nyuuzyou/wikis huggingface dataset:
Prime Decomposition of 7th Fermat Number

== Theorem ==
The prime decomposition of the $7$th Fermat number is given by:
== Also see ==
* 
== Historical Note ==
== Sources ==
* : $257$
* : $257$
Category:340,282,366,920,938,463,463,374,607,431,768,211,457
Category:59,649,589,127,497,217
Category:5,704,689,200,685,129,054,721
Category:Fermat Numbers
Prime Decomposition of 7th Fermat Number/Historical Note

== Historical Note on Prime Decomposition of 7th Fermat Number ==
In $1909$, and reported in that they had proved that $F_7$ and $F_8$ are not prime, but without having established what the prime factors are.
The actual factors were not themselves determined until the work of and in $1970$.
== Sources ==
 
* 
* 
* : $257$
* : $257$
Category:Historical Notes
Prime Decomposition of 7th Fermat Number/Mistake

== Source Work ==
:
:The Dictionary
::$257$
:
:The Dictionary
::$257$
== Mistake ==
:Thus, in $1909$, Moorhead and Western proved that $F_7$ and $F_8$ are composite, without producing any factors.
== Correction ==
It is not Moorhead, but .
== Sources ==
* : $257$
* : $257$
Category:Mistakes/Curious and Interesting Numbers
Prime Decomposition of 8th Fermat Number/Historical Note

== Historical Note on Prime Decomposition of 8th Fermat Number ==
In $1909$, and reported in that they had proved that $F_8$ is not prime, but without having established what the prime factors are.
This factorisation was accomplished by and in $1981$.
Using a Monte Carlo method, they determined the prime factors, but were unable at the time to demonstrate that the larger factor was actually prime.
They devised a mnemonic for the smaller factor:
:I am now entirely persuaded to employ the method, a handy trick, on gigantic composite numbers.
$40$ years later, a factorisation tool freely available online, running on a machine of modest specifications, can determine the primality of the larger factor (and indeed, $F_8$ itself) practically instantaneously.
== Sources ==
* 
* : $257$
* : $257$
Category:Historical Notes
Prime Decomposition of Highly Composite Number

== Theorem ==
Let $n$ be a highly composite number.
Let the prime decomposition of $n$ be expressed as:
:$n = \ds \prod_{k \mathop \in \N} {p_k}^{r_k}$
where $p_k$ denotes the $k$th prime.
Then the sequence $\sequence {r_k}$ is decreasing.
That is:
:$\forall k \in \N: r_k \ge r_{k + 1}$
== Proof ==
Let $n = \ds \prod_{k \mathop \in \N} {p_k}^{r_k}$ be highly composite.
By definition of divisor count function:
:$\map {\sigma_0} n = \ds \prod_{k \mathop \in \N} \paren {r_k + 1}$
 $r_{l + 1} > r_l$ for some $l \in \N$.
Consider $m \in \Z$ whose prime decomposition of $n$ is expressed as:
:$m = \ds \prod_{k \mathop \in \N} {p_k}^{s_k}$
where:
:$\forall j < l: s_j = r_j$
:$\forall j > l + 1: s_j = r_j$
:$s_l = r_{l + 1}$
:$s_{l + 1} = r_l$
We have that:
Thus:
:$\map {\sigma_0} m = \map {\sigma_0} n$
Now we have that:
:$r_l < r_{l + 1}$
and so:
:$s_l > s_{l + 1}$
and so:
Then:
:$p_{l + 1} > p_l$
and so:
:${p_{l + 1} }^c > {p_l}^c$
from which it follows that:
:$m < n$
while:
:$\map {\sigma_0} m = \map {\sigma_0} n$
But $n$ is highly composite.
This means that if $\map {\sigma_0} m = \map {\sigma_0} n$, then $n \le m$.
This is contradicted by $n > m$.
Thus by Proof by Contradiction:
:$\forall k \in \N: r_k \ge r_{k + 1}$
== Sources ==
* 
Category:Highly Composite Numbers
Prime Decomposition of Integer is Unique

