Euler's Totient Function

Given an integer n, find the value of Euler's Totient Function, denoted as Φ(n). The function Φ(n) represents the count of positive integers less than or equal to n that are relatively prime to n.

Euler's Totient function Φ(n) for an input n is the count of numbers in {1, 2, 3, ..., n-1} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1.

If n is a positive integer and its prime factorization is;
Where ​ are distinct prime factors of n, then:
.

Examples:

Input: n = 11
Output: 10
Explanation: From 1 to 11, 1,2,3,4,5,6,7,8,9,10 are relatively prime to 11.

Input: n = 16
Output: 8
Explanation: From 1 to 16, 1,3,5,7,9,11,13,15 are relatively prime to 16.

[Naive Approach] Iterative GCD Method

A simple solution is to iterate through all numbers from 1 to n-1 and count numbers with gcd with n as 1. Below is the implementation of the simple method to compute Euler's Totient function for an input integer n. 

10

Time Complexity: O(n log n)
Auxiliary Space: O(log min(a,b)) where a,b are the parameters of gcd function.

[Expected Approach] Euler’s Product Formula

The idea is based on Euler's product formula which states that the value of totient functions is below the product overall prime factors p of n. 

👁 Euler's-Product-Formula

1) Initialize result as n
2) Consider every number 'p' (where 'p' varies from 2 to Φ(n)).
If p divides n, then do following
a) Subtract all multiples of p from 1 to n [all multiples of p
will have gcd more than 1 (at least p) with n]
b) Update n by repeatedly dividing it by p.
3) If the reduced n is more than 1, then remove all multiples
of n from result.

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Time Complexity: O(√n)
Auxiliary Space: O(1)

Some Interesting Properties of Euler's Totient Function

1) For a prime number p, 

Proof :

, where p is any prime number
We know that where k is any random number and
Total number from 1 to p = p
Number for which is , i.e the number p itself, so subtracting 1 from p

Examples : 

.

2) For two prime numbers a and b, used in RSA Algorithm

Proof :

Let a and b be distinct primes.
Then:

• ϕ(a)=a−1, ϕ(b)=b−1

Total numbers from 1 to ab: ab
Multiples of a: b numbers
Multiples of b: a numbers
Common multiple (i.e., double-counted): only ab

So, numbers not coprime to ab:

a+b−1

Then,

ϕ(ab) = ab − (a + b − 1) = ab − a − b + 1 = (a−1)(b−1)

Hence,

ϕ(ab) = ϕ(a)* ϕ(b)

Examples :

• ϕ(5⋅7) = ϕ(5)⋅ϕ(7) = (5−1) (7−1) = 4⋅6 = 24
• ϕ(3⋅5) = ϕ(3)⋅ϕ(5) = (3−1) (5−1) = 2⋅4 = 8
• ϕ(3⋅7) = ϕ(3)⋅ϕ(7) = (3−1) (7−1) = 2⋅6 = 12

3) For a prime number p and integer k ≥ 1:

ϕ(p^k) = p^k−p^k−1

Proof : 

, where p is a prime numberTotal numbers from 1 to Total multiples of
Removing these multiples as with them

Examples : 

p = 2, k = 5, = 32
Multiples of 2 (as with them ) = 32 / 2 = 16.

4) Special Case : gcd(a, b) = 1

.

Examples :

Special Case:

,

Normal Case:

, .

5) Sum of values of totient functions of all divisors of n is equal to n. 

👁 gausss

Example :

n = 6 , factors = {1, 2, 3, 6}
n = = 1 + 1 + 2 + 2 = 6

6) The most famous and important feature is expressed in Euler's theorem : 

The theorem states that if n and a are coprime
(or relatively prime) positive integers, then

a^Φ(n) Φ 1 (mod n)

The RSA cryptosystem is based on this theorem:
In the particular case when m is prime say p, Euler's theorem turns into the so-called Fermat's little theorem : 

a^p-1 Φ 1 (mod p)

Related Article:
Euler’s Totient function for all numbers smaller than or equal to n
Optimized Euler Totient Function for Multiple Evaluations