nyuuzyou/wikis huggingface dataset:
Divisor Sum of Integer/Examples/207

== Example of Divisor Sum of Integer ==
:$\map {\sigma_1} {207} = 312$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
From Divisor Sum of Integer
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$207 = 3^2 \times 23$
Hence:
Category:Divisor Sum of Integer
Category:207
Divisor Sum of Integer/Examples/24

== Example of Divisor Sum of Integer ==
:$\map {\sigma_1} {24} = 60$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
From Divisor Sum of Integer
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$24 = 2^3 \times 3$
Hence:
Category:Divisor Sum of Integer
Category:24
Divisor Sum of Integer/Examples/40

== Example of Divisor Sum of Integer ==
:$\map {\sigma_1} {40} = 90$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$40 = 2^3 \times 5$
Hence:
Category:Divisor Sum of Integer
Category:40
Divisor Sum of Integer/Examples/44

== Example of Divisor Sum of Integer ==
:$\map {\sigma_1} {44} = 84$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
From Divisor Sum of Integer
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$44 = 2^2 \times 11$
Hence:
Category:Divisor Sum of Integer
Category:44
Divisor Sum of Integer/Examples/48

== Example of Divisor Sum of Integer ==
:$\map {\sigma_1} {48} = 124$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$48 = 2^4 \times 3$
Hence:
Category:Divisor Sum of Integer
Category:48
Divisor Sum of Non-Square Semiprime

== Theorem ==
Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.
== Proof 1 ==
== Proof 2 ==
== Examples ==
Category:Divisor Sum Function
Category:Semiprimes
Category:Divisor Sum of Non-Square Semiprime
Category:Divisor Sum of Integer
Divisor Sum of Non-Square Semiprime/Examples

== Examples of Divisor Sum of Non-Square Semiprime ==
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ denote the divisor sum function: the sum of all the positive integer divisors of $n$.
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Category:Divisor Sum of Non-Square Semiprime
Divisor Sum of Non-Square Semiprime/Examples/115

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {115} = 144$
where $\sigma_1$ denotes the divisor sum function.
== Proof 1 ==
== Proof 2 ==
Category:Divisor Sum of Non-Square Semiprime
Category:Divisor Sum of Non-Square Semiprime/Examples/115
Category:115
Divisor Sum of Non-Square Semiprime/Examples/115/Proof 1

== Example of Divisor Sum of Integer ==
== Proof ==
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$115 = 5 \times 23$
Hence:
Category:Divisor Sum of Non-Square Semiprime/Examples/115
Category:115
Divisor Sum of Non-Square Semiprime/Examples/115/Proof 2

== Example of Divisor Sum of Non-Square Semiprime ==
== Proof ==
We have that:
:$115 = 5 \times 23$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
Category:Divisor Sum of Non-Square Semiprime/Examples/115
Category:115
Divisor Sum of Non-Square Semiprime/Examples/119

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {119} = 144$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
We have that:
:$119 = 7 \times 17$
Hence:
Category:Divisor Sum of Non-Square Semiprime
Category:119
Divisor Sum of Non-Square Semiprime/Examples/14

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {14} = 24$
where $\sigma_1$ denotes the divisor sum.
== Proof 1 ==
== Proof 2 ==
Category:Divisor Sum of Non-Square Semiprime
Category:Divisor Sum of Non-Square Semiprime/Examples/14
Category:14
Divisor Sum of Non-Square Semiprime/Examples/14/Proof 1

== Example of Divisor Sum of Integer ==
== Proof ==
From Divisor Sum of Integer
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$14 = 2 \times 7$
Hence:
Category:Divisor Sum of Non-Square Semiprime/Examples/14
Divisor Sum of Non-Square Semiprime/Examples/15

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {15} = 24$
where $\sigma_1$ denotes the divisor sum function.
== Proof 1 ==
== Proof 2 ==
Category:Divisor Sum of Non-Square Semiprime
Category:Divisor Sum of Non-Square Semiprime/Examples/15
Category:15
Divisor Sum of Non-Square Semiprime/Examples/15/Proof 1

== Example of Divisor Sum of Integer ==
== Proof ==
From Divisor Sum of Integer
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$15 = 3 \times 5$
Hence:
Category:Divisor Sum of Non-Square Semiprime/Examples/15
Divisor Sum of Non-Square Semiprime/Examples/206

