nyuuzyou/wikis huggingface dataset:
Euler Phi Function/Examples

== Examples of Euler $\phi$ Function==
The values of the Euler $\phi$ function for the first few integers are as follows:
:
$n$ 	$\map \phi n$	$m$ not coprime: $1 \le m \le n$
$1$ 	$1$ 	$\O$
$2$ 	$1$ 	$2$
$3$ 	$2$ 	$3$
$4$ 	$2$ 	$2, 4$
$5$ 	$4$ 	$5$
$6$ 	$2$ 	$2, 3, 4, 6$
$7$ 	$6$ 	$7$
$8$ 	$4$ 	$2, 4, 6, 8$
$9$ 	$6$ 	$3, 6, 9$
$10$	$4$ 	$2, 4, 5, 6, 8, 10$
=== Euler Phi Function of $1$ ===
=== Euler Phi Function of $2$ ===
=== Euler Phi Function of $3$ ===
=== Euler Phi Function of $4$ ===
=== Euler Phi Function of $9$ ===
=== Numbers for which Euler Phi Function is $6$ ===
=== Euler Phi Function of $14$ ===
=== Euler Phi Function of $16$ ===
=== Euler Phi Function of $20$ ===
=== Euler Phi Function of $24$ ===
=== Euler Phi Function of $30$ ===
=== Euler Phi Function of $42$ ===
=== Euler Phi Function of $72$ ===
=== Euler Phi Function of $78$ ===
=== Euler Phi Function of $84$ ===
=== Euler Phi Function of $87$ ===
=== Euler Phi Function of $90$ ===
=== Euler Phi Function of $216$ ===
=== Euler Phi Function of $222$ ===
=== Euler Phi Function of $228$ ===
=== Euler Phi Function of $234$ ===
=== Euler Phi Function of $248$ ===
=== Euler Phi Function of $1\,000\,000$ ===
=== Successive Solutions of $\map \phi n = \map \phi {n + 2}$ ===
== Table of Values of Euler $\phi$ Function ==
== Sources ==
* : $\S 2.3$: Congruences
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $27$
Category:Examples of Euler Phi Function
Euler Phi Function/Examples/1,000,000

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {1 \, 000 \, 000} = 400 \, 000$
where $\phi$ denotes the Euler $\phi$ function.
== Proof 1 ==
== Proof 2 ==
Category:Examples of Euler Phi Function
Category:Euler Phi Function of 1,000,000
Category:1,000,000
Euler Phi Function/Examples/1,000,000/Proof 1

== Example of Use of Euler $\phi$ Function ==
== Proof ==
We have that:
From Euler Phi Function of Prime Power:
:$\map \phi {p^n} = p^n \paren {1 - \dfrac 1 p}$
Thus:
and:
From Euler Phi Function is Multiplicative:
Thus:
:$\map \phi {1 \, 000 \, 000} = 400 \, 000$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $30$
Category:Euler Phi Function of 1,000,000
Euler Phi Function/Examples/1,000,000/Proof 2

== Example of Use of Euler $\phi$ Function ==
== Proof ==
We have that:
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p$ denotes a prime number.
So:
Thus:
:$\map \phi {1\,000\,000} = 400\,000$
Category:Euler Phi Function of 1,000,000
Euler Phi Function/Examples/16

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {16} = 8$
where $\phi$ denotes the Euler $\phi$ function.
== Proof ==
From the corollary to Euler Phi Function of Prime Power:
:$\map \phi {2^k} = 2^{k - 1}$
Thus:
:$\map \phi {16} = \map \phi {2^4} = 2^3 = 8$
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Ponderable $2.1.1$
Category:Examples of Euler Phi Function
Category:16
Euler Phi Function/Examples/20

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {20} = 8$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$20 = 2^2 \times 5$
Thus:
== Sources ==
* 
* 
Category:Examples of Euler Phi Function
Category:20
Euler Phi Function/Examples/24

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {24} = 8$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$24 = 2^3 \times 3$
Thus:
Category:Examples of Euler Phi Function
Category:24
Euler Phi Function/Examples/9

