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

Last Updated : 23 Jul, 2025

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 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

n=n = n=n % $p; // One by one check all // numbers from 2 to p-1 for ($x = 2; x<x < x<p; $x++) { if (($x * x)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

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 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 {

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

y,y, y,p) { // Initialize result $res = 1; // Update x if it // is more than or // equal to p x=x = x=x % $p; while ($y > 0) { // If y is odd, multiply // x with result if ($y & 1) res=(res = (res=(res * x)x) % x)p; // y must be even now // y = y/2 y=y = y=y >> 1; x=(x = (x=(x * 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 = n=n % $p; x=power(x = power(x=power(n, ($p + 1) / 4, $p); if (($x * x)x) % x)p == $n) { echo "Square root is ", $x; return; } // Try "-(n ^ ((p + 1)/4))" x=x = x=p - $x; if (($x * x)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

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 = 4i + 3] x ? ±n(i + 1) [Taking Square root of both sides] x ? ±n(p + 1)/4 [Putting 4i + 3 = p or i = (p-3)/4]

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