== Theorem ==
Let $n$ be an integer such that $n > 1$.
Then the prime decomposition of $n$ is unique.
== Proof ==
From Integer is Expressible as Product of Primes, $n$ can be expressed as the product of one or more primes.
Let $n = q_1 q_2 \dotsm q_s$ where $q_1, q_2, \ldots, q_s$ are all primes such that:
:$(1): \quad n = q_1 \le q_2 \le \dotsb \le q_s$
From Expression for Integer as Product of Primes is Unique, the expression for $(1)$ is unique.
By the Fundamental Theorem of Equivalence Relations, we can partition $\set {q_1, q_2, \dotsc, q_s}$ in $(1)$ according to equality.
Thus the equivalence classes $\eqclass {q_j} =$ contain all repetitions of $q_j$.
Hence the contribution of $q_j$ to $n$ is:
:${q_j}^{k_j}$
where $k_j = \card {\eqclass {q_j} =}$, the cardinality of $\eqclass {q_j} =$.
Renaming the representative elements of the various $\eqclass {q_r} =$ as $p_1, p_2, \ldots, p_r$, where $r$ is the number of equivalence classes.
Hence:
:$n = {p_1}^{k_1} {p_2}^{k_2} \dotsm {p_r}^{k_r}$
As $n = q_1 \le q_2 \le \dotsb \le q_s$ is a unique representation, so is $n = {p_1}^{k_1} {p_2}^{k_2} \dotsm {p_r}^{k_r}$.
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $3$
Category:Prime Decompositions
Prime Divides Power

== Theorem ==
Let $p$ be a prime number.
Let $a, n \in \Z$ be integers.
Then $p$ divides $a^n$ $p^n$ divides $a^n$.
== Proof ==
=== Sufficient Condition ===
Let $p^n \divides a^n$.
We have $p \divides p^n$ as $p^n = p \paren {p^{n - 1} }$.
From the fact that Divisor Relation is Transitive, we have that $p \divides a^n$.
=== Necessary Condition ===
Let $p \divides a^n$.
Using Euclid's Lemma for Prime Divisors with $a_1 = a_2 = \cdots = a_n = a$ we have that:
:$p \divides a^n \implies p \divides a$
Hence $a = p r$ for some $r \in \Z$.
Raising both sides of this equation to the power $n$ we get that:
:$a^n = p^n r^n$
So:
:$p^n \divides a^n$
Category:Prime Numbers
Prime Divisors of Cyclotomic Polynomials

== Theorem ==
Let $n \ge 1$ be a positive integer.
Let $\map {\Phi_n} x$ denote the $n$th cyclotomic polynomial.
Let $a \in \Z$ be an integer such that $\map {\Phi_n} a \ne 0$.
Let $p$ be a prime divisor of $\map {\Phi_n} a$.
Then $p \equiv 1 \pmod n$ or $p \divides n$.
== Proof ==
Let $k$ be the order of $a$ modulo $p$.
By Element to Power of Multiple of Order is Identity, $k \divides p - 1$.
If $k = n$, the result follows.
Let $k < n$.
Then by Product of Cyclotomic Polynomials, there exists $d \divides k$ such that $p \divides \map {\Phi_d} a$.
Consequently, $a$ is a double root of $\Phi_d \Phi_n$ modulo $p$.
Again by Product of Cyclotomic Polynomials, $a$ is a double root of $x^n - 1$ modulo $p$.
Thus, by Double Root of Polynomial is Root of Derivative, $a$ is a root of the derivative of $x^n - 1$ modulo $p$, which is the constant polynomial $n$.
Thus $n \equiv 0 \pmod p$, for a nonzero constant polynomial has no roots.
== Also see ==
* Cyclotomic Polynomial has Integer Coefficients
Category:Number Theory
Category:Cyclotomic Polynomials
Prime Dudeney Number

== Theorem ==
The only prime Dudeney number is $17$:
== Proof ==
We have that:
From Sequence of Dudeney Numbers, the complete set of positive integers with this property are:
:$0, 1, 8, 17, 18, 26, 27$
Of these, only $17$ is prime.
== Sources ==
* : $17$
* : $17$
Category:17
Category:Prime Numbers
Category:Dudeney Numbers
Prime Element iff Complement of Lower Closure is Filter

