== Theorem ==
The quadratic form:
:$2 a^2 + p$
yields prime numbers for $a = 0, 1, \ldots, p - 1$ for values of $p$:
:$3, 5, 11, 29$
== Proof ==
=== 3 ===
=== 5 ===
=== 11 ===
=== 29 ===
When $x = p$ we have:
which has divisors $p$ and $2 p + 1$ and so is not prime.
== Sources ==
* 
* : $29$
Category:Prime Numbers
Category:Prime-Generating Quadratics of form 2 a squared plus p
Category:Polynomial Expressions for Primes
Prime-Generating Quadratics of form 2 a squared plus p/29

== Theorem ==
The quadratic form:
:$2 a^2 + 29$
yields prime numbers for $a = 0, 1, \ldots, 28$.
== Proof ==
== Sources ==
* : $29$
Category:Prime-Generating Quadratics of form 2 a squared plus p
Category:29
Prime Decomposition/Examples

== Examples of Prime Decompositions ==
The prime decompositions for the first few integers are as follows:
:
$n$ 	Prime Decomposition of $n$
$1$ 	$1$
$2$ 	$2$
$3$ 	$3$
$4$ 	$2^2$
$5$ 	$5$
$6$ 	$2 \times 3$
$7$ 	$7$
$8$ 	$2^3$
$9$ 	$3^2$
$10$	$2 \times 5$
$11$	$11$
$12$	$2^2 \times 3$
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Theorem $\text {2-5}$
Category:Prime Decompositions
Prime Decomposition of 10th Fermat Number/Historical Note

== Historical Note on Prime Decomposition of 10th Fermat Number ==
$F_{10}$ was proved composite in $1952$ by using Pépin's Test on the SWAC, but at that time the factors had yet to be determined.
 discovered the factor $45 \, 592 \, 577$ in $1953$, also using the SWAC.
At the same time he discovered the factor $825 \, 753 \, 601$ of $F_{16}$.
 discovered the factor $6 \, 487 \, 031 \, 809$ in $1962$ on an IBM 704.
 later found that the cofactor was a $291$-digit composite number.
The factors of this $291$-digit composite were finally found by in $1995$.
As he explained in his $1999$ article, the reason why the factors of $F_{11}$ were found so much earlier is that the second largest factor of $F_{11}$ had a mere $22$ digits, as opposed to the $40$ digits of that of $F_{10}$.
It may be noted that the author of this page performed this exercise in $2020$ using a factorisation tool freely available online, running on a machine of modest specifications.
It took just under $7$ hours in total.
== Sources ==
* 
* 
* 
* 
Category:Historical Notes
Prime Decomposition of 2^58+1

== Theorem ==
The number $2^{58} + 1$ has the prime decomposition:
:$2^{58} + 1 = 5 \times 107 \, 367 \, 629 \times 536 \, 903 \, 681$
== Proof ==
From Aurifeuillian Factorization of 2 Mod 4th Power of Two plus 1, we have:
:$2^{4 n + 2} + 1 = \paren {2^{2 n + 1} - 2^{n + 1} + 1} \paren {2^{2 n + 1} + 2^{n + 1} + 1}$
Setting $n = 14$:
== Historical Note ==
== Sources ==
* : $2^{58} + 1$
* : $2^{58} + 1$
Category:Prime Numbers
Category:288,230,376,151,711,745
Prime Decomposition of 2^58+1/Historical Note

== Historical Note on Prime Decomposition of $2^{58} + 1$ ==
The prime decomposition of $2^{58} + 1$ was accomplished by in $1869$.
In his words:
:No one of our numerous factorizations of $2^n \pm 1$ gave us as much trouble and labour as that of $2^{58} + 1$. This number is divisible by $5$ and if we remove this factor we obtain a number of $17$ digits whose factors have $9$ digits each. If we lose this result we shall miss patience and courage to repeat all calculations we have made and it is possible that many years will pass before someone else will discover the factorization of $2^{58} + 1$.
Then in $1871$, discovered the factorization:
:$2^{58} + 1 = \paren {2^{29} - 2^{15} + 1} \paren {2^{29} + 2^{15} + 1}$
which was subsequently generalized by .
In his of $1986$, reports that this result appears in Volume $128$ of , but as there do not appear to have been that many volumes of that publication, this statement is suspect.
== Sources ==
* : $2^{58} + 1$
* : $2^{58} + 1$
Category:Historical Notes
Prime Decomposition of 5th Fermat Number

