Given a prime . The task is to count all the primitive roots of .
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 but xp – 1 – 1 is divisible by .
Input: P = 3
The only primitive root modulo 3 is 2.
Input: P = 5
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:
# 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
result = 1
for i in range(2, p, 1):
if (gcd(i, p) == 1):
result += 1
# Driver code
if __name__ == ‘__main__’:
p = 5
print(countPrimitiveRoots(p – 1))
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