nyuuzyou/wikis huggingface dataset:
Integers are Arbitrarily Close to P-adic Integers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $x \in \Z_p$.
Then for $n \in \N$ there exists unique $\alpha \in \Z$:
:$(1): \quad 0 \le \alpha \le p^n - 1$
:$(2): \quad \norm { x -\alpha}_p \le p^{-n}$
== Proof ==
Let $n \in \N$.
From Rational Numbers are Dense Subfield of P-adic Numbers:
:the rational numbers are dense in $\Q_p$.
So there exists:
:$\dfrac a b \in \Q: \norm {x - \dfrac a b}_p \le p^{-n}$
From Unique Integer Close to Rational in Valuation Ring of P-adic Norm, there exists unique $\alpha \in \Z$ such that:
:$\norm {\dfrac a b - \alpha}_p \le p^{-n}$
:$0 \le \alpha \le p^n - 1$
Then:
Now suppose $\beta \in \Z$ also satisfies conditions $(1)$ and $(2)$, that is:
:$0 \le \beta \le p^n - 1$
:$\norm { x - \beta}_p \le p^{-n}$
Then:
From Unique Integer Close to Rational in Valuation Ring of P-adic Norm, $\alpha \in \Z$ was unique, so:
:$\beta = \alpha$
The result follows.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.4 \ \text{(ii)}$
Category:P-adic Number Theory
Integers are Coprime iff Powers are Coprime

== Theorem ==
Let $a, b \in \Z$ be integers.
Then:
:$a \perp b \iff \forall n \in \N: a^n \perp b^n$
That is, two integers are coprime all their positive integer powers are coprime.
== Proof ==
The forward implication is shown in Powers of Coprime Numbers are Coprime.
The reverse implication is shown by substituting $n = 1$.
Category:Coprime Integers
Integers are Countably Infinite

== Theorem ==
The set $\Z$ of integers is countably infinite.
== Proof ==
Define the inclusion mapping $i: \N \to \Z$.
From Inclusion Mapping is Injection, $i: \N \to \Z$ is an injection.
Thus there exists an injection from $\N$ to $\Z$.
Hence $\Z$ is infinite.
Next, let us arrange $\Z$ in the following order:
:$\Z = \set {0, 1, -1, 2, -2, 3, -3, \ldots}$
Then we can directly see that we can define a mapping $\phi: \Z \to \N$ as follows:
:$\forall x \in \Z: \map \phi x = \begin{cases} 2 x - 1 & : x > 0 \\ -2 x & : x \le 0 \end{cases}$
This is shown to be an injection as follows:
Let $\map \phi x = \map \phi y$.
Then one of the following applies:
: $(1): \quad -2 x = -2 y$ in which case $x = y$
: $(2): \quad 2 x - 1 = 2 y - 1$ in which case $2 x = 2 y$ and so $x = y$
: $(3): \quad 2 x - 1 = -2 y$ in which case $y = -x + \frac 1 2$ and therefore $y \notin \Z$
: $(4): \quad 2 y - 1 = -2 x$ in which case $x = -y + \frac 1 2$ and therefore $x \notin \Z$.
So $2 x - 1 = -2 y$ and $2 y - 1 = -2 x$ can't happen and so $x = y$.
Thus $\phi$ is injective.
The result follows from Domain of Injection to Countable Set is Countable.
== Sources ==
* : $\S 1.1$. Sets: Example $6$
* : $\S 2.2$: Countable sets: Example $1$
* : Chapter $1$: Mappings: $\S 15$
* : Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-1}$: The Real Numbers
* : $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets: Example $1$
* : $2.5 \ \text{(iii)}$
* : Appendix $\text{A}$: Set Theory: Countable Sets
* : Chapter $1$: General Background: $\S 2$ Countable or uncountable?
Category:Integers
Category:Countable Sets
Integers are Dense in P-adic Integers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $d_p$ be the metric induced by the norm $\norm {\,\cdot\,}_p$ restricted to the $p$-adic integers.
The integers $\Z$ are dense in the metric space $\struct{\Z_p, d_p}$.
=== Corollary ===
== Proof ==
From Open Ball Characterization of Denseness it is sufficient to show that every open ball of $\struct {\Z_p, d_p}$ contains an element of $\Z$.
Let $x \in \Z_p$ and $\epsilon \in \R_{>0}$.
By definition the open ball $\map {B_\epsilon} x$ is:
:$\map {B_\epsilon} x = \set {y \in \Z_p: \norm y_p < \epsilon}$
From Sequence of Powers of Number less than One then:
:$\ds \lim_{n \mathop \to \infty} p^{-n} = 0$
Hence there exists $N \in \N$:
:$\forall n \ge N: p^{-n} < \epsilon$
Consider the open ball $\map {B_{p^{-N} } } x$.
Since $0 < p^{-n} < \epsilon$ then:
:$\map {B_{p^{-N} } } x \subseteq \map {B_\epsilon} x$.
From Integers are Arbitrarily Close to P-adic Integers then:
:$\exists \alpha \in \Z: \alpha \in \map {B_{p^{-N} } } x$
Hence $\alpha \in \map {B_\epsilon} x$.
Since $x$ and $\epsilon$ were arbitrary then every open ball of $\struct {\Z_p, d_p}$ contains an element of $\Z$.
From Open Ball Characterization of Denseness then $\Z$ is dense in the metric space $\struct {\Z_p, d_p}$.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.4 \ \text {(ii)}$
Category:P-adic Number Theory
Category:Integers are Dense in P-adic Integers
Integers are Dense in P-adic Integers/Unit Ball

