Difference between revisions of "Euler's totient function"
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== Identities == | == Identities == | ||
− | For [[prime]] p, <math>\phi(p)=p-1</math>, because all numbers less than <math>{p}</math> are relatively prime to it. | + | (a) For [[prime]] p, <math>\phi(p)=p-1</math>, because all numbers less than <math>{p}</math> are relatively prime to it. |
− | For relatively prime <math>{a}, {b}</math>, <math> \phi{(a)}\phi{(b)} = \phi{(ab)} </math>. | + | (b) For relatively prime <math>{a}, {b}</math>, <math> \phi{(a)}\phi{(b)} = \phi{(ab)} </math>. |
− | In fact, we also have for any <math>{a}, {b}</math> that <math>\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})</math>. | + | (c) In fact, we also have for any <math>{a}, {b}</math> that <math>\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})</math>. |
− | For any <math>n</math>, we have <math>\sum_{d|n}\phi(d)=n</math> where the sum is taken over all divisors d of <math> n </math>. | + | (d) If <math>p</math> is prime and <math>n\ge{1}</math>,then <math>\phi(p^n)=p^n-p^{n-1}</math> |
+ | |||
+ | (e) For any <math>n</math>, we have <math>\sum_{d|n}\phi(d)=n</math> where the sum is taken over all divisors d of <math> n </math>. | ||
Proof. Split the set <math>\{1,2,\ldots,n\}</math> into disjoint sets <math>A_d</math> where for all <math>d\mid n</math> we have | Proof. Split the set <math>\{1,2,\ldots,n\}</math> into disjoint sets <math>A_d</math> where for all <math>d\mid n</math> we have |
Revision as of 22:50, 22 May 2014
Euler's totient function applied to a positive integer
is defined to be the number of positive integers less than or equal to
that are relatively prime to
.
is read "phi of n."
Contents
[hide]Formulas
To derive the formula, let us first define the prime factorization of as
where the
are distinct prime numbers. Now, we can use a PIE argument to count the number of numbers less than or equal to
that are relatively prime to it.
First, let's count the complement of what we want (i.e. all the numbers less than that share a common factor with it). There are
numbers less than
that are divisible by
. If we do the same for each
and add these up, we get
![\[\frac{n}{p_1} + \frac{n}{p_2} + \cdots + \frac{n}{p_m} = \sum^m_{i=1}\frac{n}{p_i}.\]](http://latex.artofproblemsolving.com/5/7/6/576821b22ebb5e1a9c986631792e516828e99b1c.png)
But we are obviously overcounting. We then subtract out those divisible by two of the . There are
such numbers. We continue with this PIE argument to figure out that the number of elements in the complement of what we want is
![\[\sum_{1 \le i \le m}\frac{n}{p_i} - \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}.\]](http://latex.artofproblemsolving.com/1/a/5/1a54da6486432515085420e1ba73b8010bf4351b.png)
This sum represents the number of numbers less than sharing a common factor with
, so
$\phi(n) = n - \left(\sum_{1 \le i \le m}\frac{n}{p_i}$ (Error compiling LaTeX. Unknown error_msg) $- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}\right)$ (Error compiling LaTeX. Unknown error_msg)
Given the general prime factorization of , one can compute
using the formula
Identities
(a) For prime p, , because all numbers less than
are relatively prime to it.
(b) For relatively prime ,
.
(c) In fact, we also have for any that
.
(d) If is prime and
,then
(e) For any , we have
where the sum is taken over all divisors d of
.
Proof. Split the set into disjoint sets
where for all
we have
Now
if and only if
. Furthermore,
if and only if
. Now one can see that the number of elements of
equals the number of elements of
Thus by the definition of Euler's phi we have that
. As every integer
which satisfies
belongs in exactly one of the sets
, we have that
Notation
Sometimes, instead of ,
is used. This variation of the Greek letter phi is common in textbooks, and is standard usage on the English Wikipedia