# 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++

 `// C++ program to find primitive root of a ` `// given number n ` `#include ` `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; ` `} `

/div>

## Python3

# Python3 program to find primitive root
# of a given number n
from math import sqrt

# Returns True if n is prime
def isPrime( 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 or n % 3 == 0): return False i = 5 while(i * i <= n): if (n % i == 0 or n % (i + 2) == 0) : return False i = i + 6 return True """ Iterative Function to calculate (x^n)%p in O(logy) */""" def power( x, y, p): 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
def findPrimefactors(s, n) :

# Pr the number of 2s that divide n
while (n % 2 == 0) :
n = n // 2

# n must be odd at this po. So we can
# skip one element (Note i = i +2)
for i in range(3, int(sqrt(n)), 2):

# While i divides n, pr 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
def findPrimitive( n) :
s = set()

# 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.
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 r in range(2, phi + 1):

# Iterate through all prime factors of phi.
# and check if we found a power with value 1
flag = False
for it in s:

# 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
n = 761
print(“Smallest primitive root of”,
n, “is”, findPrimitive(n))

# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

Output:

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