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
The integers $\Z$ are dense in the closed ball $\map {B^-_1} 0$.
== Proof ==
We have:
From Integers are Dense in P-adic Integers, the integers $\Z$ are dense in the closed ball $\map {B^-_1} 0$.
Category:Integers are Dense in P-adic Integers
Integers are Euclidean Domain

== Theorem ==
The integers $\Z$ with the mapping $\nu: \Z \to \Z$ defined as:
:$\forall x \in \Z: \map \nu x = \size x$
form a Euclidean domain.
== Proof ==
From Integers form Ordered Integral Domain we have that $\struct {\Z, +, \times, \le}$ forms an ordered integral domain.
For all $a \in \Z$, the absolute value of $a$ is defined as:
:$\size a = \begin{cases} a & : 0 \le a \\ -a & : a < 0 \end{cases}$
By Product of Absolute Values on Ordered Integral Domain we have:
:$\forall a, b \in \Z: \size a \cdot \size b = \size {a b}$
Since $\size b > 0$ it follows from Relation Induced by Strict Positivity Property is Compatible with Multiplication that $\size a \cdot \size b \ge \size a$.
The second criterion:
:For any $a, b \in R$, $b \ne 0$, there exist $q, r \in R$ with $\map \nu r < \map \nu b$, or $r = 0$ such that:
::$ a = q \circ b + r$
follows from the Division Theorem.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 27$. Euclidean Rings: Example $51$
Category:Integers
Category:Examples of Euclidean Domains
Integers are not Densely Ordered

== Theorem ==
The integers $\Z$ are not densely ordered.
That is:
:$\forall n \in \Z: \not \exists m \in \Z: n < m < n + 1$
== Proof ==
By definition of immediate successor element, this is equivalent to the statement:
:$\forall n \in \Z: n + 1$ is the immediate successor to $n$
We have that Integers form Ordered Integral Domain.
From One Succeeds Zero in Well-Ordered Integral Domain:
:$\not \exists r \in \Z: 0 < r < 1$
From Properties of Ordered Ring:
:$a < b \implies n + a < n + b$
Putting $a = 0, b = 1, m = n + r$:
:$\not \exists m \in \Z: n + 0 < m < n + 1$
Hence the result.
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers
Category:Integers
Category:Densely Ordered
Integers as Sum of Three Pairwise Coprime Integers

