nyuuzyou/wikis huggingface dataset:
Euler Phi Function of 72

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {72} = 24$
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:
:$72 = 2^3 \times 3^2$
Thus:
Category:Euler Phi Function
Category:72
Euler Phi Function of 735

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {735} = 336$
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:
:$735 = 3 \times 5 \times 7^2$
Thus:
Category:Euler Phi Function
Category:735
Euler Phi Function of 76,326

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {76 \, 326} = 25 \, 440$
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:
:$76 \, 326 = 2 \times 3 \times 12 \, 721$
Thus:
Category:Euler Phi Function
Category:76,326
Euler Phi Function of 76,332

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {76 \, 332} = 25 \, 440$
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:
:$76 \, 332 = 2^3 \times 3 \times 6361$
Thus:
Category:Euler Phi Function
Category:76,332
Euler Phi Function of 76,338

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {76 \, 338} = 25 \, 440$
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:
:$76 \, 338 = 2 \times 3^2 \times 4241$
Thus:
Category:Euler Phi Function
Category:76,338
Euler Phi Function of 76,344

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {76 \, 338} = 25 \, 440$
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:
:$76 \, 344 = 2^3 \times 3 \times 3181$
Thus:
Category:Euler Phi Function
Category:76,344
Euler Phi Function of 78

== Example of Euler $\phi$ Function of Square-Free Integer ==
:$\map \phi {78} = 24$
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:
:$78 = 2 \times 3 \times 13$
and so is square-free.
Thus:
Category:Euler Phi Function of Square-Free Integer
Category:78
Euler Phi Function of 824

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {824} = 408$
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:
:$824 = 2^3 \times 103$
Thus:
Category:Euler Phi Function
Category:824
Euler Phi Function of 825

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {825} = 400$
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:
:$825 = 3 \times 5^2 \times 11$
Thus:
Category:Euler Phi Function
Category:825
Euler Phi Function of 83,623,935

== Example of Euler $\phi$ Function of Square-Free Integer ==
:$\map \phi {83 \, 623 \, 935} = 41 \, 811 \, 968$
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:
:$83 \, 623 \, 935 = 3 \times 5 \times 17 \times 353 \times 929$
and so is square-free.
Thus:
Category:Euler Phi Function of Square-Free Integer
Category:83,623,935
Euler Phi Function of 84

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {84} = 24$
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:
:$84 = 2^2 \times 3 \times 7$
Thus:
Category:Euler Phi Function
Category:84
Euler Phi Function of 87

== Example of Euler $\phi$ Function of Non-Square Semiprime ==
:$\phi \left({87}\right) = 56$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$87 = 3 \times 29$
Thus:
Category:Euler Phi Function of Non-Square Semiprime
Category:87
Euler Phi Function of 90

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {90} = 24$
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:
:$90 = 2 \times 3^2 \times 5$
Thus:
Category:Euler Phi Function
Category:90
Euler Phi Function of 91

== Example of Euler $\phi$ Function of Non-Square Semiprime ==
:$\map \phi {91} = 72$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$91 = 7 \times 13$
Thus:
Category:Euler Phi Function of Non-Square Semiprime
Category:91
Euler Phi Function of 9374

== Example of Euler $\phi$ Function of Square-Free Integer ==
:$\map \phi {9374} = 4536$
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:
:$9374 = 2 \times 43 \times 109$
and so is square-free.
Thus:
Category:Euler Phi Function of Square-Free Integer
Category:9374
Euler Phi Function of 944

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {944} = 464$
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:
:$944 = 2^4 \times 59$
Thus:
Category:Euler Phi Function
Category:944
Euler Phi Function of 945

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {945} = 432$
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:
:$945 = 3^3 \times 5 \times 7$
Thus:
Category:Euler Phi Function
Category:945
Euler Phi Function of 95

== Example of Euler $\phi$ Function of Non-Square Semiprime ==
:$\phi \left({95}\right) = 72$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$91 = 5 \times 19$
Thus:
Category:Euler Phi Function of Non-Square Semiprime
Category:95
Euler Phi Function of 974

== Example of Euler Phi Function of 2 times Odd Prime ==
:$\phi \left({974}\right) = 486$
where $\phi$ denotes the Euler $\phi$ Function.
== Proof ==
We have that:
:$974 = 2 \times 487$
Thus:
Category:Euler Phi Function of 2 times Odd Prime
Category:974
Euler Phi Function of 975

