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

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; ` `} ` |

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Output:

Smallest primitive root of 761 is 6

**Reference:**

https://en.wikipedia.org/wiki/Primitive_root_modulo_n#Finding_primitive_roots

http://math.stackexchange.com/questions/124408/finding-a-primitive-root-of-a-prime-number

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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