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

12 Apr, 2021
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++
Java
Python3
C#
PHP
Javascript
// 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 existsvoid 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 testint main(){    int p = 7;    int n = 2;    squareRoot(n, p);    return 0;}
// 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
# 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 existsdef 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 testp = 7n = 2squareRoot(n, p) # This code is Contributed by Anuj_67
// 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// 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 existsfunction 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.?>
<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>
Output: 
Square root is 3
Time Complexity of this solution is O(p)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 above idea :  
C++
Java
Python3
C#
PHP
Javascript
// 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 >= 1void 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 testint main(){    int p = 7;    int n = 2;    squareRoot(n, p);    return 0;}
// 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 >= 1static 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);    }}
# 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 >= 1def 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 Codep = 7n = 2squareRoot(n, p) # This code is contributed by# Shubham Singh(SHUBHAMSINGH10)
// 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 >= 1static 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// 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 $res;} // Returns true if square root// of n under modulo p exists// Assumption: p is of the// form 3*i + 4 where i >= 1function squareRoot($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?>
<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>
Output:
Square root is 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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Modular Arithmetic
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