== Theorem ==
The prime decomposition of the $5$th Fermat number is given by:
== Proof 1 ==
== Proof 2 ==
== Also see ==
* 
== Historical Note ==
== Sources ==
* : $257$
* 
* : $257$
* 
* 
* : Chapter $7$: Patterns in Numbers: Gauss
Category:4,294,967,297
Category:641
Category:6,700,417
Category:Fermat Numbers
Category:Prime Decomposition of 5th Fermat Number
Prime Decomposition of 5th Fermat Number/Historical Note

== Historical Note on Prime Decomposition of 5th Fermat Number ==
The prime decomposition of the $5$th Fermat number was determined by in $1732$.
== Sources ==
* : $257$
* : $641$
* : $257$
* : $641$
* 
* 
* : Chapter $7$: Patterns in Numbers: Gauss
Category:Historical Notes
Prime Decomposition of 5th Fermat Number/Mistake

== Source Work ==
:
:The Dictionary
::$641$
:
:The Dictionary
::$641$
== Mistake ==
: found the first counterexample to conjecture that $2^{2^n} + 1$ is always prime, when he discovered in $1742$ that $2^{2^5} + 1$ is divisible by $641$.
The date was in fact $1732$.
In the entry for $257$, gets the date right.
== Sources ==
* : $641$
* : $641$
Category:Mistakes/Curious and Interesting Numbers
Prime Decomposition of 5th Fermat Number/Proof 1

== Theorem ==
== Proof ==
From Divisor of Fermat Number, if $2^{\paren {2^n} } + 1$ has a divisor, it will be in the form:
:$k \, 2^{n + 2} + 1$
In the case of $n = 5$, a divisor of $2^{\paren {2^n} } + 1$ is then of the form:
:$k \, 2^7 + 1 = k \times 128 + 1$
Further, such a number will (for small $k$ at least) be prime, otherwise it will itself have a divisor which is not of that form.
Thus we investigate:
We have that:
But
:$2^8 + 1 = 257$
and so:
:$2^{\paren {2^5} } - 1 \equiv 0 \pmod {257}$
which means:
:$2^{\paren {2^5} } + 1 \equiv 2 \pmod {257}$
so there is no need to do a trial division of $2^{\paren {2^5} } + 1$ by $257$.
Now dividing $4 \, 294 \, 967 \, 297$ by $641$ we find:
 6 700 417
 ---------------
641 ) 4 294 967 297
 3 846
 -----
 448 9
 448 7
 -----
 267 2
 256 4
 -----
 10 89
 6 41
 -----
 4 487
 4 487
 =====
and the divisor has been found.
== Sources ==
* : $641$
* : $641$
Category:Prime Decomposition of 5th Fermat Number
Prime Decomposition of 5th Fermat Number/Proof 2

== Theorem ==
== Proof ==
Note the remarkable coincidence that $2^4 + 5^4 = 2^7 \cdot 5 + 1 = 641$.
First we eliminate $y$ from $x^4 + y^4 = x^7 y + 1 = 0$:
Now we use the above result for $x = 2$ and $y = 4$ in modulo $641$:
Thus $2^{\paren {2^5} } + 1 = 6 \, 700 \, 417 \times 641$ and hence is not prime.
Category:Prime Decomposition of 5th Fermat Number
Prime Decomposition of 6th Fermat Number

== Theorem ==
The prime decomposition of the $6$th Fermat number is given by:
== Also see ==
* 
== Historical Note ==
== Sources ==
* : $257$
* : $257$
Category:18,446,744,073,709,551,617
Category:274,177
Category:67,280,421,310,721
Category:Fermat Numbers