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 = 2Input: 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:
// 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;
} |
// 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 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 |
// 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 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. ?> |
<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 :
// 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;
} |
// 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);
}
} |
# 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) |
// 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 // 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 ?> |
<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.