Check whether a number is Non-hypotenuse number

Given a positive integer n, the task is to check if n is a Non-hypotenuse number or not. If n is a Non-hypotenuse number then print ‘YES’ else print ‘NO’.

Non-hypotenuse number : In mathematics, a Non-hypotenuse number is a natural number whose square can not be expressed as sum of two distinct non-zero squares,i.e a non-hypotenuse number can not be put into the form of (x² + x² ) or K(x² + x² ), where K, x and y are positive integers. The number 1, 2, 3, 4 are Non-hypotenuse numbers while 5 is not a Non-hypotenuse number. A Non-hypotenuse number can not be the hypotenuse of the right-angled triangle having integer sides.

Examples:

Input: 5
Output: YES
Explanation: 5 can be expressed as 2² + 1². 

Input: 6
Output: NO
Explanation: 6 can not be expressed as sum of two different squares.

First, a few Non-hypotenuse numbers are-

1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 14, 16, 18, 19, 21, 22, 23, 24, 27, 28, 31, 32, 33, 36, 38, 42, 43, 44, 46, 47

A Simple Solution to check if the given number ‘n‘ is a Non-Hypotenuse number or not is to check if any combination of squares of x and y is equal to n or not.

An Efficient Solution is based on the fact that a non-hypotenuse number do not have any prime factor of the form 4k+1.

Example:

Input: 12
Output: YES
Explanation: Prime factors of 12 is 2 and 3. None of them is of the form 4k+1 

Input: 10
Output: NO
Explanation: Prime factors of 10 is 2 and 5. Here 5 is of the form 4k+1

Approach

• Find all prime factors of n
• Check if any prime factor is of form 4k+1 or not.
• Print ‘YES’ if none of the factors is of the form 4k+1 Else print ‘NO’

To read more about the method of calculating the prime factor of any number, refer to this. 

Below is the implementation of the above approach-

C++

// CPP program to check if
// a given number is
// Non-Hypotenuse number or not.

#include <bits/stdc++.h>
using namespace std;

// Function to find prime factor
// and check if it is of the form
// 4k+1 or not

bool isNonHypotenuse(int n)
{
    // 2 is a prime number but
    // not of the form 4k+1
    // so, keep Dividing n by 2
    // until n is divisible by 2
    while (n % 2 == 0) {
        n = n / 2;
    }

    // n must be odd at this point. So we can skip
    // one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i + 2) {

        // if i divides n
        // check if i is of the form
        // 4k+1 or not

        if (n % i == 0) {
            if ((i - 1) % 4 == 0)
                return false;

            // while i divides n
            // divide n by i
            // and update n
            while (n % i == 0) {
                n = n / i;
            }
        }
    }

    // This condition is to handle the case when n
    // is a prime number greater than 2
    if (n > 2 && (n - 1) % 4 == 0)
        return false;

    else
        return true;
}

void test(int n)
{
    cout << "Testing for "
         << n << " : ";

    if (isNonHypotenuse(n))
        cout << "YES"
             << "\n";

    else
        cout << "NO"
             << "\n";
}

// Driver code
int main()
{
    int n = 11;
    test(n);

    n = 10;
    test(n);

    return 0;
}

Java

// JAVA program to check if
// a given number is
// Non-Hypotenuse number or not.

class GFG {
    // Function to find prime factor
    // and check if it is of the form
    // 4k+1 or not

    static boolean isNonHypotenuse(int n)
    {
        // 2 is a prime number but
        // not of the form 4k+1
        // so, keep Dividing n by 2
        // until n is divisible by 2
        while (n % 2 == 0) {
            n = n / 2;
        }

        // n must be odd at this point. So we can skip
        // one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i = i + 2) {

            // if i divides n
            // check if i is of the form
            // 4k+1 or not

            if (n % i == 0) {
                if ((i - 1) % 4 == 0)
                    return false;

                // while i divides n
                // divide n by i
                // and update n
                while (n % i == 0) {
                    n = n / i;
                }
            }
        }

        // This condition is to handle the
        // case when n  is a prime number
        // greater than 2
        if (n > 2 && (n - 1) % 4 == 0)
            return false;

        else
            return true;
    }

    public static void test(int n)
    {

        System.out.println("Testing for "
                           + n + " : ");

        if (isNonHypotenuse(n))
            System.out.println("YES");

        else
            System.out.println("NO");
    }

    // Driver code
    public static void main(String args[])
    {

        int n = 11;
        test(n);

        n = 10;
        test(n);
    }
}

Python3

# Python3 program to check if
# a given number is
# Non-Hypotenuse number or not.

# From math lib import sqrt function
from math import sqrt

# Function to find prime factor
# and check if it is of the form
# 4k+1 or not
def isNonHypotenuse(n) :

    # 2 is a prime number but not of
    # the form 4k+1 so, keep Dividing
    # n by 2 until n is divisible by 2
    while (n % 2 == 0) :
        n = n // 2

