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Euler’s Totient Function

Last Updated : 15 Apr, 2024
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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.

Examples :

Φ(1) = 1 gcd(1, 1) is 1 Φ(2) = 1 gcd(1, 2) is 1, but gcd(2, 2) is 2. Φ(3) = 2 gcd(1, 3) is 1 and gcd(2, 3) is 1 Φ(4) = 2 gcd(1, 4) is 1 and gcd(3, 4) is 1 Φ(5) = 4 gcd(1, 5) is 1, gcd(2, 5) is 1, gcd(3, 5) is 1 and gcd(4, 5) is 1 Φ(6) = 2 gcd(1, 6) is 1 and gcd(5, 6) is 1,

Recommended Practice

How to compute Φ(n) for an input n?
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. 

C++

// A simple C++ program to calculate
// Euler's Totient Function 
#include <iostream>
using namespace std; 
  
// Function to return gcd of a and b 
int gcd(int a, int b) 
    if (a == 0) 
        return b; 
    return gcd(b % a, a); 
  
// A simple method to evaluate Euler Totient Function 
int phi(unsigned int n) 
    unsigned int result = 1; 
    for (int i = 2; i < n; i++) 
        if (gcd(i, n) == 1) 
            result++; 
    return result; 
  
// Driver program to test above function 
int main() 
    int n; 
    for (n = 1; n <= 10; n++) 
        cout << "phi("<<n<<") = " << phi(n) << endl; 
    return 0; 
  
// This code is contributed by SHUBHAMSINGH10

C

// A simple C program to calculate Euler's Totient Function
#include <stdio.h>
  
// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// A simple method to evaluate Euler Totient Function
int phi(unsigned int n)
{
    unsigned int result = 1;
    for (int i = 2; i < n; i++)
        if (gcd(i, n) == 1)
            result++;
    return result;
}
  
// Driver program to test above function
int main()
{
    int n;
    for (n = 1; n <= 10; n++)
        printf("phi(%d) = %d\n", n, phi(n));
    return 0;
}

Java

// A simple java program to calculate
// Euler's Totient Function
import java.io.*;
  
class GFG {
  
    // Function to return GCD of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
  
    // A simple method to evaluate
    // Euler Totient Function
    static int phi(int n)
    {
        int result = 1;
        for (int i = 2; i < n; i++)
            if (gcd(i, n) == 1)
                result++;
        return result;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int n;
  
        for (n = 1; n <= 10; n++)
            System.out.println("phi(" + n + ") = " + phi(n));
    }
}
  
// This code is contributed by sunnusingh

Python3

# A simple Python3 program 
# to calculate Euler's 
# Totient Function
  
# Function to return
# gcd of a and b
def gcd(a, b):
  
    if (a == 0):
        return b
    return gcd(b % a, a)
  
# A simple method to evaluate
# Euler Totient Function
def phi(n):
  
    result = 1
    for i in range(2, n):
        if (gcd(i, n) == 1):
            result+=1
    return result
  
# Driver Code
for n in range(1, 11):
    print("phi(",n,") = "
           phi(n), sep = "")
             
# This code is contributed
# by Smitha

C#

// A simple C# program to calculate
// Euler's Totient Function
using System;
  
class GFG {
  
    // Function to return GCD of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
  
    // A simple method to evaluate
    // Euler Totient Function
    static int phi(int n)
    {
        int result = 1;
        for (int i = 2; i < n; i++)
            if (gcd(i, n) == 1)
                result++;
        return result;
    }
  
    // Driver code
    public static void Main()
    {
        for (int n = 1; n <= 10; n++)
        Console.WriteLine("phi(" + n + ") = " + phi(n));
    }
}
  
// This code is contributed by nitin mittal

PHP

<Φphp
// PHP program to calculate 
// Euler's Totient Function
  
// Function to return 
// gcd of a and b
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
  
// A simple method to evaluate
// Euler Totient Function
function phi($n)
{
    $result = 1;
    for ($i = 2; $i < $n; $i++)
        if (gcd($i, $n) == 1)
            $result++;
    return $result;
}
  
