Check if a number is Quartan Prime or not
Last Updated :
27 Aug, 2022
Given a positive integer N, check if it is Quartan prime or not. Print ‘Yes’ if it is a Quartan prime otherwise Print ‘No’.
Quartan Prime : A prime number of the form x4 + y4 where x > 0, y > 0, and x and y are integers is a Quartan Prime.
Quartan Prime in the range 1 – 100 are:
2, 17, 97
Examples:
Input : 17
Output : Yes
Explanation : 17 is a prime number and can be
expressed in the form of:
x4 + y4 as ( 14 + 24 )
Input : 31
Output : No
Explanation: 31 is prime number but can not be
expressed in the form of x4 + y4.
A Simple Solution is to check if the given number is prime or not and then check if it can be expressed in the form of x4 + y4 or not.
An Efficient Solution is based on the fact that every Quartan Prime can also be expressed in the form 16*n + 1. So, we can check if a number is prime or not and can be expressed in the form of 16*n + 1 or not. If yes, Then the number is Quartan Prime otherwise not.
Below is the implementation of the above approach
C++
#include <bits/stdc++.h>
using namespace std;
bool isPrime( int n)
{
if (n <= 1)
return false ;
if (n <= 3)
return true ;
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n; i = i + 6) {
if (n % i == 0 || n % (i + 2) == 0) {
return false ;
}
}
return true ;
}
int main()
{
int n = 17;
if (isPrime(n) && (n % 16 == 1)) {
cout << "YES" ;
}
else {
cout << "NO" ;
}
return 0;
}
|
Java
class GFG {
static boolean isPrime( int n)
{
if (n <= 1 )
return false ;
if (n <= 3 )
return true ;
if (n % 2 == 0 || n % 3 == 0 )
return false ;
for ( int i = 5 ; i * i <= n; i = i + 6 ) {
if (n % i == 0 || n % (i + 2 ) == 0 ) {
return false ;
}
}
return true ;
}
public static void main(String[] args)
{
int n = 17 ;
if (isPrime(n) && (n % 16 == 1 )) {
System.out.println( "YES" );
}
else {
System.out.println( "NO" );
}
}
}
|
Python3
def isPrime(n) :
if (n < = 1 ) :
return False
if (n < = 3 ) :
return True
if (n % 2 = = 0 or n % 3 = = 0 ) :
return False
i = 5
while (i * i < = n) :
if (n % i = = 0 or n % (i + 2 ) = = 0 ) :
return False
i = i + 6
return True
n = 17
if (isPrime(n) and (n % 16 = = 1 ) ):
print ( "YES" )
else :
print ( "NO" )
|
C#
using System;
class GFG
{
static bool isPrime( int n)
{
if (n <= 1)
return false ;
if (n <= 3)
return true ;
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n; i = i + 6)
{
if (n % i == 0 || n % (i + 2) == 0)
{
return false ;
}
}
return true ;
}
public static void Main()
{
int n = 17;
if (isPrime(n) && (n % 16 == 1))
{
Console.WriteLine( "YES" );
}
else
{
Console.WriteLine( "NO" );
}
}
}
|
PHP
<?php
function isPrime( $n )
{
if ( $n <= 1)
return false;
if ( $n <= 3)
return true;
if ( $n % 2 == 0 || $n % 3 == 0)
return false;
for ( $i = 5; $i * $i <= $n ;
$i = $i + 6)
{
if ( $n % $i == 0 ||
$n % ( $i + 2) == 0)
{
return false;
}
}
return true;
}
$n = 17;
if (isPrime( $n ) && ( $n % 16 == 1))
{
echo "YES" ;
}
else
{
echo "NO" ;
}
?>
|
Javascript
<script>
function isPrime(n)
{
if (n <= 1)
return false ;
if (n <= 3)
return true ;
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( var i = 5; i * i <= n; i = i + 6) {
if (n % i == 0 || n % (i + 2) == 0) {
return false ;
}
}
return true ;
}
var n = 17;
if (isPrime(n) && (n % 16 == 1)) {
document.write( "YES" );
}
else {
document.write( "NO" );
}
</script>
|
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...