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded above lattice.
Let $p \in S$ such that
:$p \ne \top$
where $\top$ denotes the top of $L$.
Then:
:$p$ is a prime element
:$\relcomp S {p^\preceq}$ is filter in $L$
where
:$p^\preceq$ denotes the lower closure of $p$
:$\relcomp S {p^\preceq}$ denotes the relative complement of $p^\preceq$ relative to $S$.
== Proof ==
=== Sufficient Condition ===
Assume that $p$ is a prime element.
By definition of the greatest element:
:$p \preceq \top$
By definition of $\prec$:
:$p \prec \top$
By definition of antisymmetry:
:$\top \npreceq p$
By definition of lower closure of element:
:$\top \notin p^\preceq$
By definition of relative complement:
:$\top \in \relcomp S {p^\preceq}$
By definition
:$X := \relcomp S {p^\preceq}$ is a non-empty set
We will prove that
:$X$ is filtered.
Let $x, y \in X$.
By definition of relative complement:
:$x \notin p^\preceq$ and $y \notin p^\preceq$
By definition of lower closure of element:
:$x \npreceq p$ and $y \npreceq p$
Then by definition of prime element:
:$x \wedge y \npreceq p$
By definition of lower closure of element:
:$x \wedge y \notin p^\preceq$
Thus by definition of relative complement:
:$x \wedge y \in X$
Thus by Meet Precedes Operands:
:$x \wedge y \preceq x$ and $x \wedge y \preceq y$
Thus
:$\exists z \in X: z \preceq x \land z \preceq y$
We will prove that $X$ is an upper section.
Let $x \in X, y \in S$ such that:
:$x \preceq y$
By definition of relative complement:
:$x \notin p^\preceq$
By definition of lower closure of element:
:$x \npreceq p$
By definition of transitivity:
:$y \npreceq p$
By definition of lower closure of element:
:$y \notin p^\preceq$
Thud by definition of relative complement:
:$y \in X$
Hence, by definition, $X$ is filter in $L$.
=== Necessary Condition ===
Assume that $\relcomp S {p^\preceq}$ is filter in $L$.
Let $x, y \in S$ such that:
:$x \wedge y \preceq p$
 that:
:$x \npreceq p$ and $y \npreceq p$
By definition of lower closure of element:
$x \notin p^\preceq$ and $y \notin p^\preceq$
By definition of relative complement:
:$x \in \relcomp S {p^\preceq}$ and $y \in \relcomp S {p^\preceq}$
By definition of filtered:
:$\exists z \in \relcomp S {p^\preceq}: z \preceq x \text { and } z \preceq y$
Then:
$z \preceq x \wedge y$
By definition of transitivity:
:$z \preceq p$
By definition of upper section:
:$p \in \relcomp S {p^\preceq}$
By definition of reflexivity:
:$p \preceq p$
By definition of lower closure of element:
:$p \in p^\preceq$
This contradicts $p \in \relcomp S {p^\preceq}$.
Hence it is not the case that $x \npreceq p$ and $y \npreceq p$.
That is:
:$x \preceq p$ or $y \preceq p$
and so, by definition, $p$ is a prime element of $\struct {S, \vee, \wedge, \preceq}$.
== Sources ==
* 
* 
Category:Prime Elements
Prime Element iff Element Greater is Top