// Driver Code
for ($n = 1; $n <= 10; $n++)
    echo "phi(" .$n. ") =" . phi($n)."\n";
  
// This code is contributed by Sam007
Φ>

Javascript

<script>
// Javascript program to calculate 
// Euler's Totient Function
  
// Function to return 
// gcd of a and b
function gcd(a, b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// A simple method to evaluate
// Euler Totient Function
function phi(n)
{
    let result = 1;
    for (let i = 2; i < n; i++)
        if (gcd(i, n) == 1)
            result++;
    return result;
}
  
// Driver Code
for (let n = 1; n <= 10; n++)
    document.write(`phi(${n}) = ${phi(n)} <br>`);
  
// This code is contributed by _saurabh_jaiswal
  
</script>


Output

phi(1) = 1 phi(2) = 1 phi(3) = 2 phi(4) = 2 phi(5) = 4 phi(6) = 2 phi(7) = 6 phi(8) = 4 phi(9) = 6 phi(10) = 4


The above code calls gcd function O(n) times. The time complexity of the gcd function is O(h) where “h” is the number of digits in a smaller number of given two numbers. Therefore, an upper bound on the time complexity of the above solution is O(N^2 log N) [How Φ there can be at most Log10n digits in all numbers from 1 to n]

Auxiliary Space: O(log N)


Below is a Better Solution. 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. 

eulersproduct


The formula basically says that the value of Φ(n) is equal to n multiplied by-product of (1 – 1/p) for all prime factors p of n. For example value of Φ(6) = 6 * (1-1/2) * (1 – 1/3) = 2.
We can find all prime factors using the idea used in this post. 

1) Initialize : result = n 2) Run a loop from 'p' = 2 to sqrt(n), do following for every 'p'. a) If p divides n, then Set: result = result * (1.0 - (1.0 / (float) p)); Divide all occurrences of p in n. 3) Return result


Below is the implementation of Euler’s product formula.  

C++

// C++ program to calculate Euler's 
// Totient Function using Euler's
// product formula
#include <bits/stdc++.h>
using namespace std;
  
int phi(int n)
{
      
    // Initialize result as n
    float result = n; 
   
    // Consider all prime factors of n 
    // and for every prime factor p,
    // multiply result with (1 - 1/p)
    for(int p = 2; p * p <= n; ++p)
    {
          
        // Check if p is a prime factor.
        if (n % p == 0)
        {
              
            // If yes, then update n and result
            while (n % p == 0)
                n /= p;
                  
            result *= (1.0 - (1.0 / (float)p));
        }
    }
   
    // If n has a prime factor greater than sqrt(n)
    // (There can be at-most one such prime factor)
    if (n > 1)
        result -= result / n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number
   
    return (int)result;
}
   
// Driver code
int main()
{
    int n;
      
    for(n = 1; n <= 10; n++)
    {
        cout << "Phi" << "(" 
             << n << ")" << " = "
             << phi(n) <<endl;
    }
    return 0;
}
  
// This code is contributed by koulick_sadhu

C

// C program to calculate Euler's Totient Function
// using Euler's product formula
#include <stdio.h>
  
int phi(int n)
{
    float result = n; // Initialize result as n
  
    // Consider all prime factors of n and for every prime
    // factor p, multiply result with (1 - 1/p)
    for (int p = 2; p * p <= n; ++p) {
          
        // Check if p is a prime factor.
        if (n % p == 0) {
              
            // If yes, then update n and result
            while (n % p == 0)
                n /= p;
            result *= (1.0 - (1.0 / (float)p));
        }
    }
  
    // If n has a prime factor greater than sqrt(n)
    // (There can be at-most one such prime factor)
    if (n > 1)
        result -= result / n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number
  
    return (int)result;
}
  
// Driver program to test above function
int main()
{
    int n;
    for (n = 1; n <= 10; n++)
        printf("phi(%d) = %d\n", n, phi(n));
    return 0;
}

Java

// Java program to calculate Euler's Totient
// Function using Euler's product formula
import java.io.*;
  
class GFG {
    static int phi(int n)
    {
        // Initialize result as n
        float result = n;
  