== Theorem ==
Let $n$ be an integer greater than $17$.
Then $n$ is the sum of $3$ integers greater than $1$ which are pairwise coprime.
== Proof ==
=== Case $1$: $n$ is even ===
There is some integer $k > 2$ such that one of the following holds:
:$n = 6 k = 2 + 3 + \paren {6 k - 5}$
:$n = 6 k + 2 = 4 + 3 + \paren {6 k - 7}$
:$n = 6 k + 4 = 2 + 3 + \paren {6 k - 5}$
All terms are greater than $1$.
The first two terms are powers of $2$ and $3$, so they are coprime.
The last term is not divisible by $2$ or $3$, so the three numbers are pairwise coprime.
=== Case $2$: $n$ is odd ===
There is some integer $k \ge 1$ such that one of the following holds:
:$n = 12 k + 1 = 3 + \paren {6 k - 7} + \paren {6 k + 5}$ and $k \ge 2$
:$n = 12 k + 3 = 9 + \paren {6 k - 5} + \paren {6 k - 1}$ and $k \ge 2$
:$n = 12 k + 5 = 3 + \paren {6 k - 5} + \paren {6 k + 7}$ and $k \ge 2$
:$n = 12 k + 7 = 3 + \paren {6 k - 1} + \paren {6 k + 5}$
:$n = 12 k + 9 = 3 + \paren {6 k - 1} + \paren {6 k + 7}$
:$n = 12 k + 11 = 3 + \paren {6 k + 1} + \paren {6 k + 7}$
All terms are greater than $1$.
The first term is a power of $3$.
The last two terms are not divisible by $3$, so they are coprime with $3$.
 the last two terms are not coprime.
Then they must have a common divisor greater than $1$.
This common divisor must also divide their difference.
However their difference is only divisible by $2$ and $3$, and the terms themselves are not.
Hence they cannot have a common divisor greater than $1$.
So they must be coprime.
Hence the result.
== Sources ==
* : $48$
* : $17$
Category:Number Theory
Category:17
Integers do not form Field

== Corollary of Invertible Integers under Multiplication ==
The integers $\struct {\Z, +, \times}$ do not form a field.
== Proof ==
For $\struct {\Z, +, \times}$ to be a field, it would require that all elements of $\Z$ have an inverse.
However, from Invertible Integers under Multiplication, only $1$ and $-1$ have inverses (each other).
== Examples ==
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $1$
* : Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $17$
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $3$
* 
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers
* 
Category:Integers
Category:Examples of Fields
Category:Integers do not form Field
Integers do not form Field/Examples/2

== Examples of Use of Integers do not form Field ==
Take $2$, for example.
In the field of rational numbers, we have that $2 \times \dfrac 1 2 = 1$ and so $2$ has an inverse in $\Q$.
But that inverse is not in $\Z$.
== Sources ==
* : Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $17$
Category:Integers do not form Field
Integers for which Divisor Sum of Phi equals Divisor Sum

== Theorem ==
The following positive integers have the property that the divisor sum of their Euler $\phi$ value equals their divisor sum:
:$\map {\sigma_1} {\map \phi n} = \map {\sigma_1} n$
:$1, 87, 362, 1257, 1798, 5002, 9374, \ldots$
== Proof ==
== Sources ==
* 
* : $87$
Category:Divisor Sum Function
Category:Euler Phi Function
Category:Integers for which Divisor Sum of Phi equals Divisor Sum
Integers form Commutative Ring

== Theorem ==
The set of integers $\Z$ forms a commutative ring under addition and multiplication.
== Proof ==
We have that:
:From Integers under Addition form Abelian Group, the algebraic structure $\struct {\Z, +}$ is an abelian group.
:From Integers under Multiplication form Countably Infinite Commutative Monoid, the algebraic structure $\struct {\Z, \times}$ is a commutative monoid and therefore a commutative semigroup.
:Integer Multiplication Distributes over Addition.
Thus all the ring axioms are fulfilled, and $\struct {\Z, +, \times}$ is a commutative ring.
By Integer Multiplication has Zero, the zero is $0$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.1$
* : Chapter $1$: Rings - Definitions and Examples: $1$: The definition of a ring: Definitions $1.1 \ \text{(c)}$
* : $\S 4$: Number systems $\text{I}$: The rational integers
* : $\S 55$. Special types of ring and ring elements: $(1)$
Category:Integers
Category:Commutative Rings
Integers form Commutative Ring with Unity

== Theorem ==
The integers $\struct {\Z, +, \times}$ form a commutative ring with unity under addition and multiplication.
== Proof ==
We have that:
:$\struct {\Z, +, \times}$ form a commutative ring.
:$\struct {\Z, +, \times}$ has a unity, and the unity is $1$.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $1$
Category:Integers
Category:Commutative Algebra
Integers form Integral Domain