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a Boolean lattice.
Let $p \in S$ such that:
:$p \ne \top$
Then:
:$p$ is prime element
:
:$\forall x \in S: \paren {p \prec x \implies x = \top}$
== Proof ==
=== Sufficient Condition ===
Suppose that:
:$p$ is prime element
Let $x \in S$ such that:
:$p \prec x$
By definition of Boolean lattice:
:$L$ is complemented distributive lattice.
By definition of complemented:
:$\exists y \in S: y$ is complement of $x$.
By definition of complement:
:$x \wedge y = \bot$
By definition of smallest element:
:$x \wedge y \preceq p$
By definition of prime element:
:$x \preceq p$ or $y \preceq p$
Then by definition of antisymmetry:
:$y \prec x$
By definition of $\prec$:
:$y \preceq x$
By definition of complement:
:$x \vee y = \top$
Thus by Preceding iff Join equals Larger Operand:
:$x = \top$
=== Necessary Condition ===
Suppose:
:$\forall x \in S: \paren {p \prec x \implies x = \top}$
Let $x, y \in S$ such that:
:$x \wedge y \preceq p$
:
:$x \npreceq p$ and $y \npreceq p$
By Preceding iff Join equals Larger Operand:
:$p \ne p \vee y$ and $p \ne p \vee x$
By Join Succeeds Operands:
:$p \preceq p \vee y$ and $p \preceq p \vee x$
By definition of $\prec$:
:$p \prec p \vee y$ and $p \prec p \vee x$
By assumption:
:$p \vee y = \top$ and $p \vee x = \top$
By Meet is Idempotent:
:$p = \top$
This contradicts $p \ne \top$.
== Sources ==
* 
* 
Category:Prime Elements
Prime Element iff Meet Irreducible in Distributive Lattice

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a distributive lattice.
Let $p \in S$.
Then $p$ is a prime element $p$ is meet irreducible.
== Proof ==
By Prime Element is Meet Irreducible:
:if $p$ is a prime element, then $p$ is meet irreducible.
Assume that:
:$p$ is meet irreducible.
Let $x, y \in S$ such that:
:$x \wedge y \preceq p$
By definition of meet irreducible:
:$p = p \vee x$ or $p = p \vee y$
Thus by Preceding iff Join equals Larger Operand:
:$x \preceq p$ or $y \preceq p$
== Sources ==
* 
* 
Category:Prime Elements
Category:Meet Irreducible Elements
Prime Element iff There Exists Way Below Open Filter which Complement has Maximum

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous distributive lattice.
Let $p \in S$ such that:
:$p \ne \top$
where $\top$ denotes the top of $L$.
Then
:$p$ is a prime element
:
:there exists a way below open filter $F$ in $L$: $p = \map \max {\relcomp S F}$
== Proof ==
=== Sufficient Condition ===
Let $p$ be a prime element.
By definition of continuous:
:$\forall x \in S: x^\ll$ is directed
We will prove that:
:$\forall x \in S: \paren {x \in \relcomp S {p^\preceq} \implies \exists y \in S: y \in \relcomp S {p^\preceq} \land y \ll x}$
Let $x \in S$ such that:
:$x \in \relcomp S {p^\preceq}$
By definition of relative complement:
:$x \notin p^\preceq$
By definition of lower closure of element:
:$x \npreceq p$
By Axiom of Approximation in Up-Complete Semilattice:
:$\exists y \in S: y \ll x \land y \npreceq p$
By definition of lower closure of element:
:$y \notin p^\preceq$
By definition of relative complement:
:$y \in \relcomp S {p^\preceq}$
Thus:
:$\exists y \in S: y \in \relcomp S {p^\preceq} \land y \ll x$
By Prime Element iff Complement of Lower Closure is Filter:
:$F := \relcomp S {p^\preceq}$ is way below open filter in $L$.
By definitions of antisymmetry and lower closure of element:
:$\lnot \exists y \in S: y \in F \land y \prec p$
By Relative Complement of Relative Complement::
:$\relcomp S F = p^\preceq$
By definitions of reflexivity and lower closure of element:
:$p \in \relcomp S F$
Thus by definition of greatest element:
:$p = \map \max {\relcomp S F}$
=== Necessary Condition ===
Let:
:there exists a way below open filter $F$ in $L$: $p = \map \max {\relcomp S F}$
By Maximal Element of Complement of Filter is Meet Irreducible:
:$p$ is meet irreducible.
Thus by Prime Element iff Meet Irreducible in Distributive Lattice:
:$p$ is a prime element.
== Sources ==
* 
* 
Category:Prime Elements
Prime Element is Meet Irreducible

