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. 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 range[0, n-2] are different. Return -1 if n is a non-prime number.


Input : 7
Output : Smallest primitive root = 3
Explanation: n = 7
3^0(mod 7) = 1
3^1(mod 7) = 3
3^2(mod 7) = 2
3^3(mod 7) = 6
3^4(mod 7) = 4
3^5(mod 7) = 5

Input : 761
Output : Smallest primitive root = 6

A simple solution is to try all numbers from 2 to n-1. For every number r, compute values of r^x(mod n) where x is in range[0, n-2]. If all these values are different, then return r, else continue for next value of r. If all values of r are tried, return -1.

An efficient solution is based on below fact.
If the multiplicative order of a number r modulo n is equal to Euler Totient Function Φ(n) (Note that Euler Totient Function for a prime n is n-1), then it is a primitive root [Source : Wiki]

1- Euler Totient Function phi = n-1 [Assuming n is prime]
1- Find all prime factors of phi.
2- Calculate all powers to be calculated further 
   using (phi/prime-factors) one by one.
3- Check for all numbered for all powers from i=2 
   to n-1 i.e. (i^ powers) modulo n.
4- If it is 1 then 'i' is not a primitive root of n.
5- If it is never 1 then return i;.

Although there can be multiple primitive root for a prime number but we are only concerned for smallest one.If you want to find all roots then continue the process till p-1 instead of breaking up on finding first primitive root.





// C++ program to find primitive root of a
// given number n
using namespace std;
// Returns true if n is prime
bool isPrime(int n)
    // Corner cases
    if (n <= 1)  return false;
    if (n <= 3)  return true;
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n%2 == 0 || n%3 == 0) return false;
    for (int i=5; i*i<=n; i=i+6)
        if (n%i == 0 || n%(i+2) == 0)
            return false;
    return true;
/* Iterative Function to calculate (x^n)%p in
   O(logy) */
int power(int x, unsigned int y, int p)
    int res = 1;     // Initialize result
    x = x % p; // Update x if it is more than or
    // equal to p
    while (y > 0)
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % p;
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x*x) % p;
    return res;
// Utility function to store prime factors of a number
void findPrimefactors(unordered_set<int> &s, int n)
    // Print the number of 2s that divide n
    while (n%2 == 0)
        n = n/2;
    // n must be odd at this point. So we can skip
    // one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i+2)
        // While i divides n, print i and divide n
        while (n%i == 0)
            n = n/i;
    // This condition is to handle the case when
    // n is a prime number greater than 2
    if (n > 2)
// Function to find smallest primitive root of n
int findPrimitive(int n)
    unordered_set<int> s;
    // Check if n is prime or not
    if (isPrime(n)==false)
        return -1;
    // Find value of Euler Totient function of n
    // Since n is a prime number, the value of Euler
    // Totient function is n-1 as there are n-1
    // relatively prime numbers.
    int phi = n-1;
    // Find prime factors of phi and store in a set
    findPrimefactors(s, phi);
    // Check for every number from 2 to phi
    for (int r=2; r<=phi; r++)
        // Iterate through all prime factors of phi.
        // and check if we found a power with value 1
        bool flag = false;
        for (auto it = s.begin(); it != s.end(); it++)
            // Check if r^((phi)/primefactors) mod n
            // is 1 or not
            if (power(r, phi/(*it), n) == 1)
                flag = true;
         // If there was no power with value 1.
         if (flag == false)
           return r;
    // If no primitive root found
    return -1;
// Driver code
int main()
    int n = 761;
    cout << " Smallest primitive root of " << n
         << " is " << findPrimitive(n);
    return 0;



Smallest primitive root of 761 is 6


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