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {206} = 312$
where $\sigma_1$ denotes the divisor sum function.
== Proof 1 ==
== Proof 2 ==
Category:Divisor Sum of Non-Square Semiprime
Category:Divisor Sum of Non-Square Semiprime/Examples/206
Category:206
Divisor Sum of Non-Square Semiprime/Examples/206/Proof 1

== Example of Divisor Sum of Integer ==
== Proof ==
From Divisor Sum of Integer
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$206 = 2 \times 103$
Hence:
Category:Divisor Sum of Non-Square Semiprime/Examples/206
Category:206
Divisor Sum of Non-Square Semiprime/Examples/206/Proof 2

== Example of Divisor Sum of Non-Square Semiprime ==
== Proof ==
We have that:
:$206 = 2 \times 103$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
Category:Divisor Sum of Non-Square Semiprime/Examples/206
Category:206
Divisor Sum of Non-Square Semiprime/Examples/22

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {22} = 36$
where $\sigma_1$ denotes the divisor sum function.
== Proof 1 ==
== Proof 2 ==
Category:Divisor Sum of Non-Square Semiprime
Category:Divisor Sum of Non-Square Semiprime/Examples/22
Category:22
Divisor Sum of Non-Square Semiprime/Examples/22/Proof 1

== Example of Divisor Sum of Integer ==
== Proof ==
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$22 = 2 \times 11$
Hence:
Category:Divisor Sum of Non-Square Semiprime/Examples/22
Divisor Sum of Non-Square Semiprime/Examples/26

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {26} = 42$
where ${\sigma_1}$ denotes the divisor sum function.
== Proof 1 ==
== Proof 2 ==
Category:Divisor Sum of Non-Square Semiprime
Category:Divisor Sum of Non-Square Semiprime/Examples/26
Category:26
Divisor Sum of Non-Square Semiprime/Examples/26/Proof 1

== Example of Divisor Sum of Integer ==
== Proof ==
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$26 = 2 \times 13$
Hence:
Category:Divisor Sum of Non-Square Semiprime/Examples/26
Divisor Sum of Non-Square Semiprime/Examples/33

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {33} = 48$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
We have that:
:$33 = 3 \times 11$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
Category:Divisor Sum of Non-Square Semiprime
Category:33
Divisor Sum of Non-Square Semiprime/Examples/35

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {35} = 48$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
We have that:
:$35 = 5 \times 7$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
Category:Divisor Sum of Non-Square Semiprime
Category:35
Divisor Sum of Non-Square Semiprime/Examples/38

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {38} = 60$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
We have that:
:$38 = 2 \times 19$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
Category:Divisor Sum of Non-Square Semiprime
Category:38
Divisor Sum of Non-Square Semiprime/Examples/58

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {58} = 90$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
We have that:
:$58 = 2 \times 29$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
Category:Divisor Sum of Non-Square Semiprime
Category:58
Divisor Sum of Non-Square Semiprime/Examples/94

== Example of Divisor Sum of Non-Square Semiprime ==
:$\map {\sigma_1} {94} = 144$
where $\sigma_1$ denotes the divisor sum function.
== Proof 1 ==
== Proof 2 ==
Category:Divisor Sum of Non-Square Semiprime
Category:Divisor Sum of Non-Square Semiprime/Examples/94
Category:94
Divisor Sum of Non-Square Semiprime/Examples/94/Proof 1

== Example of Divisor Sum of Integer ==
== Proof ==
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$94 = 2 \times 47$
Hence:
Category:Divisor Sum of Non-Square Semiprime/Examples/94
Divisor Sum of Non-Square Semiprime/Proof 1

== Theorem ==
== Proof ==
As $p$ and $q$ are distinct prime numbers, it follows that $p$ and $q$ are coprime.
Thus by Divisor Sum Function is Multiplicative:
:$\map {\sigma_1} n = \map {\sigma_1} p \map {\sigma_1} q$
From Divisor Sum of Prime Number:
:$\map {\sigma_1} p = \paren {p + 1}$
:$\map {\sigma_1} q = \paren {q + 1}$
Hence the result.
== Examples ==
Category:Divisor Sum of Non-Square Semiprime
Divisor Sum of Non-Square Semiprime/Proof 2

