Find the number of primitive roots modulo prime Last Updated : 11 Jul, 2025 Summarize Comments Improve Suggest changes Share Like Article Like Report 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 <?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 Time Complexity: O(p * log(min(a, b))), where a and b are two parameters of gcd. Auxiliary Space: O(log(min(a, b))) Comment More infoAdvertise with us Next Article Compute power of power k times % m P pawan_asipu Follow Improve Article Tags : DSA Modular Arithmetic Prime Number number-theory euler-totient +1 More Practice Tags : Modular Arithmeticnumber-theoryPrime Number Similar Reads Euler Totient for Competitive Programming What is Euler Totient function(ETF)?Euler Totient Function or Phi-function for 'n', gives the count of integers in range '1' to 'n' that are co-prime to 'n'. It is denoted by \phi(n) .For example the below table shows the ETF value of first 15 positive integers: 3 Important Properties of Euler Totie 8 min read 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 re 10 min read Count of non co-prime pairs from the range [1, arr[i]] for every array element Given an array arr[] consisting of N integers, the task for every ith element of the array is to find the number of non co-prime pairs from the range [1, arr[i]]. Examples: Input: N = 2, arr[] = {3, 4}Output: 2 4Explanation: All non-co-prime pairs from the range [1, 3] are (2, 2) and (3, 3).All non- 13 min read Generate an array having sum of Euler Totient Function of all elements equal to N Given a positive integer N, the task is to generate an array such that the sum of the Euler Totient Function of each element is equal to N. Examples: Input: N = 6Output: 1 6 2 3 Input: N = 12Output: 1 12 2 6 3 4 Approach: The given problem can be solved based on the divisor sum property of the Euler 5 min read Count all possible values of K less than Y such that GCD(X, Y) = GCD(X+K, Y) Given two integers X and Y, the task is to find the number of integers, K, such that gcd(X, Y) is equal to gcd(X+K, Y), where 0 < K <Y. Examples: Input: X = 3, Y = 15Output: 4Explanation: All possible values of K are {0, 3, 6, 9} for which GCD(X, Y) = GCD(X + K, Y). Input: X = 2, Y = 12Output: 8 min read Count of integers up to N which are non divisors and non coprime with N Given an integer N, the task is to find the count of all possible integers less than N satisfying the following properties: The number is not coprime with N i.e their GCD is greater than 1.The number is not a divisor of N. Examples: Input: N = 10 Output: 3 Explanation: All possible integers which ar 5 min read 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 modul 5 min read Compute power of power k times % m Given x, k and m. Compute (xxxx...k)%m, x is in power k times. Given x is always prime and m is greater than x. Examples: Input : 2 3 3 Output : 1 Explanation : ((2 ^ 2) ^ 2) % 3 = (4 ^ 2) % 3 = 1 Input : 3 2 3 Output : 0 Explanation : (3^3)%3 = 0 A naive approach is to compute the power of x k time 15+ min read Primitive root of a prime number n modulo n Given a prime number n, the task is to find its primitive root under modulo n. The primitive root of a prime number n is an integer r between[1, n-1] such that the values of r^x(mod n) where x is in the range[0, n-2] are different. Return -1 if n is a non-prime number. Examples: Input : 7 Output : S 15 min read Euler's Totient function for all numbers smaller than or equal to n Euler's Totient function ?(n) for an input n is the count of numbers in {1, 2, 3, ..., n} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1. For example, ?(4) = 2, ?(3) = 2 and ?(5) = 4. There are 2 numbers smaller or equal to 4 that are relatively pri 13 min read Like