== Example of Use of Euler $\phi$ Function ==
:$\map \phi 9 = 6$
where $\phi$ denotes the Euler $\phi$ function.
== Proof ==
From Euler Phi Function of Prime Power:
:$\map \phi {3^k} = 2 \times 3^{k - 1}$
Thus:
:$\map \phi 9 = \map \phi {3^2} = 2 \times 3 = 6$
They can be enumerated as:
:$1, 2, 4, 5, 7, 8$
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$
Category:Examples of Euler Phi Function
Category:9
Euler Phi Function/Examples/Phi is 6

== Example of Use of Euler $\phi$ Function ==
There are $4$ numbers for which the value of the Euler $\phi$ function is $6$:
:$7, 9, 14, 18$
== Proof ==
From Euler Phi Function of $9$, we already have that $\map \phi 9 = 6$.
=== $7$ ===
From Euler Phi Function of Prime:
:$\map \phi 7 = 7 - 1 = 6$
=== $14$ ===
From Euler Phi Function of Integer:
=== $18$ ===
From Euler Phi Function of Integer:
Hence the result.
== Sources ==
* : $18$
Category:Euler Phi Function
Category:7
Category:9
Category:14
Category:18
Euler Phi Function/Table

== Table of Euler $\phi$ Function ==
The Euler $\phi$ function for the first $100$ positive integers is as follows:
$\quad \begin{array} {|r|r|}
\hline
n & \map \phi n \\
\hline
1 & 1 \\
2 & 1 \\
3 & 2 \\
4 & 2 \\
5 & 4 \\
6 & 2 \\
7 & 6 \\
8 & 4 \\
9 & 6 \\
10 & 4 \\
11 & 10 \\
12 & 4 \\
13 & 12 \\
14 & 6 \\
15 & 8 \\
16 & 8 \\
17 & 16 \\
18 & 6 \\
19 & 18 \\
20 & 8 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map \phi n \\
\hline
21 & 12 \\
22 & 10 \\
23 & 22 \\
24 & 8 \\
25 & 20 \\
26 & 12 \\
27 & 18 \\
28 & 12 \\
29 & 28 \\
30 & 8 \\
31 & 30 \\
32 & 16 \\
33 & 20 \\
34 & 16 \\
35 & 24 \\
36 & 12 \\
37 & 36 \\
38 & 18 \\
39 & 24 \\
40 & 16 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map \phi n \\
\hline
41 & 40 \\
42 & 12 \\
43 & 42 \\
44 & 20 \\
45 & 24 \\
46 & 22 \\
47 & 46 \\
48 & 16 \\
49 & 42 \\
50 & 20 \\
51 & 32 \\
52 & 24 \\
53 & 52 \\
54 & 18 \\
55 & 40 \\
56 & 24 \\
57 & 36 \\
58 & 28 \\
59 & 58 \\
60 & 16 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map \phi n \\
\hline
61 & 60 \\
62 & 30 \\
63 & 36 \\
64 & 32 \\
65 & 48 \\
66 & 20 \\
67 & 66 \\
68 & 32 \\
69 & 44 \\
70 & 24 \\
71 & 70 \\
72 & 24 \\
73 & 72 \\
74 & 36 \\
75 & 40 \\
76 & 36 \\
77 & 60 \\
78 & 24 \\
79 & 78 \\
80 & 32 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map \phi n \\
\hline
81 & 54 \\
82 & 40 \\
83 & 82 \\
84 & 24 \\
85 & 64 \\
86 & 42 \\
87 & 56 \\
88 & 40 \\
89 & 88 \\
90 & 24 \\
91 & 72 \\
92 & 44 \\
93 & 60 \\
94 & 46 \\
95 & 72 \\
96 & 32 \\
97 & 96 \\
98 & 42 \\
99 & 60 \\
100 & 40 \\
\hline
\end{array}$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $11$
* : Tables: $8$ The Proper Factors, where Composite, and the Values of the Functions $\map \phi n$, $\map d n$ and $\map \sigma n$
* : Tables: $8$ The Proper Factors, where Composite, and the Values of the Functions $\map \phi n$, $\map d n$ and $\map \sigma n$
Category:Euler Phi Function
Euler Phi Function by Argument is Injective

