Find minimum sum Pair whose GCD is greater than 1

Given two positive integers A and B, the task is to find two positive integers such that they belong in the range [A, B], such that their GCD is greater than 1 and their sum is the minimum among all possible pairs.

Input: 2, 3
Output: -1
Explanation: Clearly there doesn’t exits any pair for which gcd is greater than 1, Hence we simply return -1.

Input: 2 10
Output: 2 4
Explanation: The smallest pair for which GCD is greater than 1 is 2, 4 .

Approach: The problem can be based on the following observation:

Two consecutive even numbers have GCD greater than 1 and the difference among them is also minimum.

Based on the above observation, it is optimal to find the pair of two consecutive even numbers whose values are closest to the lower limit of the range. Follow the step below to solve this problem:

• Check the absolute difference of the given range abs(A – B)
  • If abs(A – B) <= 1, then there will not exists any pair, whose gcd is greater than 1, print -1.
• If the first value is even
  •  print (A, A+2) because these are the smallest numbers whose GCD is greater than 1.
• If the first value is odd
  • Check if (B – A) == 2 then print -1 because the GCD of two consecutive odd numbers is always 1
  • Check if (B – A) > 2 then
    •  If A is divisible by 3 then print (A, A+3),
    • Otherwise print (A+1, A +3).

Below is the implementation of the above approach:

C++

// C++ code to implement the above approach

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

// Function to find the minimum Sum pair
// such that their GCD(x, y) > 1
void minPair(int A, int B)
{
    // If absolute difference is 1
    // then no such pair exists
    if (abs(B - A) <= 1)
        cout << "-1";
    else {
        // Check if first number is even
        if ((A % 2) == 0)
            cout << A << " " << A + 2;
        else {
            if (B - A == 2) {
                cout << -1;
            }
            else {

                // Check if first number is
                // divisible by 3
                if (A % 3 == 0) {
                    cout << A << " " << A + 3;
                }
                else {
                    cout << A + 1 << " "
                         << A + 3;
                }
            }
        }
    }
}

// Drive code
int main()
{
    int A = 2;
    int B = 10;

    // Function call
    minPair(A, B);
    return 0;
}

C

// C code to implement the above approach
#include <stdio.h>

// Function to find the minimum Sum pair
// such that their GCD(x, y) > 1
void minPair(int A, int B)
{
  int check = (B-A);
  if(check<0){ check = check*(-1); }

  // If absolute difference is 1
  // then no such pair exists
  if (check <= 1)
    printf("-1");
  else {
    // Check if first number is even
    if ((A % 2) == 0)
      printf("%d %d", A, A + 2);
    else {
      if (B - A == 2) {
        printf("-1");
      }
      else {

        // Check if first number is
        // divisible by 3
        if (A % 3 == 0) {
          printf("%d %d",A, A + 3);
        }
        else {
          printf("%d %d",A + 1,
                 A + 3);
        }
      }
    }
  }
}

// Drive code
void main()
{
  int A = 2;
  int B = 10;

  // Function call
  minPair(A, B);
}

// This code is contributed by ashishsingh13122000.

Java

// Java code to implement the above approach
import java.io.*;
class GFG
{

  // Function to find the minimum Sum pair
  // such that their GCD(x, y) > 1
  public static void minPair(int A, int B)
  {

    // If absolute difference is 1
    // then no such pair exists
    if (Math.abs(B - A) <= 1)
      System.out.print("-1");
    else
    {

      // Check if first number is even
      if ((A % 2) == 0)
        System.out.print(A + " " + (A + 2));
      else {
        if (B - A == 2) {
          System.out.print("-1");
        }
        else {

          // Check if first number is
          // divisible by 3
          if (A % 3 == 0) {
            System.out.print(A + " " + (A + 3));
          }
          else {
            System.out.print((A + 1) + " "
                             + (A + 3));
          }
        }
      }
    }
  }

  // Driver Code
  public static void main(String[] args)
  {
    int A = 2;
    int B = 10;

    // Function call
    minPair(A, B);
  }
}

// This code is contributed by Rohit Pradhan

Python3

# Python3 code to implement the above approach

# Function to find the minimum Sum pair
# such that their GCD(x, y) > 1
def minPair(A, B):

    # If absolute difference is 1
    # then no such pair exists
    if (abs(B - A) <= 1):
        print("-1")
    else:
        # Check if first number is even
        if ((A % 2) == 0):
            print(f"{A} {A + 2}")
        else:
            if (B - A == 2):
                print(-1)

            else:

                # Check if first number is
                # divisible by 3
                if (A % 3 == 0):
                    print(f"{A} {A + 3}")

                else:
                    print(f"{A + 1} {A + 3}")

# Drive code
if __name__ == "__main__":

    A = 2
    B = 10

    # Function call
    minPair(A, B)

    # This code is contributed by rakeshsahni

C#

// C# to implement the above approach
using System;
class GFG{

  // Function to find the minimum Sum pair
  // such that their GCD(x, y) > 1
  static void minPair(int A, int B)
  {
    int f = (B-A);
    if(f<0) { f = f*(-1);}

    // If absolute difference is 1
    // then no such pair exists
    if (f <= 1)
      Console.WriteLine(-1);
    else
    {

      // Check if first number is even
      int z = A+2;
      if ((A % 2) == 0)
        Console.Write(A + " " + z);
      else {
        if (B - A == 2) {
          Console.Write(-1);
        }
        else {

          int y = A+3;
          // Check if first number is
          // divisible by 3
          if (A % 3 == 0) {
            Console.Write(A + " " + y);
          }
          else {
            int c = A+1;
            int d = A+3;
            Console.Write(c + " "
                          + d);
          }
        }
      }
    }
  }

  // Drive code
  public static void Main()
  {
    int A = 2;
    int B = 10;

    // Function call
    minPair(A, B);
  }
}

// This code is contributed by ashishsingh13122000.

Javascript

<script>
       // JavaScript program for the above approach

       // Function to find the minimum Sum pair
       // such that their GCD(x, y) > 1
       function minPair(A, B)
       {

           // If absolute difference is 1
           // then no such pair exists
           if (Math.abs(B - A) <= 1)
               document.write("-1");
           else
           {

               // Check if first number is even
               if ((A % 2) == 0)
                   document.write(A + " " + (A + 2));
               else {
                   if (B - A == 2) {
                       document.write(-1);
                   }
                   else {

                       // Check if first number is
                       // divisible by 3
                       if (A % 3 == 0) {
                           document.write(A + " " + (A + 3));
                       }
                       else {
                           document.write((A + 1) + " "
                               + (A + 3));
                       }
                   }
               }
           }
       }

       // Drive code
       let A = 2;
       let B = 10;

       // Function call
       minPair(A, B);

      // This code is contributed by Potta Lokesh

   </script>

Output

2 4

Time Complexity: O(1)
Auxiliary Space: O(1)