        // Consider all prime factors of n and for
        // every prime factor p, multiply result
        // with (1 - 1/p)
        for (int p = 2; p * p <= n; ++p) {
            // Check if p is a prime factor.
            if (n % p == 0) {
                // If yes, then update n and result
                while (n % p == 0)
                    n /= p;
                result *= (1.0 - (1.0 / (float)p));
            }
        }
  
        // If n has a prime factor greater than sqrt(n)
        // (There can be at-most one such prime factor)
        if (n > 1)
            result -= result / n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number
  
        return (int)result;
    }
  
    // Driver program to test above function
    public static void main(String args[])
    {
        int n;
        for (n = 1; n <= 10; n++)
            System.out.println("phi(" + n + ") = " + phi(n));
    }
}
  
// This code is contributed by Nikita Tiwari.

Python3

# Python 3 program to calculate
# Euler's Totient Function
# using Euler's product formula
  
def phi(n) :
  
    result = n   # Initialize result as n
       
    # Consider all prime factors
    # of n and for every prime
    # factor p, multiply result with (1 - 1 / p)
    p = 2
    while p * p<= n :
  
        # Check if p is a prime factor.
        if n % p == 0 :
  
            # If yes, then update n and result
            while n % p == 0 :
                n = n // p
            result = result * (1.0 - (1.0 / float(p)))
        p = p + 1
          
          
    # If n has a prime factor
    # greater than sqrt(n)
    # (There can be at-most one
    # such prime factor)
    if n > 1 :
        result -= result // n
  #Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  #if n is a prime number
   
    return int(result)
      
      
# Driver program to test above function
for n in range(1, 11) :
    print("phi(", n, ") = ", phi(n))
     
  
# This code is contributed
# by Nikita Tiwari.

C#

// C# program to calculate Euler's Totient
// Function using Euler's product formula
using System;
  
class GFG {
      
    static int phi(int n)
    {
          
        // Initialize result as n
        float result = n;
  
        // Consider all prime factors
        // of n and for every prime 
        // factor p, multiply result
        // with (1 - 1 / p)
        for (int p = 2; p * p <= n; ++p) 
        {
              
            // Check if p is a prime factor.
            if (n % p == 0) 
            {
                  
                // If yes, then update
                // n and result
                while (n % p == 0)
                    n /= p;
                result *= (float)(1.0 - (1.0 / (float)p));
            }
        }
  
        // If n has a prime factor 
        // greater than sqrt(n)
        // (There can be at-most 
        // one such prime factor)
        if (n > 1)
            result -= result / n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number
  
        return (int)result;
    }
  
    // Driver Code
    public static void Main()
    {
        int n;
        for (n = 1; n <= 10; n++)
            Console.WriteLine("phi(" + n + ") = " + phi(n));
    }
}
  
// This code is contributed by nitin mittal.

PHP

<Φphp
// PHP program to calculate 
// Euler's Totient Function 
// using Euler's product formula
function phi($n)
{
    // Initialize result as n
    $result = $n
  
    // Consider all prime factors
    // of n and for every prime
    // factor p, multiply result 
    // with (1 - 1/p)
    for ($p = 2; $p * $p <= $n; ++$p
    {
          
        // Check if p is
        // a prime factor.
        if ($n % $p == 0) 
        {
              
            // If yes, then update
            // n and result
            while ($n % $p == 0)
                $n /= $p;
            $result *= (1.0 - (1.0 / $p));
        }
    }
  
    // If n has a prime factor greater 
    // than sqrt(n) (There can be at-most
    // one such prime factor)
    if ($n > 1)
        $result -= $result / $n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number
  
    return intval($result);
}
  
// Driver Code
for ($n = 1; $n <= 10; $n++)
echo "phi(" .$n. ") =" . phi($n)."\n";
      
// This code is contributed by Sam007
Φ>

Javascript

// Javascript program to calculate 
// Euler's Totient Function 
// using Euler's product formula
function phi(n)
{
    // Initialize result as n
    let result = n; 
  
    // Consider all prime factors
    // of n and for every prime
    // factor p, multiply result 
    // with (1 - 1/p)
    for (let p = 2; p * p <= n; ++p) 
    {
          