    # n must be odd at this point. So we
    # can skip one element (Note i = i +2)
    for i in range(3, int(sqrt(n)) + 1, 2) :

        # if i divides n check if i
        # is of the form 4k+1 or not
        if (n % i == 0) :
            if ((i - 1) % 4 == 0) :
                return False

            # while i divides n divide n
            # by i and update n
            while (n % i == 0) :
                n = n // i

    # This condition is to handle the case
    # when n is a prime number greater than 2
    if (n > 2 and (n - 1) % 4 == 0) :
        return False

    else :
        return True

def test(n) :
    print("Testing for", n, ":", end = " ")

    if (isNonHypotenuse(n)) :
        print("YES")

    else :
        print("NO")

# Driver code
if __name__ == "__main__" :

    n = 11
    test(n)

    n = 10
    test(n)

# This code is contributed by Ryuga

C#

// C# program to check if
// a given number is
// Non-Hypotenuse number or not.

using System;
class GFG {
    // Function to find prime factor
    // and check if it is of the form
    // 4k+1 or not

    static bool isNonHypotenuse(int n)
    {
        // 2 is a prime number but
        // not of the form 4k+1
        // so, keep Dividing n by 2
        // until n is divisible by 2
        while (n % 2 == 0) {
            n = n / 2;
        }

        // n must be odd at this point. So we can skip
        // one element (Note i = i +2)
        for (int i = 3; i <= Math.Sqrt(n); i = i + 2) {

            // if i divides n
            // check if i is of the form
            // 4k+1 or not

            if (n % i == 0) {
                if ((i - 1) % 4 == 0)
                    return false;

                // while i divides n
                // divide n by i
                // and update n
                while (n % i == 0) {
                    n = n / i;
                }
            }
        }

        // This condition is to handle the
        // case when n is a prime number
        // greater than 2
        if (n > 2 && (n - 1) % 4 == 0)
            return false;

        else
            return true;
    }

    public static void test(int n)
    {
        Console.WriteLine("Testing for " + n + " : ");
        if (isNonHypotenuse(n))
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }

    // Driver code
    public static void Main()
    {
        int n = 11;
        test(n);

        n = 10;
        test(n);
    }
}

PHP

<?php
// PHP program to check if a given number
// is Non-Hypotenuse number or not.

// Function to find prime factor and check
// if it is of the form 4k+1 or not
function isNonHypotenuse($n)
{
    // 2 is a prime number but not of the
    // form 4k+1 so, keep Dividing n by 2
    // until n is divisible by 2
    while ($n % 2 == 0)
    {
        $n = $n / 2;
    }

    // n must be odd at this point. So we
    // can skip one element (Note i = i +2)
    for ($i = 3; $i <= sqrt($n); $i = $i + 2)
    {

        // if i divides n check if i is of
        // the form 4k+1 or not

        if ($n % $i == 0)
        {
            if (($i - 1) % 4 == 0)
                return false;

            // while i divides n divide n by i
            // and update n
            while ($n % $i == 0)
            {
                $n = $n / $i;
            }
        }
    }

    // This condition is to handle the case
    // when n is a prime number greater than 2
    if ($n > 2 && ($n - 1) % 4 == 0)
        return false;
    else
        return true;
}

function test($n)
{
    echo "Testing for ", $n , " : ";

    if (isNonHypotenuse($n))
        echo "YES". "\n";
    else
        echo "NO". "\n";
}

// Driver code
$n = 11;
test($n);

$n = 10;
test($n);

// This code is contributed by Sach_Code
?>

Javascript

// JavaScript program to check if
// a given number is
// Non-Hypotenuse number or not.

// Function to find prime factor
// and check if it is of the form
// 4k+1 or not
function isNonHypotenuse(n)
{

    // 2 is a prime number but
    // not of the form 4k+1
    // so, keep Dividing n by 2
    // until n is divisible by 2
    while (n % 2 == 0) {
        n = Math.floor(n / 2);
    }

    // n must be odd at this point. So we can skip
    // one element (Note i = i +2)
    for (var i = 3; i <= Math.sqrt(n); i = i + 2) {

        // if i divides n
        // check if i is of the form
        // 4k+1 or not

        if (n % i == 0) {
            if ((i - 1) % 4 == 0)
                return false;

            // while i divides n
            // divide n by i
            // and update n
            while (n % i == 0) {
                n = Math.floor(n / i);
            }
        }
    }

    // This condition is to handle the case when n
    // is a prime number greater than 2
    if (n > 2 && (n - 1) % 4 == 0)
        return false;

    else
        return true;
}

function test(n)
{
    process.stdout.write("Testing for " + n + " : ");

    if (isNonHypotenuse(n))
        console.log("YES");

    else
        console.log("NO");
}

// Driver code
let n = 11;
test(n);

n = 10;
test(n);

// This code is contributed by phasing17

Output:

Testing for 11 : YES
Testing for 10 : NO

Time complexity: O(sqrt(n)*logn)
Auxiliary Space: O(1)