== Theorem ==
== Proof ==
A semiprime with distinct prime factors is a square-free integer.
By Divisor Sum of Square-Free Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$
Hence the result.
== Examples ==
Category:Divisor Sum of Non-Square Semiprime
Divisor Sum of Power of Prime

== Theorem ==
Let $n = p^k$ be the power of a prime number $p$.
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
:$\map {\sigma_1} n = \dfrac {p^{k + 1} - 1} {p - 1}$
== Proof ==
From Divisors of Power of Prime, the divisors of $n = p^k$ are $1, p, p^2, \ldots, p^{k - 1}, p^k$.
Hence from Sum of Geometric Sequence:
:$\map {\sigma_1} {p^k} = 1 + p + p^2 + \cdots + p^{k - 1} + p^k = \dfrac {p^{k + 1} - 1} {p - 1}$
== Examples ==
== Sources ==
* 
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
Category:Prime Numbers
Category:Divisor Sum Function
Category:Divisor Sum of Power of Prime
Divisor Sum of Power of Prime/Examples/81

== Example of Divisor Sum of Power of Prime ==
:$\map {\sigma_1} {81} = 121$
where $\sigma_1$ denotes the divisor sum function.
== Proof ==
From Divisor Sum of Power of Prime:
:$\map {\sigma_1} {p^k} = \dfrac {p^{k + 1} - 1} {p_i - 1}$
We have that:
:$81 = 3^4$
Hence:
Category:Divisor Sum of Power of Prime
Category:81
Divisor Sum of Prime Number

== Theorem ==
Let $n$ be a positive integer.
Let $\map {\sigma_1} n$ be the divisor sum function of $n$.
Then $\map {\sigma_1} n = n + 1$ $n$ is prime.
== Proof ==
From Rule of Transposition, we may replace the only if statement by its contrapositive.
Therefore, the following suffices:
=== Sufficient Condition ===
Suppose $n$ is a prime.
By definition, the only positive divisors of $n$ are $1$ and $n$ itself.
Therefore $\map {\sigma_1} n$, defined as the sum of the divisors of $n$, equals $n + 1$.
=== Necessary Condition ===
Suppose $n$ is not a prime.
From One Divides all Integers and Integer Divides Itself, both $1$ and $n$ are divisors of $n$.
As $n$ is composite:
:$\exists r, s \in \N: r, s > 1: r s = n$
Trivially, both $r$ and $s$ are divisors of $n$.
Hence:
:$\map {\sigma_1} n \ge n + 1 + r + s > n + 1$
== Also see ==
* Divisor Sum of Power of Prime
== Sources ==
* : $1$
* : $1$
Category:Prime Numbers
Category:Divisor Sum Function
Category:Divisor Sum of Prime Number
Divisor Sum of Square-Free Integer

== Theorem ==
Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i = p_1 p_2 \cdots p_r$
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$
== Proof 1 ==
== Proof 2 ==
== Examples ==
Category:Divisor Sum Function
Category:Divisor Sum of Integer
Category:Divisor Sum of Square-Free Integer
Divisor Sum of Square-Free Integer/Examples/170

== Example of Divisor Sum of Square-Free Integer ==
:$\map {\sigma_1} {170} = 324$
where $\sigma_1$ denotes the divisor sum.
== Proof ==
We have that:
:$170 = 2 \times 5 \times 17$
Hence:
Category:Divisor Sum of Square-Free Integer
Category:170
Divisor Sum of Square-Free Integer/Examples/66

== Example of Divisor Sum of Square-Free Integer ==
:$\map {\sigma_1} {66} = 144$
where $\sigma_1$ denotes the divisor sum.
== Proof 1 ==
== Proof 2 ==
Category:Divisor Sum of Square-Free Integer
Category:66
Category:Divisor Sum of Square-Free Integer/Examples/66
Divisor Sum of Square-Free Integer/Examples/66/Proof 1

== Example of Divisor Sum of Square-Free Integer ==
== Proof ==
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$66 = 2 \times 3 \times 11$
Hence:
Category:Divisor Sum of Square-Free Integer/Examples/66
Divisor Sum of Square-Free Integer/Examples/70