        // Check if p is
        // a prime factor.
        if (n % p == 0) 
        {
              
            // If yes, then update
            // n and result
            while (n % p == 0)
                n /= p;
            result *= (1.0 - (1.0 / p));
        }
    }
  
    // If n has a prime factor greater 
    // than sqrt(n) (There can be at-most
    // one such prime factor)
    if (n > 1)
        result -= result / n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number
  
    return parseInt(result);
}
  
// Driver Code
for (let n = 1; n <= 10; n++)
 document.write(`phi(${n}) = ${phi(n)} <br>`);
      
// This code is contributed by _saurabh_jaiswal


Output

Phi(1) = 1 Phi(2) = 1 Phi(3) = 2 Phi(4) = 2 Phi(5) = 4 Phi(6) = 2 Phi(7) = 6 Phi(8) = 4 Phi(9) = 6 Phi(10) = 4

Time Complexity: O(Φ n log n)
Auxiliary Space: O(1)

We can avoid floating-point calculations in the above method. The idea is to count all prime factors and their multiples and subtract this count from n to get the totient function value (Prime factors and multiples of prime factors won’t have gcd as 1) 

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.

Below is the implementation of the above algorithm. 

C++

// C++ program to calculate Euler's
// Totient Function
#include <bits/stdc++.h>
using namespace std;
  
int phi(int n)
{
    // Initialize result as n
    int result = n; 
   
    // Consider all prime factors of n 
    // and subtract their multiples 
    // from result
    for(int p = 2; p * p <= n; ++p)
    {
          
        // Check if p is a prime factor.
        if (n % p == 0) 
        {
              
            // If yes, then update n and result
            while (n % p == 0)
                n /= p;
                  
            result -= result / p;
        }
    }
   
    // If n has a prime factor greater than sqrt(n)
    // (There can be at-most one such prime factor)
    if (n > 1)
        result -= result / n;
          
    return result;
}
   
// Driver code
int main()
{
    int n;
    for(n = 1; n <= 10; n++)
    {
        cout << "Phi" << "(" 
             << n << ")" << " = "
             << phi(n) << endl;
    }
    return 0;
}
  
// This code is contributed by koulick_sadhu

C

// C program to calculate Euler's Totient Function
#include <stdio.h>
  
int phi(int n)
{
    int result = n; // Initialize result as n
  
    // Consider all prime factors of n and subtract their
    // multiples from result
    for (int p = 2; p * p <= n; ++p) {
          
        // Check if p is a prime factor.
        if (n % p == 0) {
              
            // If yes, then update n and result
            while (n % p == 0)
                n /= p;
            result -= result / p;
        }
    }
  
    // If n has a prime factor greater than sqrt(n)
    // (There can be at-most one such prime factor)
    if (n > 1)
        result -= result / n;
    return result;
}
  
// Driver program to test above function
int main()
{
    int n;
    for (n = 1; n <= 10; n++)
        printf("phi(%d) = %d\n", n, phi(n));
    return 0;
}

Java

// Java program to calculate 
// Euler's Totient Function
import java.io.*;
  
class GFG 
{
static int phi(int n)
{
    // Initialize result as n
    int result = n; 
  
    // Consider all prime factors 
    // of n and subtract their
    // multiples from result
    for (int p = 2; p * p <= n; ++p)
    {
          
        // Check if p is 
        // a prime factor.
        if (n % p == 0
        {
              
            // If yes, then update
            // n and result
            while (n % p == 0)
                n /= p;
            result -= result / p;
        }
    }
  
    // If n has a prime factor
    // greater than sqrt(n)
    // (There can be at-most 
    // one such prime factor)
    if (n > 1)
        result -= result / n;
    return result;
}
  
// Driver Code
public static void main (String[] args)
{
    int n;
    for (n = 1; n <= 10; n++)
        System.out.println("phi(" + n + 
                           ") = " + phi(n));
}
}
  
// This code is contributed by ajit

Python3

# Python3 program to calculate 
# Euler's Totient Function
def phi(n):
      
    # Initialize result as n
    result = n; 
  
    # Consider all prime factors
    # of n and subtract their
    # multiples from result
    p = 2
    while(p * p <= n):
          
        # Check if p is a 
        # prime factor.
        if (n % p == 0): 
              
            # If yes, then 
            # update n and result
            while (n % p == 0):
                n = int(n / p);
            result -= int(result / p);
        p += 1;
  