== Theorem ==
The set of integers $\Z$ form an integral domain under addition and multiplication.
== Proof ==
First we note that the integers form a commutative ring with unity whose zero is $0$ and whose unity is $1$.
Next we see that the $\struct {\Z, +, \times}$ has no divisors of zero.
So, by definition, the algebraic structure $\struct {\Z, +, \times}$ is an integral domain whose zero is $0$ and whose unity is $1$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains
* : Chapter $1$: Integral Domains: $\S 3$. Definition of an Integral Domain
* : $\S 55$. Special types of ring and ring elements: $(5)$
* : $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization
* 
* 
* 
* 
* 
Category:Integers
Category:Examples of Integral Domains
Integers form Ordered Integral Domain

== Theorem ==
The integers $\Z$ form an ordered integral domain under addition and multiplication.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $9$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
Category:Integers
Category:Examples of Ordered Integral Domains
Category:Integers form Ordered Integral Domain
Integers form Ordered Integral Domain/Proof 1

== Theorem ==
== Proof ==
From Integers form Integral Domain we have that $\struct {\Z, +, \times}$ forms an integral domain.
From Natural Numbers are Non-Negative Integers we have that the set $\N$ can be considered as a subset of the integers.
Then we have that $\struct {\N_{>0}, +, \times}$ is a (commutative) semiring.
So it follows by definition of semiring, in particular the fact that $+$ and $\times$ on $\N_{> 0}$ are closed, that:
:$\forall a, b \in \N_{> 0}: a + b \in \N_{>0}$
:$\forall a, b \in \N_{> 0}: a \times b \in \N_{>0}$
It follows that we can define a property $P$ on $\Z$ such that:
:$\forall x \in \Z: \map P x \iff x \in \N_{>0}$
Checking that $P$ fulfils the conditions for it to be the (strict) positivity property:
:$(1): \quad \forall a, b \in \Z: \map P a \land \map P b \implies \map P {a + b}$
Follows directly from the fact that $\map P x \iff x \in \N_{> 0}$.
:$(2): \quad \forall a, b \in \Z: \map P a \land \map P b \implies \map P {a \times b}$
Follows directly from the fact that $\map P x \iff x \in \N_{> 0}$.
:$(3): \quad \forall a \in \Z: \map P a \lor \map P {-a} \lor a = 0$
This follows from the definition of the integers as the Inverse Completion of Natural Numbers.
If $a \in \Z - \N$ then $\exists b \in \N: a + b = 0$ and so $a = -b$.
Hence the conditions are fulfilled and $\struct {\Z, +, \times}$ forms an ordered integral domain.
Category:Integers form Ordered Integral Domain
Integers form Subdomain of Rationals

== Theorem ==
The integral domain of integers $\left({\Z, +, \times}\right)$ forms a subdomain of the field of rational numbers.
== Proof ==
The rational numbers are defined as the quotient field of the integers.
From its method of construction, it follows that the integers $\Z$ are a subset of the rational numbers $\Q$.
Hence the result, from the definition of subdomain.
== Sources ==
* : $\S 2.1$: Subrings: Examples $1$
Category:Integers
Category:Rational Numbers
Integers form Subdomain of Reals

== Theorem ==
The integral domain of integers $\struct {\Z, +, \times}$ forms a subdomain of the field of real numbers.
== Proof ==
We have that Integers form Subdomain of Rationals.
We have that Rational Numbers form Subfield of Real Numbers.
Hence the result, from the definition of subdomain.
Category:Integers
Category:Real Numbers
Integers form Subring of Reals

== Theorem ==
The ring of integers $\struct {\Z, +, \times}$ forms a subring of the field of real numbers.
== Proof ==
We have that the set of integers $\Z$ are a subset of the real numbers $\R$.
The field of real numbers is, a fortiori, also a ring.
Hence the result, by definition of subring.
== Sources ==
* : $\S 56$. Subrings and Subfields
* 
Category:Integers
Category:Real Numbers
Category:Subrings
Integers form Totally Ordered Ring

