There are several formulae for computing \varphi(n).
Euler's product formula It states :\varphi(n) =n \prod_{p\mid n} \left(1-\frac{1}{p}\right), where the product is over the distinct
prime numbers dividing . An equivalent formulation is :\varphi(n) = p_1^{k_1-1}(p_1{-}1)\,p_2^{k_2-1}(p_2{-}1)\cdots p_r^{k_r-1}(p_r{-}1), where n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} is the
prime factorization of n (that is, p_1, p_2,\ldots, p_r are distinct prime numbers). The proof of these formulae depends on two important facts.
Phi is a multiplicative function This means that if \gcd(m,n) = 1, then \varphi(m) \varphi(n) = \varphi(mn).
Proof outline: Let A,B,C be the sets of positive integers which are
coprime to and less than , , , respectively, so that |A| = \varphi(m), etc. Then there is a
bijection between A\times B and by the
Chinese remainder theorem.
Value of phi for a prime power argument If is prime and k\geq1, then :\varphi \left(p^k\right) = p^k-p^{k-1} = p^{k-1}(p-1) = p^k \left( 1 - \tfrac{1}{p} \right).
Proof: Since is a prime number, the only possible values of \gcd(p^{k},m) are 1,p,p^{2},\dots,p^{k}, and the only way to have \gcd(p^{k},m)>1 is if is a multiple of , that is, m \in \{ p, 2p, 3p, \ldots, p^{k-1} p= p^{k}\}, and there are p^{k-1} such multiples not greater than p^{k}. Therefore, the other p^{k}-p^{k-1} numbers are all relatively prime to p^{k}.
Proof of Euler's product formula The
fundamental theorem of arithmetic states that if there is a unique expression n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}, where are
prime numbers and each . (The case corresponds to the
empty product.) Repeatedly using the multiplicative property of and the formula for gives :\begin{array} {rcl} \varphi(n)&=& \varphi(p_1^{k_1})\, \varphi(p_2^{k_2}) \cdots\varphi(p_r^{k_r})\\[.1em] &=& p_1^{k_1} \left(1- \frac{1}{p_1} \right) p_2^{k_2} \left(1- \frac{1}{p_2} \right) \cdots p_r^{k_r}\left(1- \frac{1}{p_r} \right)\\[.1em] &=& p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} \left(1- \frac{1}{p_1} \right) \left(1- \frac{1}{p_2} \right) \cdots \left(1- \frac{1}{p_r} \right)\\[.1em] &=&n \left(1- \frac{1}{p_1} \right)\left(1- \frac{1}{p_2} \right) \cdots\left(1- \frac{1}{p_r} \right). \end{array} This gives both versions of Euler's product formula. An alternative proof that does not require the multiplicative property instead uses the
inclusion-exclusion principle applied to the set \{1,2,\ldots,n\}, excluding the sets of integers divisible by the prime divisors.
Example :\varphi(20)=\varphi(2^2 5)=20\,(1-\tfrac12)\,(1-\tfrac15) =20\cdot\tfrac12\cdot\tfrac45=8. In words: the distinct prime factors of 20 are 2 and 5; half of the twenty integers from 1 to 20 are divisible by 2, leaving ten; a fifth of those are divisible by 5, leaving eight numbers coprime to 20; these are: 1, 3, 7, 9, 11, 13, 17, 19. The alternative formula uses only integers:\varphi(20) = \varphi(2^2 5^1)= 2^{2-1}(2{-}1)\,5^{1-1}(5{-}1) = 2\cdot 1\cdot 1\cdot 4 = 8.