    # If n has a prime factor
    # greater than sqrt(n)
    # (There can be at-most 
    # one such prime factor)
    if (n > 1):
        result -= int(result / n);
    return result;
  
# Driver Code
for n in range(1, 11):
    print("phi(",n,") =", phi(n));
      
# This code is contributed 
# by mits

C#

// C# program to calculate 
// Euler's Totient Function
using System;
  
class GFG
{
      
static int phi(int n)
{
// Initialize result as n
int result = n; 
  
// Consider all prime  
// factors of n and 
// subtract their 
// multiples from result
for (int p = 2;
         p * p <= n; ++p)
{
      
    // Check if p is 
    // a prime factor.
    if (n % p == 0) 
    {
          
        // If yes, then update
        // n and result
        while (n % p == 0)
            n /= p;
        result -= result / p;
    }
}
  
// If n has a prime factor
// greater than sqrt(n)
// (There can be at-most 
// one such prime factor)
if (n > 1)
    result -= result / n;
return result;
}
  
// Driver Code
static public void Main ()
{
    int n;
    for (n = 1; n <= 10; n++)
        Console.WriteLine("phi(" + n + 
                              ") = " +
                              phi(n));
}
}
  
// This code is contributed 
// by akt_mit

PHP

<Φphp
// PHP program to calculate 
// Euler's Totient Function
  
function phi($n)
{
    // Initialize 
    // result as n
    $result = $n
  
    // Consider all prime 
    // factors of n and subtract 
    // their multiples from result
    for ($p = 2; 
         $p * $p <= $n; ++$p)
    {
          
        // Check if p is 
        // a prime factor.
        if ($n % $p == 0) 
        {
              
            // If yes, then 
            // update n and result
            while ($n % $p == 0)
                $n = (int)$n / $p;
            $result -= (int)$result / $p;
        }
    }
  
    // If n has a prime factor
    // greater than sqrt(n)
    // (There can be at-most 
    // one such prime factor)
    if ($n > 1)
        $result -= (int)$result / $n;
    return $result;
}
  
// Driver Code
for ($n = 1; $n <= 10; $n++)
    echo "phi(", $n,") ="
          phi($n), "\n";
      
// This code is contributed 
// by ajit
Φ>

Javascript

// Javascript program to calculate 
// Euler's Totient Function
  
function phi(n)
{
    // Initialize 
    // result as n
    let result = n; 
  
    // Consider all prime 
    // factors of n and subtract 
    // their multiples from result
    for (let p = 2; 
         p * p <= n; ++p)
    {
          
        // Check if p is 
        // a prime factor.
        if (n % p == 0) 
        {
              
            // If yes, then 
            // update n and result
            while (n % p == 0)
                n = parseInt(n / p);
            result -= parseInt(result / p);
        }
    }
  
    // If n has a prime factor
    // greater than sqrt(n)
    // (There can be at-most 
    // one such prime factor)
    if (n > 1)
        result -= parseInt(result / n);
    return result;
}
  
// Driver Code
for (let n = 1; n <= 10; n++)
    document.write(`phi(${n}) = ${phi(n)} <br>`);
      
// This code is contributed 
// by _saurabh_jaiswal


Output

Phi(1) = 1 Phi(2) = 1 Phi(3) = 2 Phi(4) = 2 Phi(5) = 4 Phi(6) = 2 Phi(7) = 6 Phi(8) = 4 Phi(9) = 6 Phi(10) = 4

Time Complexity: O(Φ n log n)
Auxiliary Space: O(1)

Let us take an example to understand the above algorithm. 

n = 10. Initialize: result = 10 2 is a prime factor, so n = n/i = 5, result = 5 3 is not a prime factor. The for loop stops after 3 as 4*4 is not less than or equal to 10. After for loop, result = 5, n = 5 Since n > 1, result = result - result/n = 4

Some Interesting Properties of Euler’s Totient Function 


1) For a prime number p, [Tex]\phi(p) = p – 1[/Tex]