== Example of Use of Euler $\phi$ Function ==
:$\map \phi {975} = 480$
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:
:$975 = 3 \times 5^2 \times 13$
Thus:
Category:Euler Phi Function
Category:975
Euler Phi Function of Integer

== Theorem ==
Let $n \in \Z_{>0}$, that is, a (strictly) positive integer.
Let $\phi: \Z_{>0} \to \Z_{>0}$ be the Euler $\phi$ function.
Then for any $n \in \Z_{>0}$, we have:
:$\map \phi n = n \paren {1 - \dfrac 1 {p_1} } \paren {1 - \dfrac 1 {p_2} } \cdots \paren {1 - \dfrac 1 {p_r} }$
where $p_1, p_2, \ldots, p_r$ are the distinct primes dividing $n$.
Or, more compactly:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
=== Corollary ===
== Proof ==
If $n = 1$ the result holds by inspection.
Let $n \ge 2$.
We express $n$ in its prime decomposition:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}, p_1 < p_2 < \cdots < p_r$
as it is always possible to do.
By definition, all primes are coprime to each other.
Hence from Euler Phi Function is Multiplicative:
:$\map \phi n = \map \phi {p_1^{k_1} } \map \phi {p_2^{k_2} } \cdots \map \phi {p_r^{k_r} }$
and from Euler Phi Function of Prime Power:
:$\map \phi {p^k} = p^k \paren {1 - \dfrac 1 p}$
So:
:$\map \phi n = p_1^{k_1} \paren {1 - \dfrac 1 {p_1} } p_2^{k_2} \paren {1 - \dfrac 1 {p_2} } \cdots p_r^{k_r} \paren {1 - \dfrac 1 {p_r} }$
and the result follows directly from:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
== Sources ==
* : Introduction: Notation
* : Chapter $1$: Properties of the Natural Numbers: $\S 25$
* : $2.5$: A product formula for $\varphi$
* : $\S 2.3$: Congruences
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $30$
* 
* 
Category:Euler Phi Function
Euler Phi Function of Integer/Corollary

== Corollary to Euler Phi Function of Integer ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $p$ be a prime number.
Let $\map \phi n$ denote the Euler $\phi$ function of $n$.
Let $p$ be a divisor of $n$.
Then $p - 1$ is a divisor of $\map \phi n$.
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
Let $n$ be expressed as its prime decomposition:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}, p_1 < p_2 < \ldots < p_r$
Then:
Hence the result.
Category:Euler Phi Function
Euler Phi Function of Non-Square Semiprime

== Theorem ==
Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Let $\map \phi n$ denote the Euler $\phi$ function.
Then:
:$\map \phi n = \paren {p - 1} \paren {q - 1}$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
Category:Euler Phi Function
Category:Semiprimes
Category:Euler Phi Function of Non-Square Semiprime
Euler Phi Function of Non-Square Semiprime/Examples

== Examples of Euler Phi Function of Non-Square Semiprime==
=== Euler Phi Function of $87$ ===
=== Euler Phi Function of $91$ ===
=== Euler Phi Function of $95$ ===
=== Euler Phi Function of $111$ ===
=== Euler Phi Function of $1257$ ===
Category:Euler Phi Function of Non-Square Semiprime
Euler Phi Function 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 Euler Phi Function is Multiplicative:
:$\map \phi n = \map \phi p \, \map \phi q$
From Euler Phi Function of Prime:
:$\map \phi p = p - 1$
:$\map \phi q = q - 1$
Hence the result.
== Examples ==
Category:Euler Phi Function of Non-Square Semiprime
Euler Phi Function of Non-Square Semiprime/Proof 2

== Theorem ==
== Proof ==
A semiprime with distinct prime factors is a square-free integer.
:$\map \phi n = \ds \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.
As there are $2$ prime factors: $p$ and $q$, this devolves to:
:$\map \phi n = \paren {p - 1} \paren {q - 1}$
except when $p = 2$, in which case:
:$\map \phi n = q - 1$
But when $p = 2$, we have that $p - 1 = 1$ and so:
:$\paren {p - 1} \paren {q - 1} = q - 1$
Hence the result.
== Examples ==
Category:Euler Phi Function of Non-Square Semiprime
Euler Phi Function of Prime

