# 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.

Examples:

```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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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
#include<bits/stdc++.h>
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)
{
s.insert(2);
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)
{
s.insert(i);
n = n/i;
}
}

// This condition is to handle the case when
// n is a prime number greater than 2
if (n > 2)
s.insert(n);
}

// 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;
break;
}
}

// 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;
}
```

Output:

```Smallest primitive root of 761 is 6
```

This article is contributed by Niteesh kumar and Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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