Find the number of primitive roots modulo prime

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: 1
The only primitive root modulo 3 is 2.

Input: P = 5
Output: 2
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++

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// 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;
}

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Java

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// 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..

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

2


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