# 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 number 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 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

A 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 of is of the form 4k+1 or not.
• Print ‘YES’ if none of the factor is of the form 4k+1
Else print ‘NO’

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

Below is the implementation of above approach-

## C++

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

 ` 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 ` `?> `

Output:

```Testing for 11 : YES
Testing for 10 : NO
```

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