== Theorem ==
The structure $\struct {\Z, +, \times, \le}$ is a totally ordered ring.
== Proof ==
From Integers form Commutative Ring, $\struct {\Z, +, \times}$ is a commutative ring.
We need to show that $\le$ is a compatible ordering on $\Z$.
That is, that:
:$(1): \quad \le$ is compatible with $+$
:$(2): \quad \forall x, y \in \Z: 0 \le x, 0 \le y \implies 0 \le x \times y$
The first one follows from the fact that Integers under Addition form Totally Ordered Group.
Now, from Multiplicative Ordering on Integers, we have:
:$\forall z, x, y \in \Z, 0 < y: z \le x \iff y \times z \le y \times x$
Let $z = 0, 0 \le x$.
Then:
:$0 = x \times z \le x \times y = y \times x$
So if $y \ne 0$, the condition $0 \le x, 0 \le y \implies 0 \le x \times y$ holds.
Now if $y = 0$, we have $x \times y = 0$ and thus $0 \le x \times y$.
Thus $\le$ is compatible with $\Z$, and the result follows.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
Category:Integers
Category:Totally Ordered Rings
Integers form Unique Factorization Domain

== Theorem ==
The integers $\struct {\Z, +, \times}$ form a unique factorization domain.
== Proof ==
Follows directly from:
:$(1) \quad$ the fundamental theorem of arithmetic
:$(2) \quad$ the fact that $\struct {\Z, +, \times}$ is an integral domain
:$(3) \quad$ the definitions complete factorization and equivalent factorizations.
Category:Integral Domains
Category:Integers
Category:Factorization
Category:Unique Factorization Domains
Integers not Expressible as Sum of Distinct Primes of form 6n-1

== Theorem ==
$161$ is the largest integer that cannot be expressed as the sum of distinct primes of the form $6 n - 1$.
The following integers cannot be expressed as the sum of distinct primes of the form $6 n - 1$:
:$1, 2, 3, 4, 6, 7, 8, 9, 10, 12, 13, 14, 15, 18, 19, 20, 21, 24, 25, 26, 27, 30, 31, 32, 35, 36, 37, 38, 42, 43, 44, 48, 49, 50,$
:$54, 55, 60, 61, 65, 66, 67, 72, 73, 77, 78, 79, 84, 90, 91, 95, 96, 102, 108, 114, 119, 120, 125, 143, 155, 161$
== Proof ==
== Sources ==
* : $161$
* : $161$
Category:Prime Numbers
Category:161
Category:Integers not Expressible as Sum of Distinct Primes of form 6n-1
Integers not Sum of Distinct Triangular Numbers

== Theorem ==
The sequence of integers which cannot be expressed as the sum of distinct triangular numbers is:
:$2, 5, 8, 12, 23, 33$
== Proof ==
It will be proved that the largest integer which cannot be expressed as the sum of distinct triangular numbers is $33$.
The remaining integers in the sequence can be identified by inspection.
We prove this using a variant of Second Principle of Mathematical Induction.
Let $\map P n$ be the proposition:
:$n$ can be expressed as the sum of distinct triangular numbers.
=== Basis for the induction ===
We verify the result up to $T_{13} = 91$.
Then we conclude all $n$ with $34 \le n \le 91$ can be expressed as the sum of distinct triangular numbers.
So $\map P n$ is true for all $34 \le n \le 91$.
This is the basis for the induction.
=== Induction Hypothesis ===
Suppose for some $k > 91$, $\map P j$ is true for all $34 \le j < k$.
That is, every integer between $34$ and $k - 1$ can be expressed as the sum of distinct triangular numbers.
This is the induction hypothesis.
=== Induction Step ===
This is the induction step:
We find the largest integer $i$ such that $T_i < k \le T_{i + 1}$.
We show that $k - T_{i - 3}$ can be expressed as the sum of distinct triangular numbers, and the sum does not involve $T_{i - 3}$.
Then $k$ can be expressed as the sum of distinct triangular numbers.
Since $k > 91 = T_{13}$, we must have $i \ge 13$.
Hence $k > k - T_{i - 3} > T_i - T_{i - 3} = i + \paren {i - 1} + \paren {i - 2} \ge 13 + 12 + 11 = 36$.
Thus $k - T_{i - 3}$ can be expressed as the sum of distinct triangular numbers by Induction Hypothesis.
A sufficient condition such that the sum does not involve $T_{i - 3}$ is $T_{i - 3} > k - T_{i - 3}$.
We have:
Therefore we have $T_{i - 3} > k - T_{i - 3}$.
Hence $\map P k$ is true.
By the Second Principle of Mathematical Induction, $\map P n$ is true for all $n \ge 34$.
Thus every integer greater than $33$ can be expressed as the sum of distinct triangular numbers.
== Sources ==
* : $33$
* : $33$
Category:Triangular Numbers
Category:33
Integers such that Difference with Power of 2 is always Prime