== Theorem ==
Let $p$ be a prime number.
Then:
:$\map \phi p = p - 1$
where $\phi: \Z_{>0} \to \Z_{>0}$ is the Euler $\phi$ function.
== Proof ==
From the definition of a prime number, the only (strictly) positive integer less than or equal to a prime $p$ which is not prime to $p$ is $p$ itself.
Thus it follows directly that:
:$\map \phi p = p - 1$
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$: $(1.27)$
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $27$
Category:Euler Phi Function
Category:Prime Numbers
Euler Phi Function of Prime Power

== Theorem ==
Let $p^n$ be a prime power for some prime number $p > 1$.
Then:
:$\map \phi {p^n} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$
where $\phi: \Z_{>0} \to \Z_{>0}$ is the Euler $\phi$ function.
=== Corollary ===
When $p = 2$, the formula is exceptionally simple:
== Proof ==
From Euler Phi Function of Prime:
:$\map \phi p = p - 1$
From Prime not Divisor implies Coprime:
:$k \perp p^n \iff p \nmid k$
There are $p^{n - 1}$ integers $k$ such that $1 \le k \le p^n$ which are divisible by $p$:
:$k \in \set {p, 2 p, 3 p, \ldots, \paren {p^{n - 1} } p}$
Therefore:
:$\map \phi {p^n} = p^n - p^{n - 1} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$
== Also see ==
* Euler Phi Function of Prime: When $n = 1$ the expression degenerates to $\map \phi p = p - 1$.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$: $(1.28)$
* : Chapter $1$: Properties of the Natural Numbers: $\S 25$
* : $2.5$: A product formula for $\varphi$
* : $\S 2.3$: Congruences
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $27$
Category:Euler Phi Function
Category:Prime Numbers
Euler Phi Function of Prime Power/Corollary

== Corollary to Euler Phi Function of Prime Power ==
Let $\phi: \Z_{>0} \to \Z_{>0}$ be the Euler $\phi$ function.
Then:
:$\map \phi {2^k} = 2^{k - 1}$
== Proof ==
We have that:
:$1 - \dfrac 1 2 = \dfrac {2 - 1} 2 = \dfrac 1 2$
It follows from Euler Phi Function of Prime Power:
:$\map \phi {2^k} = \paren {\dfrac 1 2} 2^k = 2^{k - 1}$
Category:Euler Phi Function
Category:Prime Numbers
Euler Phi Function of Product

== Theorem ==
Let $m$ and $n$ be positive integers.
Let $d$ be the greatest common divisor of $m$ and $n$.
Then:
:$\map \phi {m n} = \map \phi m \map \phi n \paren {\dfrac d {\map \phi d} }$
where $\phi$ is the Euler $\phi$ function.
== Proof ==
From Euler Phi Function of Integer, we have:
:$\ds \frac {\map \phi m} m = \prod_{p \mathop \divides m} \paren {1 - \frac 1 p}$
:$\ds \frac {\map \phi n} n = \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
:$\ds \frac {\map \phi {m n} } {m n} = \prod_{p \mathop \divides m n} \paren {1 - \frac 1 p}$
where the products are understood to run over the primes.
We consider:
:$\ds \paren {\prod_{p \mathop \divides m} \paren {1 - \frac 1 p} } \paren {\prod_{p \mathop \divides n} \paren {1 - \frac 1 p} }$
and its relation to:
:$\ds \prod_{p \mathop \divides m n} \paren {1 - \frac 1 p}$
From Euclid's Lemma for Prime Divisors, a prime $p$ divides $m n$ $p$ divides at least one of $m$ or $n$.
So every term in the product:
:$\ds \prod_{p \mathop \divides m n} \paren {1 - \frac 1 p}$
is included within:
:$\ds \paren {\prod_{p \mathop \divides m} \paren {1 - \frac 1 p} } \paren {\prod_{p \mathop \divides n} \paren {1 - \frac 1 p} } = \Pi$
If $p$ divides exactly one of $m$ or $n$, $\paren {1 - \dfrac 1 p}$ will appear exactly once in the product $\Pi$.
If $p$ divides both of $m$ and $n$, then $\paren {1 - \dfrac 1 p}$ will appear twice in the product $\Pi$.
From the definition of the greatest common divisor, $p$ divides both of $m$ and $n$ it divides $d$.
Hence, removing the extraneous terms from $\Pi$ we obtain:
:$\ds \prod_{p \mathop \divides m n} \paren {1 - \frac 1 p} = \frac { \paren {\prod_{p \mathop \divides m} \paren {1 - \frac 1 p} } \paren {\prod_{p \mathop \divides n} \paren {1 - \frac 1 p} } } {\prod_{p \mathop \divides d} \paren {1 - \frac 1 p} }$
From Euler Phi Function of Integer, we have:
:$\ds \prod_{p \mathop \divides d} \paren {1 - \frac 1 p} = \frac {\map \phi d} d$
So, we obtain:
:$\dfrac {\map \phi {m n} } {m n} = \dfrac {\frac {\map \phi m \map \phi n} {m n} } {\frac {\map \phi d} d} = \dfrac 1 {m n} \paren {\map \phi m \map \phi n \paren {\dfrac d {\map \phi d} } }$
Hence:
:$\map \phi {m n} = \map \phi m \map \phi n \paren {\dfrac d {\map \phi d} }$
as required.
Category:Euler Phi Function
Euler Phi Function of Product with Prime

