求两个整数 X 和 Y，给定 GCD P 和它们的平方差 Q

给定两个整数P和Q ，任务是找到最大公约数（GCD）为P且平方差为 Q 的任意两个整数。如果不存在任何这样的整数，则打印“-1” 。

例子：

Input: P = 3, Q = 27
Output:  6 3
Explanation:
Consider the two number as 6, 3. Now, the GCD(6, 3) = 3 and 6*6 – 3*3 = 27 which satisfies the condition.

Input: P = 1, Q = 100
Output: -1

方法：给定的问题可以使用基于以下观察来解决：

给定的方程也可以写成：

=> $x^{2} - y^{2} = Q$
=> $(x + y)*(x - y) = Q$ 

现在对于给定方程的积分解：

(x+y)(x-y)  is always an integer 
=> (x+y)(x-y)  are divisors of Q 
  

让 (x + y) = p1 和 (x + y) = p2
是两个方程，其中 p1 和 p2 是Q的除数
这样p1 * p2 = Q 。

求解上述两个方程，我们有：

=> $x = \frac{(p1 + p2)}{2}$and   $y = \frac{(p1 - p2)}{2}$

从以上计算可知，要使x 和 y为整数，则除数之和必须为偶数。由于 x 和 y 的两个值有 4 个可能的值，即(+x, +y)、(+x, -y)、 (-x, +y) 和 (-x, -y) 。
因此，可能解决方案的总数由4*(count of divisors with even sum)给出。

现在在这些对中，找到 GCD 为 P 的对并打印该对。如果不存在这样的对，则打印 -1。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to print a valid pair with
// the given criteria
int printValidPair(int P, int Q)
{
 
    // Iterate over the divisors of Q
    for (int i = 1; i * i <= Q; i++) {
 
        // check if Q is a multiple of i
        if (Q % i == 0) {
 
            // L = (A - B) <- 1st equation
            // R = (A + B) <- 2nd equation
            int L = i;
            int R = Q / i;
 
            // Calculate value of A
            int A = (L + R) / 2;
 
            // Calculate value of B
            int B = (R - L) / 2;
 
            // As A and B both are integers
            // so the parity of L and R
            // should be the same
            if (L % 2 != R % 2) {
                continue;
            }
 
            // Check the first condition
            if (__gcd(A, B) == P) {
                cout << A << " " << B;
                return 0;
            }
        }
    }
 
    // If no such A, B exist
    cout << -1;
 
    return 0;
}
 
// Driver Code
int main()
{
    int P = 3, Q = 27;
    printValidPair(P, Q);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to print a valid pair with
// the given criteria
static int printValidPair(int P, int Q)
{
 
    // Iterate over the divisors of Q
    for (int i = 1; i * i <= Q; i++) {
 
        // check if Q is a multiple of i
        if (Q % i == 0) {
 
            // L = (A - B) <- 1st equation
            // R = (A + B) <- 2nd equation
            int L = i;
            int R = Q / i;
 
            // Calculate value of A
            int A = (L + R) / 2;
 
            // Calculate value of B
            int B = (R - L) / 2;
 
            // As A and B both are integers
            // so the parity of L and R
            // should be the same
            if (L % 2 != R % 2) {
                continue;
            }
 
            // Check the first condition
            if (__gcd(A, B) == P) {
                System.out.print(A+ " " +  B);
                return 0;
            }
        }
    }
 
    // If no such A, B exist
    System.out.print(-1);
    return 0;
 
}
static int __gcd(int a, int b) 
{ 
    return b == 0? a:__gcd(b, a % b);    
}
   
// Driver Code
public static void main(String[] args)
{
    int P = 3, Q = 27;
    printValidPair(P, Q);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# python program for the above approach
import math
 
# Function to print a valid pair with
# the given criteria
def printValidPair(P, Q):
 
    # Iterate over the divisors of Q
    for i in range(1, int(math.sqrt(Q)) + 1):
 
        # check if Q is a multiple of i
        if (Q % i == 0):
 
            # L = (A - B) <- 1st equation
            # R = (A + B) <- 2nd equation
            L = i
            R = Q // i
 
            # Calculate value of A
            A = (L + R) // 2
 
            # Calculate value of B
            B = (R - L) // 2
 
            # As A and B both are integers
            # so the parity of L and R
            # should be the same
            if (L % 2 != R % 2):
                continue
 
            # Check the first condition
            if (math.gcd(A, B) == P):
                print(f"{A} {B}")
                return 0
 
    # If no such A, B exist
    print(-1)
 
    return 0
 
# Driver Code
if __name__ == "__main__":
 
    P = 3
    Q = 27
    printValidPair(P, Q)
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
class GFG
{
 
    // Function to print a valid pair with
    // the given criteria
    static int printValidPair(int P, int Q)
    {
 
        // Iterate over the divisors of Q
        for (int i = 1; i * i <= Q; i++)
        {
 
            // check if Q is a multiple of i
            if (Q % i == 0)
            {
 
                // L = (A - B) <- 1st equation
                // R = (A + B) <- 2nd equation
                int L = i;
                int R = Q / i;
 
                // Calculate value of A
                int A = (L + R) / 2;
 
                // Calculate value of B
                int B = (R - L) / 2;
 
                // As A and B both are integers
                // so the parity of L and R
                // should be the same
                if (L % 2 != R % 2)
                {
                    continue;
                }
 
                // Check the first condition
                if (__gcd(A, B) == P)
                {
                    Console.Write(A + " " + B);
                    return 0;
                }
            }
        }
 
        // If no such A, B exist
        Console.Write(-1);
        return 0;
 
    }
    static int __gcd(int a, int b)
    {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Driver Code
    public static void Main()
    {
        int P = 3, Q = 27;
        printValidPair(P, Q);
    }
}
 
// This code is contributed by gfgking

Javascript

       // JavaScript code for the above approach
       // Recursive function to return gcd of a and b
       function __gcd(a, b) {
           // Everything divides 0
           if (a == 0)
               return b;
           if (b == 0)
               return a;
 
           // base case
           if (a == b)
               return a;
 
           // a is greater
           if (a > b)
               return __gcd(a - b, b);
           return __gcd(a, b - a);
       }
 
       // Function to print a valid pair with
       // the given criteria
       function printValidPair(P, Q) {
 
           // Iterate over the divisors of Q
           for (let i = 1; i * i <= Q; i++) {
 
               // check if Q is a multiple of i
               if (Q % i == 0) {
 
                   // L = (A - B) <- 1st equation
                   // R = (A + B) <- 2nd equation
                   let L = i;
                   let R = Q / i;
 
                   // Calculate value of A
                   let A = (L + R) / 2;
 
                   // Calculate value of B
                   let B = (R - L) / 2;
 
                   // As A and B both are integers
                   // so the parity of L and R
                   // should be the same
                   if (L % 2 != R % 2) {
                       continue;
                   }
 
                   // Check the first condition
                   if (__gcd(A, B) == P) {
                       document.write(A + " " + B);
                       return 0;
                   }
               }
           }
 
           // If no such A, B exist
           document.write(-1);
 
           return 0;
       }
 
       // Driver Code
       let P = 3, Q = 27;
       printValidPair(P, Q);
 
 // This code is contributed by Potta Lokesh
   

输出：

6 3

时间复杂度： O(sqrt(Q))
辅助空间： O(1)