== Example of Divisor Sum of Square-Free Integer ==
:$\map {\sigma_1} {70} = 144$
where $\sigma_1$ denotes the divisor sum.
== Proof 1 ==
== Proof 2 ==
Category:Divisor Sum of Square-Free Integer
Category:70
Category:Divisor Sum of Square-Free Integer/Examples/70
Divisor Sum of Square-Free Integer/Examples/70/Proof 1

== Example of Divisor Sum of Integer ==
== Proof ==
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$70 = 2 \times 5 \times 7$
Hence:
Category:Divisor Sum of Square-Free Integer/Examples/70
Divisor Sum of Square-Free Integer/Proof 1

== Theorem ==
== Proof ==
We have that the Divisor Sum Function is Multiplicative.
From the definition of prime number, each of the prime factors of $n$ is coprime to all other divisors of $n$.
From Divisor Sum of Prime Number, we have:
:$\map {\sigma_1} {p_i} = p_i + 1$
Hence the result.
== Examples ==
Category:Divisor Sum of Square-Free Integer
Divisor Sum of Square-Free Integer/Proof 2

== Theorem ==
== Proof ==
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where each of the $k_i$s are equal to $1$;
Hence:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^2 - 1} {p_i - 1}$
But from Difference of Two Squares:
:$\dfrac {p_i^2 - 1} {p_i - 1} = \dfrac {\paren {p_i + 1} \paren {p_i i 1} } {p_i - 1} = p_i + 1$
Hence the result.
== Examples ==
Category:Divisor Sum of Square-Free Integer
Divisor is Reciprocal of Divisor of Integer

== Theorem ==
Let $a, b, c \in \Z_{>0}$.
Then:
:$b = \dfrac 1 c \times a \implies c \divides a$
where $\divides$ denotes divisibilty.
== Proof ==
Let $a$ have an aliquot part $b$.
Let $c$ be an integer called by the same name as the aliquot part $b$.
Then:
:$1 = \dfrac 1 c \times c$
and so by :
:$ 1 : c = b : a$
Hence the result.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Divisors
Divisor of Deficient Number is Deficient

== Theorem ==
Let $n$ be a perfect number.
Let $n = k d$ where $r$ is a positive integer.
Then $k$ is deficient.
== Proof ==
We have by definition of divisor sum function and perfect number that:
:$\dfrac {\map {\sigma_1} {k d} } {k d} < 2$
But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors:
:$\dfrac {\map {\sigma_1} {k d} } {k d} > \dfrac {\map {\sigma_1} k} k$
That is:
:$\dfrac {\map {\sigma_1} k} k < 2$
Hence the result by definition of deficient.
== Sources ==
* : $12$
* : $12$
Category:Deficient Numbers
Divisor of Fermat Number/Euler's Result

== Theorem ==
Let $F_n$ be a Fermat number.
Let $m$ be divisor of $F_n$.
Then $m$ is in the form:
:$k \, 2^{n + 1} + 1$
where $k \in \Z_{>0}$ is an integer.
== Proof ==
It is sufficient to prove the result for prime divisors.
The general argument for all divisors follows from the argument:
:$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$
So the product of two factors of the form preserves that form.
Let $n \in \N$.
Let $p$ be a prime divisor of $F_n = 2^{2^n} + 1$.
Then we have:
:$2^{2^n} \equiv -1 \pmod p$
Squaring both sides:
:$2^{2^{n + 1}} \equiv 1 \pmod p$
From Integer to Power of Multiple of Order, the order of $2$ modulo $p$ divides $2^{n + 1}$ but not $2^n$.
Therefore it must be $2^{n + 1}$.
Hence:
== Historical Note ==
== Sources ==
* : $257$
* : $641$
* : $257$
* : $641$
Category:Divisor of Fermat Number
Divisor of Fermat Number/Historical Note

== Historical Note on Divisor of Fermat Number ==
In $1747$, proved that a divisor of a Fermat number $F_n$ is always in the form $k \, 2^{n + 1} + 1$.
This was later refined to $k \, 2^{n + 2} + 1$ by .
== Sources ==
* : $257$
* : $257$
Category:Historical Notes
Divisor of Fermat Number/Refinement by Lucas