Proof :

[Tex]\phi(p) = p - 1[/Tex] , where p is any prime numberWe know that [Tex]gcd(p, k) = 1[/Tex] where k is any random number and [Tex]k \neq p[/Tex][Tex]\\[/Tex]Total number from 1 to p = p Number for which [Tex]gcd(p, k) = 1[/Tex] is [Tex]1[/Tex], i.e the number p itself, so subtracting 1 from p [Tex]\phi(p) = p - 1[/Tex]

Examples :  

[Tex]\phi(5) = 5 - 1 = 4[/Tex][Tex]\\[/Tex][Tex]\phi(13) = 13 - 1 = 12[/Tex][Tex]\\[/Tex][Tex]\phi(29) = 29 - 1 = 28[/Tex]


2) For two prime numbers a and b[Tex] \phi(a \cdot b) = \phi(a) \cdot \phi(b) = (a – 1) \cdot (b – 1)           [/Tex], used in RSA Algorithm

Proof :

[Tex]\phi(a\cdot b) = \phi(a) \cdot \phi(b)[/Tex], where a and b are prime numbers[Tex]\phi(a) = a - 1[/Tex] , [Tex]\phi(b) = b - 1[/Tex][Tex]\\[/Tex]Total number from 1 to ab = ab Total multiples of a from 1 to ab = [Tex]\frac{a \cdot b} {a}[/Tex] = [Tex]b[/Tex]Total multiples of b from 1 to ab = [Tex]\frac{a \cdot b} {b}[/Tex] = [Tex]a[/Tex]Example:a = 5, b = 7, ab = 35Multiples of a = [Tex]\frac {35} {5}[/Tex] = 7 {5, 10, 15, 20, 25, 30, 35}Multiples of b = [Tex]\frac {35} {7}[/Tex] = 5 {7, 14, 21, 28, 35}[Tex]\\[/Tex]Can there be any double counting ?(watch above example carefully, try with other prime numbers also for more grasp)Ofcourse, we have counted [Tex]ab[/Tex] twice in multiples of a and multiples of b so, Total multiples = a + b - 1 (with which [Tex]gcd \neq 1[/Tex] with [Tex]ab[/Tex])[Tex]\\[/Tex][Tex]\phi(ab) = ab - (a + b - 1)[/Tex] , removing all number with [Tex]gcd \neq 1[/Tex] with [Tex]ab[/Tex] [Tex]\phi(ab) = a(b - 1) - (b - 1)[/Tex][Tex]\phi(ab) = (a - 1) \cdot (b - 1)[/Tex][Tex]\phi(ab) = \phi(a) \cdot \phi(b)[/Tex]

Examples :

[Tex]\phi(5 \cdot 7) = \phi(5) \cdot \phi(7) = (5 - 1) \cdot (7 - 1) = 24[/Tex][Tex]\\[/Tex][Tex]\phi(3 \cdot 5) = \phi(3) \cdot \phi(5) = (3 - 1) \cdot (5 - 1) = 8[/Tex][Tex]\\[/Tex][Tex]\phi(3 \cdot 7) = \phi(3) \cdot \phi(7) = (3 - 1) \cdot (7 - 1) = 12[/Tex]


3) For a prime number p, [Tex]\phi(p ^ k) = p ^ k – p ^ {k – 1}[/Tex]

Proof : 

[Tex]\phi(p^k) = p ^ k - p ^{k - 1}[/Tex] , where p is a prime number[Tex]\\[/Tex]Total numbers from 1 to [Tex]p ^ k = p ^ k[/Tex] Total multiples of [Tex]p = \frac {p ^ k} {p} = p ^ {k - 1}[/Tex]Removing these multiples as with them [Tex]gcd \neq 1[/Tex][Tex]\\[/Tex]Example : p = 2, k = 5, [Tex]p ^ k[/Tex] = 32Multiples of 2 (as with them [Tex]gcd \neq 1[/Tex]) = 32 / 2 = 16 {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32}[Tex]\\[/Tex][Tex]\phi(p ^ k) = p ^ k - p ^ {k - 1}[/Tex]

