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

C++

// 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;
}

Java

// A Simple Java program to find square
// root under modulo p when p is 7, 
// 11, 19, 23, 31, ... etc,
import java .io.*;
  
class GFG {
  
    // Returns true if square root of n
    // under modulo p exists
    static 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) {
                System.out.println("Square "
                    + "root is " + x);
                return;
            }
        }
        System.out.println("Square root "
                + "doesn't exist");
    }
      
    // Driver Code
    public static void main(String[] args) 
    {
        int p = 7;
        int n = 2;
        squareRoot(n, p);
    
}
  
// This code is contributed by Anuj_67

Python3

# A Simple Python program to find square 
# root under modulo p when p is 7, 11, 
# 19, 23, 31, ... etc,
  
# Returns true if square root of n under
# modulo p exists
def squareRoot(n, p):
  
    n = n % p
      
    # One by one check all numbers from 
    # 2 to p-1
    for x in range (2, p):
        if ((x * x) % p == n) :
            print( "Square root is ", x)
            return
  
    print( "Square root doesn't exist")
  
# Driver program to test
p = 7
n = 2
squareRoot(n, p)
  
# This code is Contributed by Anuj_67

C#

// A Simple C# program to find square
// root under modulo p when p is 7, 
// 11, 19, 23, 31, ... etc,
using System;
  
class GFG {
  
    // Returns true if square root of n
    // under modulo p exists
    static 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) {
                Console.Write("Square "
                     + "root is " + x);
                return;
            }
        }
        Console.Write("Square root "
                   + "doesn't exist");
    }
      
    // Driver Code
    static void Main() 
    {
        int p = 7;
        int n = 2;
        squareRoot(n, p);
    
}
  
// This code is contributed by Anuj_67

PHP

<?php
// A Simple PHP program to find 
// square root under modulo p
// when p is 7, 11, 19, 23, 31, 
// ... etc,
  
// Returns true if square
// root of n under modulo
// p exists
function squareRoot($n, $p)
{
    $n = $n % $p;
  
    // One by one check all
    // numbers from 2 to p-1
    for ($x = 2; $x < $p; $x++)
    {
        if (($x * $x) % $p == $n)
        {
            echo("Square root is " . $x);
            return;
        }
    }
    echo("Square root doesn't exist");
}
  
// Driver Code
$p = 7;
$n = 2;
squareRoot($n, $p);
  
// This code is contributed by Ajit.
?>


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 the implementation of above idea :

C++

// 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;
}

PHP

<?php
// An efficient PHP program 
// to find square root under 
// modulo p when p is 7, 11, 
// 19, 23, 31, ... etc.
  
// Utility function to do 
// modular exponentiation.
// It returns (x^y) % p.
function power($x, $y, $p)
{
      
    // Initialize result
    $res = 1;     
      
    // Update x if it 
    // is more than or
    // equal to p
    $x = $x % $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/2
        $y = $y >> 1; 
        $x = ($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
function squareRoot($n, $p)
{
    if ($p % 4 != 3)
    {
        echo "Invalid Input";
        return
    }
  
    // Try "+(n^((p + 1)/4))"
    $n = $n % $p;
    $x = power($n, ($p + 1) / 4, $p);
    if (($x * $x) % $p == $n)
    {
        echo "Square root is ", $x;
        return;
    }
  
    // Try "-(n ^ ((p + 1)/4))"
    $x = $p - $x;
    if (($x * $x) % $p == $n)
    {
        echo "Square root is ", $x;
        return;
    }
  
    // If none of the above 
    // two work, then square
    // root doesn't exist
    echo "Square root doesn't exist ";
}
  
    // Driver Code
    $p = 7;
    $n = 2;
    squareRoot($n, $p);
  
// This code is contributed by ajit
?>


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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Improved By : jit_t, vt_m