== Theorem ==
Let $F_n$ be a Fermat number.
Let $m$ be divisor of $F_n$.
Let $n \ge 2$.
Then $m$ is in the form:
:$k \, 2^{n + 2} + 1$
== Proof ==
It is sufficient to prove the result for prime divisors.
The general argument for all divisors follows from the argument:
:$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$
So the product of two factors of the form preserves that form.
Let $p$ be a prime divisor of $F_n = 2^{2^n} + 1$.
From Euler's Result:
:$\exists q \in \Z: p = q \, 2^{n + 1} + 1$
Since $n \ge 2$, $q \, 2^{n + 1}$ is divisible by $2^{2 + 1} = 8$.
Hence:
:$p \equiv 1 \pmod 8$
By Second Supplement to Law of Quadratic Reciprocity:
:$\paren {\dfrac 2 p} = 1$
so $2$ is a quadratic residue modulo $p$.
Hence:
:$\exists x \in \Z: x^2 = 2 \pmod p$
We have shown $2^{2^n} \equiv -1 \pmod p$ and $2^{2^{n + 1} } \equiv 1 \pmod p$.
By Congruence of Powers:
:$x^{2^{n + 1} } \equiv 2^{2^n} \equiv -1 \pmod p$
:$x^{2^{n + 2} } \equiv 2^{2^{n + 1}} \equiv 1 \pmod p$
From Integer to Power of Multiple of Order, the order of $x$ modulo $p$ divides $2^{n + 2}$ but not $2^{n + 1}$.
Therefore it must be $2^{n + 2}$.
Hence:
== Historical Note ==
Category:Divisor of Fermat Number
Divisor of Integer/Examples

== Examples of Divisors of Integers ==
=== $2$ divides $4$ ===
=== $3$ divides $12$ ===
=== $4$ divides $\paren {-12}$ ===
=== $2$ does not divide $5$ ===
=== $3$ does not divide $4$ ===
=== $3$ does not divide $10$ ===
=== Divisors of $6$ ===
=== $2$ divides $n \paren {n + 1}$ ===
=== $3$ divides $n \paren {n + 1} \paren {n + 2}$ ===
=== $6$ divides $n \paren {n + 1} \paren {n + 2}$ ===
=== $6$ divides $7^n - 1$ ===
=== $63$ divides $8^{2 n} - 1$ ===
=== $80$ divides $9^{2 n} - 1$ ===
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/2 divides 4

== Example of Divisor of Integer ==
:$2 \divides 4$
where $\divides$ indicates divisibility.
== Proof ==
:$4 = 2 \times 2$
== Sources ==
* : $\S 0.1$. Arithmetic
* : $\text {2-2}$ Divisibility: Example $\text {2-1}$
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/2 divides n(n+1)

== Theorem ==
Let $n$ be an integer.
Then:
:$2 \divides n \paren {n + 1}$
where $\divides$ indicates divisibility.
== Proof ==
Suppose $n$ is even.
Then $2 \divides n$ by definition.
Hence from Divisor Divides Multiple:
:$2 \divides n \paren {n + 1}$
Suppose $n$ is odd.
Then $n + 1$ is even
Then $2 \divides n + 1$ by definition.
Hence from Divisor Divides Multiple:
:$2 \divides n \paren {n + 1}$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $5 \ \text {(a)}$
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/2 does not divide 5

== Example of Non-Divisor of Integer ==
:$2 \nmid 5$
where $\nmid$ denotes non-divisibility.
== Proof ==
:$5 = 2 \times 2 + 1$
As the remainder is not $0$, the result follows.
== Sources ==
* : $\S 0.1$. Arithmetic
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/3 divides 2^n + (-1)^(n+1)

== Theorem ==
Let $n$ be an integer such that $n \ge 1$.
Then:
:$3 \divides 2^n + \paren {-1}^{n + 1}$
where $\divides$ indicates divisibility .
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$3 \divides 2^n + \paren {-1}^{n + 1}$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$3 \divides 2^k + \paren {-1}^{k + 1}$
from which it is to be shown that:
:$3 \divides 2^{k + 1} + \paren {-1}^{k + 2}$
=== Induction Step ===
This is the induction step:
Suppose $k$ is integer.
Then $k + 1$ is odd, and:
Similarly suppose $k$ is odd.
Then $k + 1$ is integer, and:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 1}: 3 \divides 2^n + \paren {-1}^{n + 1}$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $6 \ \text {(b)}$
Category:Examples of Divisors of Integers