Find the number of primitive roots modulo prime (original) (raw)

Last Updated : 11 Jul, 2025

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++ `

// 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

JavaScript

`

Time Complexity: O(p * log(min(a, b))), where a and b are two parameters of gcd.

Auxiliary Space: O(log(min(a, b)))