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Find the number of primitive roots modulo prime

  • Last Updated : 10 May, 2021

Given a prime p . The task is to count all the primitive roots of p .
A primitive root is an integer x (1 <= x < p) such that none of the integers x – 1, x2 – 1, …., xp – 2 – 1 are divisible by p but xp – 1 – 1 is divisible by p
Examples: 
 

Input: P = 3 
Output:
The only primitive root modulo 3 is 2. 
Input: P = 5 
Output:
Primitive roots modulo 5 are 2 and 3. 
 

 

Approach: There is always at least one primitive root for all primes. So, using Eulers totient function we can say that f(p-1) is the required answer where f(n) is euler totient function.
Below is the implementation of the above approach: 
 

C++




// CPP program to find the number of
// primitive roots modulo prime
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of
// primitive roots modulo p
int countPrimitiveRoots(int p)
{
    int result = 1;
    for (int i = 2; i < p; i++)
        if (__gcd(i, p) == 1)
            result++;
 
    return result;
}
 
// Driver code
int main()
{
    int p = 5;
 
    cout << countPrimitiveRoots(p - 1);
 
    return 0;
}

Java




// Java program to find the number of
// primitive roots modulo prime
 
import java.io.*;
 
class GFG {
 // Recursive function to return gcd of a and b
    static int __gcd(int a, int b)
    {
        // Everything divides 0 
        if (a == 0)
          return b;
        if (b == 0)
          return a;
        
        // base case
        if (a == b)
            return a;
        
        // a is greater
        if (a > b)
            return __gcd(a-b, b);
        return __gcd(a, b-a);
    }
 
// Function to return the count of
// primitive roots modulo p
static int countPrimitiveRoots(int p)
{
    int result = 1;
    for (int i = 2; i < p; i++)
        if (__gcd(i, p) == 1)
            result++;
 
    return result;
}
 
// Driver code
    public static void main (String[] args) {
            int p = 5;
 
    System.out.println( countPrimitiveRoots(p - 1));
    }
}
// This code is contributed by anuj_67..

Python3




# Python 3 program to find the number
# of primitive roots modulo prime
from math import gcd
 
# Function to return the count of
# primitive roots modulo p
def countPrimitiveRoots(p):
    result = 1
    for i in range(2, p, 1):
        if (gcd(i, p) == 1):
            result += 1
 
    return result
 
# Driver code
if __name__ == '__main__':
    p = 5
 
    print(countPrimitiveRoots(p - 1))
 
# This code is contributed by
# Surendra_Gangwar

C#




// C# program to find the number of
// primitive roots modulo prime
   
using System;
   
class GFG {
 // Recursive function to return gcd of a and b 
    static int __gcd(int a, int b) 
    
        // Everything divides 0  
        if (a == 0) 
          return b; 
        if (b == 0) 
          return a; 
          
        // base case 
        if (a == b) 
            return a; 
          
        // a is greater 
        if (a > b) 
            return __gcd(a-b, b); 
        return __gcd(a, b-a); 
    
   
// Function to return the count of
// primitive roots modulo p
static int countPrimitiveRoots(int p)
{
    int result = 1;
    for (int i = 2; i < p; i++)
        if (__gcd(i, p) == 1)
            result++;
   
    return result;
}
   
// Driver code
     static public void Main (String []args) {
            int p = 5;
   
    Console.WriteLine( countPrimitiveRoots(p - 1));
    }
}
// This code is contributed by Arnab Kundu

PHP




<?php
// PHP program to find the number of
// primitive roots modulo prime
 
// Recursive function to return
// gcd of a and b
function __gcd($a, $b)
{
    // Everything divides 0
    if ($a == 0)
    return b;
     
    if ($b == 0)
    return $a;
     
    // base case
    if ($a == $b)
        return $a;
     
    // a is greater
    if ($a > $b)
        return __gcd($a - $b, $b);
    return __gcd($a, $b - $a);
}
 
// Function to return the count of
// primitive roots modulo p
function countPrimitiveRoots($p)
{
    $result = 1;
    for ($i = 2; $i < $p; $i++)
        if (__gcd($i, $p) == 1)
            $result++;
 
    return $result;
}
 
// Driver code
$p = 5;
 
echo countPrimitiveRoots($p - 1);
 
// This code is contributed by anuj_67
?>

Javascript




<script>
 
// Javascript program to find the number of
// primitive roots modulo prime
 
 // Recursive function to return gcd of a and b 
 function __gcd( a,  b) 
 
     // Everything divides 0  
     if (a == 0) 
       return b; 
     if (b == 0) 
       return a; 
       
     // base case 
     if (a == b) 
         return a; 
       
     // a is greater 
     if (a > b) 
         return __gcd(a-b, b); 
     return __gcd(a, b-a); 
 
   
 
// Function to return the count of
// primitive roots modulo p
function countPrimitiveRoots(p)
{
    var result = 1;
    for (var i = 2; i < p; i++)
        if (__gcd(i, p) == 1)
            result++;
 
    return result;
}
 
// Driver code
var p = 5;
document.write( countPrimitiveRoots(p - 1));
 
</script>
Output: 
2

 




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