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Find Square Root under Modulo p (When p is in form of 4*i + 3) in C++
In this problem, we are given two values n and a prime number p. Our task is to find Square Root under Modulo p (When p is in form of 4*i + 3). Here, p is of the form (4*i + 3) i.e. p % 4 = 3 for i > 1 and p being a prime number.
Here are some numbers, 7, 11, 19, 23, 31...
Let's take an example to understand the problem,
Input : n = 3, p = 7 Output :
Solution Approach
A simple solution to the problem is using a loop. We will loop from 2 to (p - 1). And for every value, check if its square is square root under modulo p is n.
Example
Program to illustrate the working of our solution
#include <iostream> using namespace std; void findSquareRootMod(int n, int p) { n = n % p; for (int i = 2; i < p; i++) { if ((i * i) % p == n) { cout<<"Square root under modulo is "<<i; return; } } cout<<"Square root doesn't exist"; } int main(){ int p = 11; int n = 3; findSquareRootMod(n, p); return 0; }
Output
Square root under modulo is 5
One more method is directly using the formula,
If p is of the form (4*i + 3) and for square root to exist it will be $+/-n^{(p+1)/4}$
Example
Program to illustrate the working of our solution
#include <iostream> using namespace std; int calcPowerVal(int x, int y, int p) { int res = 1; x = x % p; while (y > 0) { if (y & 1) res = (res * x) % p; y /= 2; x = (x * x) % p; } return res; } void squareRoot(int n, int p) { if (p % 4 != 3) { cout << "Invalid Input"; return; } n = n % p; int sr = calcPowerVal(n, (p + 1) / 4, p); if ((sr * sr) % p == n) { cout<<"Square root under modulo is "<<sr; return; } sr = p - sr; if ((sr * sr) % p == n) { cout << "Square root is "<<sr; return; } cout<<"Square root doesn't exist "; } int main() { int p = 11; int n = 4; squareRoot(n, p); return 0; }
Output
Square root under modulo is 9
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