== Theorem ==
Let $f: \N \to \N$ be the arithmetic function defined as:
:$\map f n = n \map \phi n$ 
where $\map \phi n$ is the Euler Phi function (which is not injective).
Then $f$ is injective.
== Proof ==
By Product of Multiplicative Functions is Multiplicative and Euler Phi Function is Multiplicative, $f$ is a multiplicative function.
Suppose that $\map f {n_1} = \map f {n_2}$.
The case $n_1 = 1$ is trivial.
Let $n_1 > 1$.
Let $p$ be the largest prime factor of $n_1$.
By Largest Prime Factor of Euler Phi Function, the largest prime factor of $\map \phi {n_1}$ is less than $p$.
Thus $p$ is the largest prime factor of $\map f {n_1} = n_1 \map \phi {n_1}$.
But:
:$\map f {n_1} = \map f {n_2}$
Then $p$ is the largest prime factor of $\map f {n_2}$.
Then $p$ is the largest prime factor of $n_2$.
Let $p^k$ be the largest power of $p$ dividing $n_1$.
Then $p^{k - 1}$ is the largest power of $p$ dividing $\map \phi {n_1}$.
Then $p^{2 k - 1}$ is the largest power of $p$ dividing $\map f {n_1}$.
But:
:$\map f {n_1} = \map f {n_2}$
Then $p^{2 k - 1}$ is the largest power of $p$ dividing $\map f {n_2}$.
Then $p^k$ is the largest power of $p$ dividing $n_2$.
Thus, by the multiplicativity of $f$, we also have:
:$\map f {\dfrac {n_1} {p^k} } = \map f {\dfrac {n_2} {p^k} }$
Continuing, after having eliminated all prime factors of $n_1$, we have:
:$\map f 1 = \map f {\dfrac {n_2} {n_1} }$
hence:
:$\dfrac {n_2} {n_1} = 1$
and so:
:$n_2 = n_1$
== Source ==
* 
Category:Euler Phi Function
Euler Phi Function in terms of Möbius Function

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Then:
:$\ds \sum_{d \mathop \divides n} \map \mu d \frac n d = \map \phi n$
where:
:$\ds \sum_{d \mathop \divides n}$ denotes the sum over all of the divisors of $n$
:$\map \phi n$ is the Euler $\phi$ function, the number of integers less than $n$ that are prime to $n$
:$\map \mu d$ is the Möbius function.
Equivalently, this says that:
:$\phi = \mu * I_{\Z_{>0} }$
where:
:$*$ denotes Dirichlet convolution
:$I_{\Z_{>0} }$ denotes the identity mapping on $\Z_{>0}$, that is:
::$\forall n \in \Z_{>0}: I_{\Z_{>0} }: n \mapsto n$
== Proof ==
Sum of Möbius Function over Divisors says:
Let $\map 1 k = 1$ be the constant mapping.
Then $\phi$ is defined as:
:$\ds \map \phi n = \sum_{\substack {k \mathop \perp n \\ 1 \mathop \le k \mathop \le n}} \map 1 k$
We have that $\floor {\dfrac 1 {\gcd \set {n, k} } }$ is $1$ if $k \perp n$ and $0$ otherwise.
Thus we can rewrite the above summation as:
:$\ds \sum_{k \mathop = 1}^n \floor {\frac 1 {\gcd \set {n, k} } }$
Now we may use Sum of Möbius Function over Divisors, with $\gcd \set {n, k}$ replacing $n$, to get:
For a fixed divisor $d$ of $n$, we must sum over all those $k$ in the range $1 \le k \le n$ which are multiples of $d$.
If we write $k = q d$, then $1 \le k \le n$ $1 \le q \le \dfrac n d$. 
Hence the last sum for $\map \phi n$ can be written as:
== Sources ==
* : Chapter $1$: Properties of the Natural Numbers: $\S 25 \beta$
* : $2.4$: A relation connected $\varphi$ and $\mu$
Category:Euler Phi Function
Category:Möbius Function
Euler Phi Function is Even for Argument greater than 2

