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

Last Updated : 03 Oct, 2022
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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
Explanation: 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 the square root of n under modulo p. 

 Below is the implementation of the above approach:

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.
?>
JavaScript 
<script>
// A Simple Javascript 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 (let x = 2; x < p; x++) {
            if ((x * x) % p == n) {
                document.write("Square "
                    + "root is " + x);
                return;
            }
        }
        document.write("Square root "
                + "doesn't exist");
    }
    
    // Driver Code
    let p = 7;
    let n = 2;
    squareRoot(n, p);
    
    // This code is contributed by rag2127
    
</script>

Time Complexity: O(p)
Auxiliary Space: O(1)


Direct Method: If p is in the form of 4*i + 3, 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 the 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;
}
Java 
// An efficient Java program to find square root under 
// modulo p when p is 7, 11, 19, 23, 31, ... etc. 
public class GFG {


// Utility function to do modular exponentiation. 
// It returns (x^y) % p. 
static 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 %2== 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 
static void squareRoot(int n, int p) 
{ 
    if (p % 4 != 3) { 
        System.out.print("Invalid Input"); 
        return; 
    } 

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

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

    // If none of the above two work, then 
    // square root doesn't exist 
    System.out.print("Square root doesn't exist "); 
} 

// Driver program to test 
   static public void main(String[] args) {
       int p = 7; 
    int n = 2; 
    squareRoot(n, p); 
    }
}
Python3 
# An efficient python3 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.
def power(x, y, p) :

    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 
def squareRoot(n, p): 

    if (p % 4 != 3) : 
        print( "Invalid Input" )
        return


    # Try "+(n^((p + 1)/4))" 
    n = n % p 
    x = power(n, (p + 1) // 4, p) 
    if ((x * x) % p == n): 
        print( "Square root is ", x) 
        return

    # Try "-(n ^ ((p + 1)/4))" 
    x = p - x 
    if ((x * x) % p == n): 
        print( "Square root is ", x )
        return

    # If none of the above two work, then 
    # square root doesn't exist 
    print( "Square root doesn't exist " )

# Driver Code
p = 7
n = 2
squareRoot(n, p) 

# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)
C# 
// An efficient C# program to find square root under 
// modulo p when p is 7, 11, 19, 23, 31, ... etc. 

using System;
public class GFG {
 
// Utility function to do modular exponentiation. 
// It returns (x^y) % p. 
static 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 %2 == 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 
static void squareRoot(int n, int p) 
{ 
    if (p % 4 != 3) { 
        Console.Write("Invalid Input"); 
        return; 
    } 
 
    // Try "+(n^((p + 1)/4))" 
    n = n % p; 
    int x = power(n, (p + 1) / 4, p); 
    if ((x * x) % p == n) { 
        Console.Write("Square root is " + x); 
        return; 
    } 
 
    // Try "-(n ^ ((p + 1)/4))" 
    x = p - x; 
    if ((x * x) % p == n) { 
        Console.Write("Square root is " + x); 
        return; 
    } 
 
    // If none of the above two work, then 
    // square root doesn't exist 
    Console.Write("Square root doesn't exist "); 
} 
 
// Driver program to test 
   static public void Main() {
       int p = 7; 
    int n = 2; 
    squareRoot(n, p); 
    }
}
// This code is contributed by Ita_c.
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
?>
JavaScript 
<script>
// An efficient Javascript 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)
    {
        let 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 %2== 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
    function squareRoot(n, p)
    {
        if (p % 4 != 3)
        {
            document.write("Invalid Input");
            return;
        }
     
        // Try "+(n^((p + 1)/4))"
        n = n % p;
        let x = power(n, Math.floor((p + 1) / 4), p);
        if ((x * x) % p == n) {
            document.write("Square root is " + x);
            return;
        }
     
        // Try "-(n ^ ((p + 1)/4))"
        x = p - x;
        if ((x * x) % p == n) {
            document.write("Square root is " + x);
            return;
        }
     
        // If none of the above two work, then
        // square root doesn't exist
        document.write("Square root doesn't exist ");
    }
    
    // Driver program to test
    let p = 7;
    let n = 2;
    squareRoot(n, p);
    
    // This code is contributed by avanitrachhadiya2155
</script>

Time Complexity: O(Log p)
Auxiliary Space: O(1)


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.


Next Article
Find Square Root under Modulo p | Set 2 (Shanks Tonelli algorithm)

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