Euler's totient function

Euler's totient function $\phi(n)$ applied to a positive integer $n$ is defined to be the number of positive integers less than or equal to $n$ that are relatively prime to $n$ . $\phi(n)$ is read "phi of n."

Video

https://youtu.be/T7INnve59JM

Formula

Given the general prime factorization of ${n} = {p}_1^{e_1}{p}_2^{e_2} \cdots {p}_m^{e_m}$ , one can compute $\phi(n)$ using the formula $\phi(n)= n\left(1-\frac{1}{p_1} \right) \left(1-\frac{1}{p_2} \right)\cdots \left(1-\frac{1}{p_m}\right).$

Derivation

To derive the formula, let us first define the prime factorization of $n$ as $n =\prod_{i=1}^{m}p_i^{e_i} =p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$ where the $p_i$ are distinct prime numbers. Now, we can use a PIE argument to count the number of numbers less than or equal to $n$ that are relatively prime to it.

First, let's count the complement of what we want (i.e. all the numbers less than or equal to $n$ that share a common factor with it). There are $\frac{n}{p_1}$ positive integers less than or equal to $n$ that are divisible by $p_1$ . If we do the same for each $p_i$ and add these up, we get

$\frac{n}{p_1} + \frac{n}{p_2} + \cdots + \frac{n}{p_m} = \sum^m_{i=1}\frac{n}{p_i}.$

But we are obviously overcounting. We then subtract out those divisible by two of the $p_i$ . There are $\sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}$ such numbers. We continue with this PIE argument to figure out that the number of elements in the complement of what we want is

$\sum_{1 \le i \le m}\frac{n}{p_i} - \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}.$

This sum represents the number of numbers less than $n$ sharing a common factor with $n$ , so

$\phi(n) = n - \left(\sum_{1 \le i \le m}\frac{n}{p_i}- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}\right)$

$\phi(n)= n\left(1 - \sum_{1 \le i \le m}\frac{1}{p_i} + \sum_{1 \le i_1 < i_2 \le m}\frac{1}{p_{i_1}p_{i_2}} - \cdots + (-1)^{m}\frac{1}{p_1p_2\ldots p_m}\right)$

$\phi(n)= n\left(1-\frac{1}{p_1} \right) \left(1-\frac{1}{p_2} \right)\cdots \left(1-\frac{1}{p_m}\right).$

Note: Another way to find the closed form for $\phi(n)$ is to show that the function is multiplicative, and then breaking up $\phi(n)$ into its prime factorization.

Identities

(a) For prime $p$ , $\phi(p)=p-1$ , because all numbers less than ${p}$ are relatively prime to it.

(b) For relatively prime ${a}, {b}$ , $\phi{(a)}\phi{(b)} = \phi{(ab)}$ .

(c) In fact, we also have for any ${a}, {b}$ that $\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})$ .

(d) If $p$ is prime and $n\ge{1},$ then $\phi(p^n)=p^n-p^{n-1}$

(e) For any $n$ , we have $\sum_{d|n}\phi(d)=n$ where the sum is taken over all divisors $d$ of $n$ .

Proof. Split the set $\{1,2,\ldots,n\}$ into disjoint sets $A_d$ where for all $d\mid n$ we have $A_d=\{x:1\leq x\leq n\quad\text{and}\quad \operatorname{syt}(x,n)=d \}.$ Now $\operatorname{gcd}(dx,n)=d$ if and only if $\operatorname{gcd}(x,n/d)=1$ . Furthermore, $1\leq dx\leq n$ if and only if $1\leq x\leq n/d$ . Now one can see that the number of elements of $A_d$ equals the number of elements of $A_d^\prime=\{x:1\leq x \leq n/d\quad\text{and}\quad \operatorname{gcd}(x,n/d)=1 \}.$ Thus by the definition of Euler's phi we have that $|A_d^\prime|=\phi (n/d)$ . As every integer $i$ which satisfies $1\leq i\leq n$ belongs in exactly one of the sets $A_d$ , we have that $n=\sum_{d \mid n}\phi \left (\frac{n}{d} \right )=\sum_{d \mid n}\phi (d).$

(f) Another interesting thing to note is that $\phi(n)\leq n - 1$ . This does seem very obvious but is helpful in solving many problems.

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Notation

Sometimes, instead of $\phi$ , $\varphi$ is used. This variation of the Greek letter phi is common in textbooks, and is standard usage on the English Wikipedia

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