== Theorem ==
Let $n \in \Z: n \ge 1$.
Let $\map \phi n$ be the Euler $\phi$ function of $n$.
Then $\map \phi n$ is even $n > 2$.
== Proof ==
We have from the definition of Euler $\phi$ function:
:$\map \phi 1 = 1$
and from Euler Phi Function of Prime Power: Corollary:
:$\map \phi 2 = 1$
Now let $n \ge 3$.
There are two possibilities:
=== Odd Prime Divisor ===
$n$ has (at least one) odd prime factor: $p$, say.
From the corollary to Euler Phi Function of Integer, it follows that:
:$p - 1$ divides $\map \phi n$
But as $p$ is odd, $p - 1$ is even and hence:
:$2 \divides \paren {p - 1} \divides \map \phi n$
and so $\map \phi n$ is even.
=== No Odd Prime Divisor ===
Now suppose $n$ has no odd prime factors.
Then its only prime factor must be $2$.
Thus:
:$n = 2^k$
where $k > 1$.
Then from Euler Phi Function of Prime Power: Corollary:
:$\map \phi n = 2^k \paren {1 - \frac 1 2} = 2^{k - 1}$
where $k-1 > 0$.
Hence $\map \phi n$ is even.
Category:Euler Phi Function
Euler Phi Function is Multiplicative

== Theorem ==
The Euler $\phi$ function is a multiplicative function:
:$m \perp n \implies \map \phi {m n} = \map \phi m \map \phi n$
where $m, n \in \Z_{>0}$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $30$
* 
* 
* 
* : Chapter $1$: Properties of the Natural Numbers: $\S 25$
* : $\S 2.3$: Congruences
Category:Euler Phi Function is Multiplicative
Category:Euler Phi Function
Category:Multiplicative Functions
Euler Phi Function is Multiplicative/Proof 1

== Theorem ==
== Proof ==
Let $R = \set {r_1, r_2, \ldots, r_{\map \phi m} }$ and $S = \set {s_1, s_2, \ldots, s_{\map \phi n} }$ be the reduced residue systems for the respective moduli $m$ and $n$.
We are to show that the set of $\map \phi m \map \phi n$ integers:
:$T = \set {n r + m s: r \in R, s \in S}$
is a reduced residue system for modulus $m n$.
We need to establish the following:
:Each integer in $T$ is prime to $m n$
:No two integers in $T$ is congruent modulo $m n$
:Each integer prime to $m n$ is congruent modulo $m n$ to one of these integers in $T$.
We prove each in turn:
 $p$ is a prime divisor of $\gcd \set {n r + m s, m n}$ where $r \in R, s \in S$.
We have that $p$ divides $m n$ but $m \perp n$.
Hence from Divisors of Product of Coprime Integers, $p$ either divides $m$ or $n$ but not both.
, suppose $p \divides m$.
Then as $p \divides n r + m s$, we have $p \divides n r$ and hence $p \divides r$.
But then $p \divides \gcd \set {m, r} = 1$ which is a contradiction.
Similarly if $p \divides n$.
So there is no such prime and hence $n r + m s \perp m n$.
Let $n r + m s = n r' + m s' \pmod {m n}$, where $r, r' \in R, s, s' \in S$.
Then:
:$n \paren {r - r'} + m \paren {s - s'} = k \paren {m n}$ for some $k \in \Z$.
As $m$ divides two of these terms it must divide the third, so:
:$m \divides n \paren {r - r'}$
Now $m \perp n$ so by Euclid's Lemma:
:$m \divides \paren {r - r'}$
or:
:$r \equiv r' \pmod m$
But $r$ and $r'$ are part of the same reduced residue system modulo $m$, so:
:$r = r'$
Similarly for $n$: we get:
:$s = s'$
Hence distinct elements of $T$ can not be congruent modulo $m n$.
Let $k \in \Z: k \perp m n$.
Since $m \perp n$, from Set of Integer Combinations equals Set of Multiples of GCD we can write $k = n r' + m s'$ for some $r', s' \in \Z$.
Suppose there exists some prime number $p$ such that $p \divides m$ and $p \divides r'$.
Such a prime would be a common divisor of both $k$ and $m n$, contradicting $k \perp m n$.
Hence $r' \perp m$ and so is congruent modulo $m $ to one of these integers in $R$.
By the same argument, $s' \perp n$ and so is congruent modulo $n$ to one of these integers in $S$.
Writing $r' = r + a m, \, s' = s + b n$ we have:
:$k = n r' + m s' = n r + m s + m n \paren {a + b} \equiv n r + m s \pmod {m n}$
Hence the result.
Category:Euler Phi Function is Multiplicative
Euler Phi Function is Multiplicative/Proof 2