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $p \in S$.
Let $p$ be a prime element of $L$.
Then $p$ is meet irreducible in $L$.
== Proof ==
Let $p$ be a prime element.
Let $x, y \in S$ such that
:$p = x \wedge y$
By definition of reflexivity:
:$x \wedge y \preceq p$
By definition of prime element:
:$x \preceq p$ or $y \preceq p$
By Meet Precedes Operands:
:$p \preceq x$ and $p \preceq y$
Thus by definition of antisymmetry:
:$p = x$ or $p = y$
== Sources ==
* 
* 
Category:Prime Elements
Category:Meet Irreducible Elements
Prime Element of Integral Domain is Irreducible

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Let $p$ be a prime element of $\struct {D, +, \circ}$.
Then $p$ is an irreducible element of $\struct {D, +, \circ}$.
== Proof ==
By definition of prime element, $p$ is neither zero nor a unit of $\struct {D, +, \circ}$.
:
:$p = a \circ b$
for some non-units $a, b \in D$.
From Element of Integral Domain is Divisor of Itself:
:$p \divides a \circ b$
By definition of prime element:
:$p \divides a$ or $p \divides b$
, suppose $p \divides a$.
That is:
:$a = p \circ c$
for some $c \in D$.
Thus $b$ is a unit of $\struct {D, +, \circ}$.
This contradicts the assertion that $a$ and $b$ are not units of $\struct {D, +, \circ}$.
Thus by Proof by Contradiction $p$ is an irreducible element of $\struct {D, +, \circ}$.
Category:Integral Domains
Category:Irreducible Elements of Rings
Category:Prime Elements of Rings
Prime Enumeration Function is Primitive Recursive

== Theorem ==
Let the function $p: \N \to \N$ be the prime enumeration function, defined as:
:$\map p n = \begin{cases}
1 & : n = 0 \\
\mbox{the } n \mbox{th prime number} & : n > 0
\end{cases}$
Then $p$ is primitive recursive.
== Proof ==
We can define $p$ recursively by:
:$\map p {n + 1} = \text{the smallest } y \in \N \text { such that } y \text { is prime and } \map p n < y$
Hence we can express it as:
:$\map p {n + 1} = \map {\mu y} {\map {\chi_\Bbb P} y = 1 \land \map p n < y}$
where:
* $\map {\chi_\Bbb P} y$ is the characteristic function of the set of prime numbers $\Bbb P$
* $\map {\mu y} \RR$ means the smallest $y \in \N$ such that the relation $\RR$ holds.
Now consider the relation $\SS$ given by:
:$\map \SS {m, y} \iff \map {\chi_\Bbb P} y = 1$.
We have a reason for making $\SS$ a binary relation, even though $m$ is not actually invoked in its definition.
Then we have:
:$\map {\chi_\SS} {m, y} = \map {\chi_{\operatorname{eq} } } {\map {\chi_\Bbb P} y, 1}$.
So $\chi_\SS$ is primitive recursive as it is obtained by substitution from:
* the primitive recursive function $\chi_{\operatorname{eq} }$
* the primitive recursive function $\chi_\Bbb P$.
Then we have that $<$ is primitive recursive.
So we define the relation $\RR$ by:
:$\map \RR {m, y} \iff \map \SS {m, y} \land m < y \iff \map {\chi_\Bbb P} y = 1 \land m < y$.
This is primitive recursive from the above and Set Operations on Primitive Recursive Relations.
Now let us define the function $g: \N^2 \to \N$ as:
:$\map g {m, z} = \mu y \le \map z {\map {\chi_\Bbb P} y = 1 \land m < y}$
which is primitive recursive by Bounded Minimization is Primitive Recursive.
We note that $\map g {\map p n, z} = \map p {n + 1}$ as long as $\map p {n + 1} \le z$.
Next, let $h: \N \to \N$ be defined as $\map h n = \map \exp {2, \map \exp {2, n} }$.
Then $h$ is primitive recursive since it is obtained by substitution from:
* the primitive recursive function $\exp$;
* the constant $2$.
From Upper Bounds for Prime Numbers, we have $\map p {n + 1} \le 2^{2^n} = \map h n$.
It follows that:
:$\map p {n + 1} = \map g {\map p n, \map h n}$
where $g$ and $h$ are both primitive recursive.
So using the definition of $p$ as given above, we have:
:$\map p 0 = 1$
:$\map p {n + 1} = \map g {\map p n, \map h n}$.
So $p$ is defined by primitive recursion from the constant $1$ and the primitive recursive functions $g$ and $h$.
== Note ==
Note that we use the extravagantly large upper bound $2^{2^n}$ for the prime number $\map p {n + 1}$ because it is convenient in this context. Smaller ones would of course do the same job.
Category:Primitive Recursive Functions
Prime Exponent Function is Primitive Recursive

