Difference between revisions of "Euler's totient function"
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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>. | 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 | ||
+ | <cmath>A_d=\{x:1\leq x\leq n\quad\text{and}\quad \operatorname{syt}(x,n)=d \}.</cmath> | ||
+ | Now <math>\operatorname{gcd}(dx,n)=d</math> if and only if <math>\operatorname{gcd}(x,n/d)=1</math>. Furthermore, <math>1\leq dx\leq n</math> if and only if <math>1\leq x\leq n/d</math>. Now one can see that the number of elements of <math>A_d</math> equals the number of elements of | ||
+ | <cmath>A_d^\prime=\{x:1\leq x \leq n/d\quad\text{and}\quad \operatorname{gcd}(x,n/d)=1 \}.</cmath> | ||
+ | Thus by the definition of Euler's phi we have that <math>|A_d^\prime|=\phi (n/d)</math>. As every integer <math>i</math> which satisfies <math>1\leq i\leq n</math> belongs in exactly one of the sets <math>A_d</math>, we have that | ||
+ | <cmath>n=\sum_{d \mid n}\varphi \left (\frac{n}{d} \right )=\sum_{d \mid n}\phi (d).</cmath> | ||
==Notation== | ==Notation== |
Revision as of 15:17, 25 July 2008
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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
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
We can factor out, though:
But we are obviously overcounting. We then subtract out those divisible by two of the . We continue with this PIE argument to figure out that the number of elements in the complement of what we want is
which we can factor further as
Making one small adjustment, we write this as
Given the general prime factorization of , one can compute using the formula
Identities
For prime p, , because all numbers less than are relatively prime to it.
For relatively prime , .
In fact, we also have for any that .
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