== Theorem ==
== Proof ==
From Euler Phi Function of Product, we have:
:$\map \phi {m n} = \map \phi m \map \phi n \paren {\dfrac {\map \gcd {m, n} } {\map \phi {\map \gcd {m, n} } } }$
From hypothesis, we have:
:$\map \gcd {m, n} = 1$
From Euler Phi Function of 1, we have $\map \phi 1 = 1$, giving the result.
Category:Euler Phi Function is Multiplicative
Euler Phi Function is not Completely Multiplicative

== Theorem ==
The Euler $\phi$ function is not a completely multiplicative function.
That is, it is not always the case that:
:$\map \phi {m n} = \map \phi m \map \phi n$
where $m, n \in \Z_{>0}$.
== Proof ==
;Proof by Counterexample
Hence we see:
:$6 \times 10 = 60$
but:
:$\map \phi 6 \times \map \phi {10} \ne \map \phi {60}$
== Sources ==
* : Chapter $1$: Properties of the Natural Numbers: $\S 25$
* : $\S 2.3$: Congruences
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $30$
* 
Category:Completely Multiplicative Functions
Category:Euler Phi Function
Euler Phi Function of 1

== Theorem ==
:$\map \phi 1 = 1$
where $\phi$ denotes the Euler $\phi$ function.
== Proof ==
The only (strictly) positive integer less than or equal to $1$ is $1$ itself.
By Integer is Coprime to 1, $1$ is coprime to itself.
Hence, by definition, there is exactly $1$ integer less than or equal to $1$ which is coprime with $1$.
Hence the result.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$: $(1.29)$
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $27$
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Example $2.1.1$
Category:Examples of Euler Phi Function
Category:1
Euler Phi Function of 10

== Example of Euler $\phi$ Function of Square-Free Integer ==
:$\map \phi {10} = 4$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Square-Free Integer:
:$\ds \map \phi n = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$10 = 2 \times 5$
and so is square-free.
Thus:
== Sources ==
* 
* 
Category:Examples of Euler Phi Function
Category:10
Euler Phi Function of 104

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {104} = 48$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$104 = 2^3 \times 13$
Thus:
Category:Examples of Euler Phi Function
Category:104
Euler Phi Function of 105

== Example of Euler $\phi$ Function of Square-Free Integer ==
:$\map \phi {105} = 48$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Square-Free Integer:
:$\ds \map \phi n = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$105 = 3 \times 5 \times 7$
and so is square-free.
Thus:
Category:Euler Phi Function of Square-Free Integer
Category:105
Euler Phi Function of 106

== Example of Euler Phi Function of 2 times Odd Prime ==
:$\phi \left({106}\right) = 52$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$106 = 2 \times 53$
Thus:
Category:Euler Phi Function of 2 times Odd Prime
Category:106
Euler Phi Function of 111

== Example of Euler $\phi$ Function of Non-Square Semiprime ==
:$\phi \left({111}\right) = 72$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$111 = 3 \times 37$
Thus:
Category:Euler Phi Function of Non-Square Semiprime
Category:111
Euler Phi Function of 1154

