Find Square Root under Modulo p | Set 1 (When p is in form of 4*i + 3)

3

Given a number ‘n’ and a prime ‘p’, find square root of n under modulo p if it exists. It may be given that p is in the form for 4*i + 3 (OR p % 4 = 3) where i is an integer. Examples of such primes are 7, 11, 19, 23, 31, … etc,

Examples:

Input:  n = 2, p = 7
Output: 3 or 4
3 and 4 both are square roots of 2 under modulo
7 because (3*3) % 7 = 2 and (4*4) % 7 = 2

Input:  n = 2, p = 5
Output: Square root doesn't exist

Naive Solution : Try all numbers from 2 to p-1. And for every number x, check if x is square root of n under modulo p.

// A Simple C++ program to find square root under modulo p
// when p is 7, 11, 19, 23, 31, ... etc,
#include<iostream>
using namespace std;

// Returns true if square root of n under modulo p exists
void squareRoot(int n, int p)
{
    n = n%p;

    // One by one check all numbers from 2 to p-1
    for (int x=2; x<p; x++)
    {
        if ((x*x)%p == n)
        {
           cout << "Square root is " << x;
           return;
        }
    }
    cout << "Square root doesn't exist";
}

// Driver program to test
int main()
{
   int p = 7;
   int n = 2;
   squareRoot(n, p);
   return 0;
}

Output:

Square root is 3

Time Complexity of this solution is O(p)

Direct Method : If p is in the form of 3*i + 4, then there exist a Quick way of finding square root.

If n is in the form 4*i + 3 with i >= 1 (OR p % 4 = 3)
And 
If Square root of n exists, then it must be
        ±n(p + 1)/4

Below is C++ implementation.

// An efficient C++ program to find square root under 
// modulo p when p is 7, 11, 19, 23, 31, ... etc.
#include<iostream>
using namespace std;

// Utility function to do modular exponentiation.
// It returns (x^y) % p.
int power(int x, 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;
}

// Returns true if square root of n under modulo p exists
// Assumption: p is of the form 3*i + 4 where i >= 1
void squareRoot(int n, int p)
{
    if (p % 4 != 3)
    {
       cout << "Invalid Input";
       return;  
    }

    // Try "+(n^((p + 1)/4))"
    n = n % p;
    int x = power(n, (p+1)/4, p);
    if ((x*x)%p == n)
    {
        cout << "Square root is " << x;
        return;
    }

    // Try "-(n ^ ((p + 1)/4))"
    x = p - x;
    if ((x*x)%p == n)
    {
        cout << "Square root is " << x;
        return;
    }

    // If none of the above two work, then
    // square root doesn't exist
    cout << "Square root doesn't exist ";
}

// Driver program to test
int main()
{
   int p = 7;
   int n = 2;
   squareRoot(n, p);
   return 0;
}

Output:

4

Time Complexity of this solution is O(Log p)

How does this work?
We have discussed Euler’s Criterion in the previous post.

As per Euler's criterion, if square root exists, then 
following condition is true
 n(p-1)/2 % p = 1

Multiplying both sides with n, we get
 n(p+1)/2 % p = n % p  ------ (1)

Let x be the modulo square root. We can write,
  (x * x) ≡ n mod p
  (x * x) ≡ n(p+1)/2  [Using (1) given above]
  (x * x) ≡ n(2i + 2) [Replacing n = 4*i + 3]
        x ≡ ±n(i + 1)  [Taking Square root of both sides]
        x ≡ ±n(p + 1)/4 [Putting 4*i + 3 = p or i = (p-3)/4]

We will soon be discussing methods when p is not in above form.

This article is contributed by Shivam Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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