== Theorem ==
Let $n \in \N$ be a natural number.
Let $\tuple {n, j}: \N^2 \to \N$ be defined as:
:$\tuple {n, j} = \paren n_j$
where $\paren n_j$ is the prime exponent function.
Then $\tuple {n, j}$ is primitive recursive.
== Proof ==
Let $\map p j$ be the prime enumeration function.
For $n \ne 0$ and $j \ne 0$, we see that $\paren n_j$ is the largest value of $k$ for which $\map p j^k$ is a divisor of $n$.
Thus $\paren n_j$ is the smallest value of $k$ for which $\map p j^{k + 1}$ is not a divisor of $n$.
We note that if $r \ge n$ and $j \ne 0$, we have $\map p j^r \ge 2^r \ge 2^n > n$.
Thus $n$ is a (generous) upper bound of $\paren n_j$.
The condition that $\map p j^{k + 1}$ is not a divisor of $n$ can be expressed as:
:$\map {\operatorname{div} } {n, \map p j^{k + 1} } = 0$
where:
:$\operatorname{div}$ is primitive recursive
:The Equality Relation is Primitive Recursive
:$\map p j$ is primitive recursive
:Exponentiation is Primitive Recursive
:Addition is Primitive Recursive.
So we see that the relation:
:$\map \RR {n, j, k} \iff \map {\operatorname{div} } {n, \map p j^{k + 1} } = 0$
is primitive recursive.
From Bounded Minimization is Primitive Recursive, we also see that:
:$\paren n_j = \begin{cases}
\mu k \le n \map \RR {n, j, k} & : n \ne 0 \land j \ne 0 \\
0 & : \text{otherwise}
\end{cases}$
is primitive recursive.
The result follows.
Category:Primitive Recursive Functions
Prime Factors of 1000 Factorial

== Example of Factorial ==
The prime decomposition of $1000!$ is given as:
== Proof ==
For each prime factor $p$ of $1000!$, let $a_p$ be the integer such that:
:$p^{a_p} \divides 1000!$
:$p^{a_p + 1} \nmid 1000!$
Taking the prime factors in turn:
=== Multiplicity of $2$ ===
=== Multiplicity of $3$ ===
=== Multiplicity of $5$ ===
=== Multiplicity of $7$ ===
=== Multiplicity of $11$ ===
=== Multiplicity of $13$ ===
And so on: for the remaining factors:
The pattern is clear: the remaining indices are calculated in the same way.
Category:Factorials/Examples
Category:1000
Prime Factors of 13 Factorial

== Example of Factorial ==
The prime decomposition of $13!$ is given as:
:$13! = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \times 13$
== Proof ==
For each prime factor $p$ of $13!$, let $a_p$ be the integer such that:
:$p^{a_p} \divides 13!$
:$p^{a_p + 1} \nmid 13!$
Taking the prime factors in turn:
Similarly:
:$a_{11} = 1$
:$a_{13} = 1$
Hence the result.
Category:Factorials/Examples
Category:13
Prime Factors of 20 Factorial