Examples : 

[Tex]\phi(2 ^ 5) = 2 ^ 5 - 2 ^ {5 - 1} = 32 - 16 = 16[/Tex][Tex]\\[/Tex][Tex]\phi(5 ^ 3) = 5 ^ 3 - 5 ^ {3 - 1} = 125 - 25 = 100[/Tex][Tex]\\[/Tex][Tex]\phi(3 ^ 5) = 3 ^ 5 - 3 ^ {5 - 1} = 243 - 81 = 162[/Tex]


4) For two number a and b [Tex]\phi(a \cdot b)           [/Tex] [Tex]= \phi(a) \cdot \phi(b)           [/Tex] [Tex]\cdot \frac {gcd(a, b)} {\phi(gcd(a, b))}[/Tex]

Special Case : gcd(a, b) = 1

[Tex]\phi(a \cdot b) = \phi(a) \cdot \phi(b) \cdot \frac {1} {\phi(1)} = \phi(a) \cdot \phi(b)[/Tex]

Examples :

Special Case : [Tex]gcd(a, b) = 1[/Tex], [Tex]\phi(a \cdot b) = \phi(a) \cdot \phi(b)[/Tex] [Tex]\phi(2 \cdot 9) = \phi(2) \cdot \phi(9) = 1 \cdot 6 = 6[/Tex][Tex]\\[/Tex][Tex]\phi(8 \cdot 9) = \phi(8) \cdot \phi(9) = 4 \cdot 6 = 24[/Tex][Tex]\\[/Tex][Tex]\phi(5 \cdot 6) = \phi(5) \cdot \phi(6) = 4 \cdot 2 = 8[/Tex] [Tex]\\[/Tex][Tex]\\[/Tex]Normal Case : [Tex]gcd(a, b) \neq 1[/Tex], [Tex]\phi(a \cdot b) = \phi(a) \cdot \phi(b) \cdot \frac {gcd(a, b)} {\phi(gcd(a, b))}[/Tex][Tex]\\[/Tex][Tex]\phi(4 \cdot 6) = \phi(4) \cdot \phi(6) \cdot \frac {gcd(4, 6)} {\phi(gcd(4, 6))}[/Tex] [Tex]= 2 \cdot 2 \cdot \frac{2}{1}[/Tex] [Tex]= 2 \cdot 2 \cdot 2 = 8[/Tex][Tex]\\[/Tex][Tex]\phi(4 \cdot 8) = \phi(4) \cdot \phi(8) \cdot \frac {gcd(4, 8)} {\phi(gcd(4, 8))} = 2 \cdot 4 \cdot \frac{4}{2} = 2 \cdot 4 \cdot 2 = 16[/Tex][Tex]\\[/Tex][Tex]\phi(6 \cdot 8) = \phi(6) \cdot \phi(8) \cdot \frac {gcd(6, 8)} {\phi(gcd(6, 8))} = 2 \cdot 4 \cdot \frac{2}{1} = 2 \cdot 4 \cdot 2 = 16[/Tex]

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

gausss


Examples :

n = 6 factors = {1, 2, 3, 6} n = [Tex]\phi(1) + \phi(2) + \phi(3) + \phi(6)[/Tex] = 1 + 1 + 2 + 2 = 6[Tex]\\[/Tex]n = 8factors = {1, 2, 4, 8}n = [Tex]\phi(1) + \phi(2) + \phi(4) + \phi(8)[/Tex] = 1 + 1 + 2 + 4 = 8[Tex]\\[/Tex]n = 10factors = {1, 2, 5, 10}n = [Tex]\phi(1) + \phi(2) + \phi(5) + \phi(10)[/Tex] = 1 + 1 + 4 + 4 = 10

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 aΦ(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

ap-1 Φ 1 (mod p)

7) Number of generators of a finite cyclic group under modulo n addition is Φ(n).

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

References: 
http://e-maxx.ru/algo/euler_function
http://en.wikipedia.org/wiki/Euler%27s_totient_function

https://cp-algorithms.com/algebra/phi-function.html

http://mathcenter.oxford.memory.edu/site/math125/chineseRemainderTheorem/

 



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