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+1Input: 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 notbool 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 codeint 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 functionfrom 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 Truedef 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 notfunction 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 notfunction 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 codelet 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)
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