== Theorem ==
Let $p$ be prime and $n \in \Z: n \ge 1$.
Then:
:$\map \phi {p n} = \begin{cases} \paren {p - 1} \map \phi n & : p \nmid n \\ p \map \phi n & : p \divides n \end{cases}$
where:
:$\map \phi n$ denotes the Euler $\phi$ function of $n$
:$\divides$ denotes divisibility.
Thus for all $n \ge 1$ and for any prime $p$, we have that $\map \phi n$ divides $\map \phi {p n}$.
== Proof ==
First suppose that $p \nmid n$.
Then by Prime not Divisor implies Coprime, $p \perp n$.
So by Euler Phi Function is Multiplicative, $\map \phi {p n} = \map \phi p \map \phi n$.
It follows from Euler Phi Function of Prime that $\map \phi {p n} = \paren {p - 1} \map \phi n$.
Now suppose that $p \divides n$.
Then $n = p^k m$ for some $k, m \in \Z: k, m \ge 1$ such that $p \perp m$.
Then:
At the same time:
Category:Euler Phi Function
Euler Phi Function of Square-Free Integer

== Theorem ==
Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let $\map \phi n$ be the Euler $\phi$ function of $n$.
That is, let $\map \phi n$ be the count of strictly positive integers less than or equal to $n$ which are prime to $n$.
Then:
:$\map \phi n = \ds \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.
== Proof 1 ==
== Proof 2 ==
== Examples ==
Category:Euler Phi Function
Category:Euler Phi Function of Square-Free Integer
Euler Phi Function of Square-Free Integer/Examples

== Examples of Euler Phi Function of Square-Free Integer==
=== Euler Phi Function of $6$ ===
=== Euler Phi Function of $30$ ===
=== Euler Phi Function of $42$ ===
=== Euler Phi Function of $78$ ===
=== Euler Phi Function of $182$ ===
=== Euler Phi Function of $190$ ===
=== Euler Phi Function of $222$ ===
=== Euler Phi Function of $1798$ ===
=== Euler Phi Function of $5002$ ===
=== Euler Phi Function of $9374$ ===
Category:Euler Phi Function of Square-Free Integer
Euler Phi Function of Square-Free Integer/Proof 1

== Theorem ==
== Proof ==
We have that the Euler Phi Function is Multiplicative.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i = p_1 p_2 \cdots p_r$
From the definition of prime number, each of the prime factors of $n$ is coprime to all other divisors of $n$.
From Euler Phi Function of Prime, we have:
:$\map \phi {p_i} = \paren {p_i - 1}$
Thus:
:$\map \phi n = \ds \prod_{1 \mathop \le i \mathop \le r} \paren {p_i - 1}$
or:
:$\map \phi n = \ds \prod_{p \mathop \divides n} \paren {p - 1}$
where $p$ ranges over all prime numbers
When $p = 2$ we have that:
:$p - 1 = 1$
and so:
:$\ds \prod_{p \mathop \divides n} \paren {p - 1} = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
Hence the result.
== Examples ==
Category:Euler Phi Function of Square-Free Integer
Euler Phi Function of Square-Free Integer/Proof 2

== Theorem ==
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
As $n$ is square-free:
:$\ds n = \prod_{p \mathop \divides n} p$
Hence:
:$\ds \map \phi n = \prod_{p \mathop \divides n} p \paren {1 - \frac 1 p}$
and so:
:$\ds \map \phi n = \prod_{p \mathop \divides n} \paren {p - 1}$
When $p = 2$ we have that:
:$p - 1 = 1$
and so:
:$\ds \prod_{p \mathop \divides n} \paren {p - 1} = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
Hence the result.
== Examples ==
Category:Euler Phi Function of Square-Free Integer
Euler Phi Function of n equal to Euler Phi Function of n+3