Fourier transform The totient is the
discrete Fourier transform of the
gcd, evaluated at 1. Let : \mathcal{F} \{ \mathbf{x} \}[m] = \sum\limits_{k=1}^n x_k \cdot e^{{-2\pi i}\frac{mk}{n}} where for {{math|
k ∈ {1, ...,
n}}}. Then :\varphi (n) = \mathcal{F} \{ \mathbf{x} \}[1] = \sum\limits_{k=1}^n \gcd(k,n) e^{-2\pi i\frac{k}{n}}. The real part of this formula is :\varphi (n)=\sum\limits_{k=1}^n \gcd(k,n) \cos {\tfrac{2\pi k}{n}} . For example, using \cos\tfrac{\pi}5 = \tfrac{\sqrt 5+1}4 and \cos\tfrac{2\pi}5 = \tfrac{\sqrt 5-1}4 :\begin{array}{rcl} \varphi(10) &=& \gcd(1,10)\cos\tfrac{2\pi}{10} + \gcd(2,10)\cos\tfrac{4\pi}{10} + \gcd(3,10)\cos\tfrac{6\pi}{10}+\cdots+\gcd(10,10)\cos\tfrac{20\pi}{10}\\ &=& 1\cdot(\tfrac{\sqrt5+1}4) + 2\cdot(\tfrac{\sqrt5-1}4) + 1\cdot(-\tfrac{\sqrt5-1}4) + 2\cdot(-\tfrac{\sqrt5+1}4) + 5\cdot (-1) \\ && +\ 2\cdot(-\tfrac{\sqrt5+1}4) + 1\cdot(-\tfrac{\sqrt5-1}4) + 2\cdot(\tfrac{\sqrt5-1}4) + 1\cdot(\tfrac{\sqrt5+1}4) + 10 \cdot (1) \\ &=& 4 . \end{array} Unlike the
Euler product and the divisor sum formula, this one does not require knowing the factors of . However, it does involve the calculation of the greatest common divisor of and every positive integer less than , which suffices to provide the factorization anyway.
Divisor sum The property established by Gauss, that :\sum_{d\mid n}\varphi(d)=n, where the sum is over all positive divisors of , can be proven in several ways. (See
Arithmetical function for notational conventions.) One proof is to note that is also equal to the number of possible generators of the
cyclic group ; specifically, if with , then is a generator for every coprime to . Since every element of generates a cyclic
subgroup, and each subgroup is generated by precisely elements of , the formula follows. Equivalently, the formula can be derived by the same argument applied to the
multiplicative group of the th roots of unity and the
primitive th roots of unity. The formula can also be derived from
elementary arithmetic. For example, let and consider the positive fractions up to 1 with denominator 20: : \tfrac{ 1}{20},\,\tfrac{ 2}{20},\,\tfrac{ 3}{20},\,\tfrac{ 4}{20},\, \tfrac{ 5}{20},\,\tfrac{ 6}{20},\,\tfrac{ 7}{20},\,\tfrac{ 8}{20},\, \tfrac{ 9}{20},\,\tfrac{10}{20},\,\tfrac{11}{20},\,\tfrac{12}{20},\, \tfrac{13}{20},\,\tfrac{14}{20},\,\tfrac{15}{20},\,\tfrac{16}{20},\, \tfrac{17}{20},\,\tfrac{18}{20},\,\tfrac{19}{20},\,\tfrac{20}{20}. Put them into lowest terms: : \tfrac{ 1}{20},\,\tfrac{ 1}{10},\,\tfrac{ 3}{20},\,\tfrac{ 1}{ 5},\, \tfrac{ 1}{ 4},\,\tfrac{ 3}{10},\,\tfrac{ 7}{20},\,\tfrac{ 2}{ 5},\, \tfrac{ 9}{20},\,\tfrac{ 1}{ 2},\,\tfrac{11}{20},\,\tfrac{ 3}{ 5},\, \tfrac{13}{20},\,\tfrac{ 7}{10},\,\tfrac{ 3}{ 4},\,\tfrac{ 4}{ 5},\, \tfrac{17}{20},\,\tfrac{ 9}{10},\,\tfrac{19}{20},\,\tfrac{1}{1} These twenty fractions are all the positive ≤ 1 whose denominators are the divisors . The fractions with 20 as denominator are those with numerators relatively prime to 20, namely , , , , , , , ; by definition this is fractions. Similarly, there are fractions with denominator 10, and fractions with denominator 5, etc. Thus the set of twenty fractions is split into subsets of size for each dividing 20. A similar argument applies for any
n. Möbius inversion applied to the divisor sum formula gives : \varphi(n) = \sum_{d\mid n} \mu\left( d \right) \cdot \frac{n}{d} = n\sum_{d\mid n} \frac{\mu (d)}{d}, where is the
Möbius function, the
multiplicative function defined by \mu(p) = -1 and \mu(p^k) = 0 for each prime and . This formula may also be derived from the product formula by multiplying out \prod_{p\mid n} (1 - \frac{1}{p}) to get \sum_{d \mid n} \frac{\mu (d)}{d}. An example: \begin{align} \varphi(20) &= \mu(1)\cdot 20 + \mu(2)\cdot 10 +\mu(4)\cdot 5 +\mu(5)\cdot 4 + \mu(10)\cdot 2+\mu(20)\cdot 1\\[.5em] &= 1\cdot 20 - 1\cdot 10 + 0\cdot 5 - 1\cdot 4 + 1\cdot 2 + 0\cdot 1 = 8. \end{align} ==Some values==