== Example of Factorial ==
The prime decomposition of $20!$ is given as:
:$20! = 2^{18} \times 3^8 \times 5^4 \times 7^2 \times 11 \times 13 \times 17 \times 19$
== Proof ==
For each prime factor $p$ of $20!$, let $a_p$ be the integer such that:
:$p^{a_p} \divides 20!$
:$p^{a_p + 1} \nmid 20!$
Taking the prime factors in turn:
Similarly:
:$a_{11} = 1$
:$a_{13} = 1$
:$a_{17} = 1$
:$a_{19} = 1$
Hence the result.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $6$
Category:Factorials/Examples
Category:Prime Decompositions
Category:20
Prime Factors of 2^64 - 1

== Theorem ==
The prime decomposition of $2^{64} - 1$ is given by:
:$2^{64} - 1 = 3 \times 5 \times 17 \times 257 \times 641 \times 65 \, 537 \times 6 \, 700 \, 417$
== Proof ==
From Sequence of Fermat Primes:
all of which are Fermat primes.
Then:
Hence the result.
Category:18,446,744,073,709,551,615
Prime Factors of 35, 36, 4734 and 4735

== Theorem ==
The integers:
:$35, 4374$
have the same prime factors between them as the integers:
:$36, 4375$
== Proof ==
We have:
Thus both pairs of integers can be seen to have the same prime factors:
:$2, 3, 5, 7$
== Sources ==
* : $35$
:::but there is a mistake in his exposition: the pairings are the wrong way round.
Category:Factorization
Category:35
Category:36
Category:4374
Category:4375
Prime Factors of 35, 36, 4734 and 4735/Mistake

== Source Work ==
:
:The Dictionary
::$35$
== Mistake ==
:$35$ and $4375$ have the same prime factors between them (namely $2$, $3$, $5$ and $7$) as have $36$ and $4374$.
The pairings are the wrong way round. It should say:
:$35$ and $4374$ have the same prime factors between them (namely $2$, $3$, $5$ and $7$) as have $36$ and $4375$.
== Sources ==
* : $35$
Category:Mistakes/Curious and Interesting Numbers
Prime Factors of 39 Factorial

== Example of Factorial ==
The prime decomposition of $39!$ is given as:
:$39! = 2^{35} \times 3^{18} \times 5^8 \times 7^5 \times 11^3 \times 13^3 \times 17^2 \times 19^2 \times 23 \times 29 \times 31 \times 37$
== Proof ==
For each prime factor $p$ of $39!$, let $a_p$ be the integer such that:
:$p^{a_p} \divides 39!$
:$p^{a_p + 1} \nmid 39!$
Taking the prime factors in turn:
Similarly:
:$a_{23} = 1$
:$a_{29} = 1$
:$a_{31} = 1$
:$a_{37} = 1$
Hence the result.
Category:Factorials/Examples
Category:39
Prime Factors of 52 Factorial

== Example of Factorial ==
The prime decomposition of $52!$ is given as:
:$52! = 2^{49} \times 3^{23} \times 5^{12} \times 7^8 \times 11^4 \times 13^4 \times 17^3 \times 19^2 \times 23^2 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47$
== Proof ==
For each prime factor $p$ of $52!$, let $a_p$ be the integer such that:
:$p^{a_p} \divides 52!$
:$p^{a_p + 1} \nmid 52!$
Taking the prime factors in turn:
Similarly:
:$a_{29} = 1$
:$a_{31} = 1$
:$a_{37} = 1$
:$a_{41} = 1$
:$a_{43} = 1$
:$a_{47} = 1$
Hence the result.
Category:Factorials/Examples
Category:52
Prime Filter is Prime Ideal in Dual Lattice

== Theorem ==
Let $L = \struct {S, \preceq}$ be a lattice.
Let $X$ be a subset of $S$.
Then
:$X$ is a prime filter in $L$
:
:$X$ is a prime ideal in $L^{-1}$
where $L^{-1} = \struct {S, \succeq}$ denotes the dual of $L$.
== Proof ==
By Dual of Dual Ordering:
:dual of $L^{-1}$ is $L$.
Hence by Prime Ideal is Prime Filter in Dual Lattice:
:the result follows.
== Sources ==
* 
* 
Category:Prime Ideals (Order Theory)
Category:Dual Orderings