== Unproven Hypothesis ==
The following positive integers are believed to be the only values of $n$ such that:
:$\forall k > 0: n - 2^k$
is prime whenever $n - 2^k$ is (strictly) positive:
:$4, 7, 15, 21, 45, 75, 105$
== Examples ==
== Historical Note ==
== Sources ==
* : $105$
* 
* : $105$
Category:Unproven Hypotheses
Category:Integers such that Difference with Power of 2 is always Prime
Integers such that Difference with Power of 2 is always Prime/Examples

== Examples of Integers such that Difference with Power of 2 is always Prime ==
The following positive integers are believed to be the only values of $n$ such that:
:$\forall k > 0: n - 2^k$
is prime whenever it is (strictly) positive:
:$4, 7, 15, 21, 45, 75, 105$
=== 4 ===
=== 7 ===
=== 15 ===
=== 21 ===
=== 45 ===
=== 75 ===
=== 105 ===
Category:Integers such that Difference with Power of 2 is always Prime
Integers such that Difference with Power of 2 is always Prime/Examples/105

== Example of Integers such that Difference with Power of 2 is always Prime ==
The positive integer $105$ has the property that such that:
:$\forall k > 0: 105 - 2^k$
is prime whenever it is (strictly) positive.
== Proof ==
Category:Integers such that Difference with Power of 2 is always Prime
Category:105
Integers such that Difference with Power of 2 is always Prime/Examples/15

== Example of Integers such that Difference with Power of 2 is always Prime ==
The positive integer $15$ has the property that such that:
:$\forall k > 0: 15 - 2^k$
is prime whenever it is (strictly) positive.
== Proof ==
Category:Integers such that Difference with Power of 2 is always Prime
Category:15
Integers such that Difference with Power of 2 is always Prime/Examples/21

== Example of Integers such that Difference with Power of 2 is always Prime ==
The positive integer $21$ has the property that such that:
:$\forall k > 0: 21 - 2^k$
is prime whenever it is (strictly) positive.
== Proof ==
Category:Integers such that Difference with Power of 2 is always Prime
Category:21
Integers such that Difference with Power of 2 is always Prime/Examples/4

== Example of Integers such that Difference with Power of 2 is always Prime ==
The positive integer $4$ has the property that such that:
:$\forall k > 0: 4 - 2^k$
is prime whenever it is (strictly) positive.
== Proof ==
Category:Integers such that Difference with Power of 2 is always Prime
Category:4
Integers such that Difference with Power of 2 is always Prime/Examples/45

== Example of Integers such that Difference with Power of 2 is always Prime ==
The positive integer $45$ has the property that such that:
:$\forall k > 0: 45 - 2^k$
is prime whenever it is (strictly) positive.
== Proof ==
Category:Integers such that Difference with Power of 2 is always Prime
Category:45
Integers such that Difference with Power of 2 is always Prime/Examples/7

== Example of Integers such that Difference with Power of 2 is always Prime ==
The positive integer $7$ has the property that such that:
:$\forall k > 0: 7 - 2^k$
is prime whenever it is (strictly) positive.
== Proof ==
Category:Integers such that Difference with Power of 2 is always Prime
Category:7
Integers such that Difference with Power of 2 is always Prime/Examples/75

== Example of Integers such that Difference with Power of 2 is always Prime ==
The positive integer $75$ has the property that such that:
:$\forall k > 0: 75 - 2^k$
is prime whenever it is (strictly) positive.
== Proof ==
Category:Integers such that Difference with Power of 2 is always Prime
Category:75
Integers such that Difference with Power of 2 is always Prime/Mistake

== Source Work ==
:
:The Dictionary
::$105$
:
:The Dictionary
::$105$
== Mistake ==
: conjectured that [105] is the largest number $n$ such that the positive values of $n - 2^k$ are all prime. The only other known numbers with this property are $7$, $15$, $21$, $45$ and $75$.
== Correction ==
First it should be noted that it should be specified that $k \ge 1$. Otherwise the statement is false: for example $7 - 2^0 = 6$ is composite.
Secondly, the list should also include $4$.
== Sources ==
* : $105$
* : $105$
Category:Mistakes/Curious and Interesting Numbers