== Theorem ==
Let $\phi$ denote the Euler $\phi$ function.
The only solutions to the equation:
:$\map \phi n = \map \phi {n + 3}$
less than $1 \, 000 \, 000$ are:
:$\map \phi 3 = \map \phi 6 = 2$
:$\map \phi 5 = \map \phi 8 = 4$
== Proof ==
== Sources ==
* : $3$
Category:Euler Phi Function
Euler Phi Function preserves Divisibility

== Theorem ==
Let $n \in \Z_{>0}$ be a positive integer.
Let $\map \phi n$ denote the Euler $\phi$ function of $n$.
Then:
:$d \divides n \implies \map \phi d \divides \map \phi n$
where $d \divides n$ denotes that $d$ is a divisor of $n$.
== Proof ==
Let $d \divides n$.
We can write $n$ as $n = d p_1 p_2 p_3 \cdots p_r$, where $p_1, p_2, \ldots, p_r$ are all the primes (not necessarily distinct) which divide $\dfrac n d$.
Thus, repeatedly using Euler Phi Function of Product with Prime:
As the last expression is $\map \phi n$, the result follows from Divisor Relation on Positive Integers is Partial Ordering.
Category:Euler Phi Function
Euler Polyhedron Formula

== Theorem ==
For any convex polyhedron with $V$ vertices, $E$ edges, and $F$ faces: 
:$V - E + F = 2$
== Proof ==
== Also known as ==
== Examples ==
== Also see ==
* Definition:Euler Characteristic of Surface
== Sources ==
* : $\S 3$: Appendix $\text A$: Euler
* : $2$
* : Chapter $\text {A}.21$: Euler ()
* : $2$
* 
* 
* 
* 
* 
Category:Euler Polyhedron Formula
Category:Graph Theory
Euler Quartic Conjecture

== Famous False Conjecture ==
The Diophantine equation:
:$A^4 = B^4 + C^4 + D^4$
has no solutions.
That is, there exist no integers $A, B, C, D \in \Z$ that satisfy the above.
== Refutation ==
The following counterexamples exist:
== Also known as ==
== Historical Note ==
== Sources ==
* 
* : $20,615,673$
* 
* 
* 
Category:Euler Quartic Conjecture
Category:Fourth Powers
Category:Famous False Conjectures
Euler Quartic Conjecture/Historical Note

== Historical Note on Euler Quartic Conjecture ==
 put forward this conjecture in $1772$.
Nobody made any progress on proving it one way or another until discovered the counterexample $20 \, 615 \, 673$ in $1987$, and proved that there exists an infinite number of such solutions.
Soon after that, found the smaller counterexample $422 \, 481$, and demonstrated that there were none smaller.
Both counterexamples were published in the cited article in by in $1988$.
This discovery was subsequently reported in the New York Times.
 found the solution $638 \, 523 \, 249$ in $1997$.
== Sources ==
* 
* : $20,615,673$
* 
* 
* 
Category:Historical Notes
Euler Triangle Formula

:
== Theorem ==
Let $d$ be the distance between the incenter and the circumcenter of a triangle.
Then:
:$d^2 = R \paren {R - 2 \rho}$
where:
:$R$ is the circumradius
:$\rho$ is the inradius.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
Category:Euler Triangle Formula
Category:Circumcenters
Category:Incircles of Triangles
Category:Triangles
Euler Triangle Formula/Also presented as

:
== Euler Triangle Formula: Also presented as ==
Some writers prefer the following form:
Let:
:$d$ be the distance between the incenter and the circumcenter of a triangle
:$R$ be the circumradius
:$\rho$ be the inradius
of a triangle.
Then:
:$d^2 + 2 \rho R = R^2$
Category:Euler Triangle Formula
Euler Triangle Formula/Corollary

== Theorem ==
Let $\rho$ be the inradius of a triangle.
Then:
:$\dfrac 1 \rho = \dfrac 1 {R - d} + \dfrac 1 {R + d}$
where:
:$R$ is the circumradius
:$d$ is the distance between the incenter and the circumcenter.
== Proof ==
Category:Euler Triangle Formula