== Example of Euler Phi Function of 2 times Odd Prime ==
:$\phi \left({1154}\right) = 576$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$1154 = 2 \times 577$
Thus:
Category:Euler Phi Function of 2 times Odd Prime
Category:1154
Euler Phi Function of 1155

== Example of Euler $\phi$ Function of Square-Free Integer ==
:$\map \phi {1155} = 480$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Square-Free Integer:
:$\ds \map \phi n = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$1155 = 3 \times 5 \times 7 \times 11$
and so is square-free.
Thus:
Category:Euler Phi Function of Square-Free Integer
Category:1155
Euler Phi Function of 117

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {117} = 72$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$117 = 3^2 \times 13$
Thus:
Category:Euler Phi Function
Category:117
Euler Phi Function of 1257

== Example of Euler $\phi$ Function of Non-Square Semiprime ==
:$\phi \left({1257}\right) = 836$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$1257 = 3 \times 419$
Thus:
Category:Euler Phi Function of Non-Square Semiprime
Category:1257
Euler Phi Function of 135

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {135} = 72$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$135 = 3^3 \times 5$
Thus:
Category:Euler Phi Function
Category:135
Euler Phi Function of 1364

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {1364} = 600$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$1364 = 2^2 \times 11 \times 31$
Thus:
Category:Euler Phi Function
Category:1364
Euler Phi Function of 1365

== Example of Euler $\phi$ Function of Square-Free Integer ==
:$\map \phi {1365} = 576$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Square-Free Integer:
:$\ds \map \phi n = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$1365 = 3 \times 5 \times 7 \times 13$
and so is square-free.
Thus:
Category:Euler Phi Function of Square-Free Integer
Category:1365
Euler Phi Function of 14

== Example of Euler Phi Function of 2 times Odd Prime ==
:$\map \phi {14} = 6$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$14 = 2 \times 7$
Thus:
Category:Euler Phi Function of 2 times Odd Prime
Category:14
Euler Phi Function of 140

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {140} = 48$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$140 = 2^2 \times 5 \times 7$
Thus:
Category:Euler Phi Function
Category:140
Euler Phi Function of 146

== Example of Euler Phi Function of 2 times Odd Prime ==
:$\map \phi {146} = 72$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$146 = 2 \times 73$
Thus:
Category:Euler Phi Function of 2 times Odd Prime
Category:146
Euler Phi Function of 148

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {148} = 72$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$148 = 2^2 \times 37$
Thus:
Category:Euler Phi Function
Category:148
Euler Phi Function of 1485

== Example of Euler $\phi$ Function of Square-Free Integer ==
:$\map \phi {1485} = 720$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$1485 = 3^3 \times 5 \times 11$
Thus:
Category:Euler Phi Function of Square-Free Integer
Category:1485
Euler Phi Function of 15

== Example of Euler $\phi$ Function of Non-Square Semiprime ==
:$\map \phi {15} = 8$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$15 = 3 \times 5$
Thus:
Category:Euler Phi Function of Non-Square Semiprime
Category:15
Euler Phi Function of 152

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {152} = 72$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$152 = 2^3 \times 19$
Thus:
Category:Euler Phi Function
Category:152
Euler Phi Function of 164

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {164} = 80$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$164 = 2^2 \times 41$
Thus:
Category:Euler Phi Function
Category:164
Euler Phi Function of 165

== Example of Euler $\phi$ Function of Square-Free Integer ==
:$\map \phi {165} = 80$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
From Euler Phi Function of Square-Free Integer:
:$\ds \map \phi n = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$165 = 3 \times 5 \times 11$
and so is square-free.
Thus:
Category:Euler Phi Function of Square-Free Integer
Category:165
Euler Phi Function of 166

== Example of Euler Phi Function of 2 times Odd Prime ==
:$\phi \left({166}\right) = 82$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$166 = 2 \times 83$
Thus:
Category:Euler Phi Function of 2 times Odd Prime
Category:166