找出两个数字，使它们的平方差等于N

给定一个整数N ，任务是找到两个非负整数A和B ，使得A ² -B ² = N。如果不存在这样的整数，则打印-1 。

例子：

Input: N = 7
Output: 4 3
Explanation:
The two integers 4 and 3 can be represented as 4² – 3² = 7.

Input: N = 6
Output: -1
Explanation:
No pair of (A, B) exists that satisfies the required condition.

方法：

• A ² – B ²可以表示为(A – B)*(A + B) 。

  A² – B² = (A – B) * (A + B)

• 因此，对于A ² -B ²等于N ， (A + B)和(A-B)都应为N的因数。
• 考虑到A + B和A – B分别等于C和D ，则C和D必须是N的因数，这样C <= D且C和D应该具有相同的奇偶性。
• 因此，为了解决这个问题，我们只需要找到满足上述条件的任何一对C和D即可。如果不存在这样的C＆D ，则打印-1 。

下面是上述方法的实现：

C++

// C++ Program to find two numbers
// with difference of their
// squares equal to N
  
#include 
using namespace std;
  
// Function to check and print
// the required two positive integers
void solve(int n)
{
  
    // Iterate till sqrt(n) to find
    // factors of N
    for (int x = 1; x <= sqrt(n); x++) {
  
        // Check if x is one
        // of the factors of N
        if (n % x == 0) {
  
            // Store the factor
            int small = x;
  
            // Compute the other factor
            int big = n / x;
  
            // Check if the two factors
            // are of the same parity
            if (small % 2 == big % 2) {
  
                // Compute a and b
                int a = (small + big) / 2;
                int b = (big - small) / 2;
  
                cout << a << " "
                     << b << endl;
                return;
            }
        }
    }
  
    // If no pair exists
    cout << -1 << endl;
}
  
// Driver Code
int main()
{
    int n = 7;
  
    solve(n);
  
    return 0;
}

Java

// Java Program to find two numbers
// with difference of their
// squares equal to N
import java.util.*;
class GFG{
  
// Function to check and print
// the required two positive integers
static void solve(int n)
{
  
    // Iterate till sqrt(n) to find
    // factors of N
    for (int x = 1; x <= Math.sqrt(n); x++) 
    {
  
        // Check if x is one
        // of the factors of N
        if (n % x == 0) 
        {
  
            // Store the factor
            int small = x;
  
            // Compute the other factor
            int big = n / x;
  
            // Check if the two factors
            // are of the same parity
            if (small % 2 == big % 2) 
            {
  
                // Compute a and b
                int a = (small + big) / 2;
                int b = (big - small) / 2;
  
                System.out.print(a + " " + b);
                return;
            }
        }
    }
  
    // If no pair exists
    System.out.print(-1);
}
  
// Driver Code
public static void main(String args[])
{
    int n = 7;
  
    solve(n);
}
}
  
// This code is contributed by Code_Mech

Python3

# Python3 Program to find two numbers 
# with difference of their 
# squares equal to N 
from math import sqrt
  
# Function to check and print 
# the required two positive integers 
def solve(n) :
      
    # Iterate till sqrt(n) to find
    # factors of N
    for x in range(1, int(sqrt(n)) + 1) :
          
        # Check if x is one
        # of the factors of N
        if (n % x == 0) :
              
            # Store the factor 
            small = x;
              
            # Compute the other factor
            big = n // x;
              
            # Check if the two factors
            # are of the same parity
            if (small % 2 == big % 2) :
                  
                # Compute a and b
                a = (small + big) // 2;
                b = (big - small) // 2;
                print(a, b) ;
                return;
                  
    # If no pair exists
    print(-1); 
  
# Driver Code
if __name__ == "__main__" :
    n = 7;
    solve(n); 
  
# This code is contributed by AnkitRai01

C#

// C# Program to find two numbers
// with difference of their
// squares equal to N
using System;
class GFG{
  
// Function to check and print
// the required two positive integers
static void solve(int n)
{
  
    // Iterate till sqrt(n) to find
    // factors of N
    for (int x = 1; x <= Math.Sqrt(n); x++) 
    {
  
        // Check if x is one
        // of the factors of N
        if (n % x == 0) 
        {
  
            // Store the factor
            int small = x;
  
            // Compute the other factor
            int big = n / x;
  
            // Check if the two factors
            // are of the same parity
            if (small % 2 == big % 2) 
            {
  
                // Compute a and b
                int a = (small + big) / 2;
                int b = (big - small) / 2;
  
                Console.WriteLine(a + " " + b);
                return;
            }
        }
    }
  
    // If no pair exists
    Console.WriteLine(-1);
}
  
// Driver Code
public static void Main()
{
    int n = 7;
  
    solve(n);
}
}
  
// This code is contributed by Code_Mech

输出：

4 3

时间复杂度： O(sqrt(N))
辅助空间： O(1)