Coprime divisors of a number

Given an integer N. The task is to find a pair of co-prime divisors of N, greater than 1. If such divisors don’t exists then print ‘-1’. 

Examples:

Input: N = 45 
Output: 3 5 
Explanation: Since 3 and 5 are divisors of 45 and gcd( 3, 5 ) = 1 . 
Hence, they satisfy the condition.

Input: N = 25 
Output: -1 
Explanation: No pair of divisors of 25 satisfy the condition such 
that their gcd is 1.

Approach: 

1. Iterate from x = 2 to sqrt(N), to find all divisors of N
2. For any value x, check if it divides N
3. If it divides, then keep dividing N by x as long as it is divisible.
4. Now, check if N > 1, then the pair of divisors (x, N) will have 
   gcd(x, N) == 1, since all the factors of ‘x’ has been eliminated from N.
5. Similarly, check for all values of x.

Below is the implementation of the above approach:

C++

// C++ program to find two coprime
// divisors of a given number
// such that both are greater
// than 1

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

// Function which finds the
// required pair of divisors
// of N
void findCoprimePair(int N)
{
    // We iterate upto sqrt(N)
    // as we can find all the
    // divisors of N in this
    // time
    for (int x = 2; x <= sqrt(N); x++) {
        if (N % x == 0) {
            // If x is a divisor of N
            // keep dividing as long
            // as possible
            while (N % x == 0) {
                N /= x;
            }
            if (N > 1) {
                // We have found a
                // required pair
                cout << x << " "
                     << N << endl;
                return;
            }
        }
    }
    // No such pair of divisors
    // of N was found, hence
    // print -1
    cout << -1 << endl;
}

// Driver Code
int main()
{
    // Sample example 1
    int N = 45;
    findCoprimePair(N);

    // Sample example 2
    N = 25;
    findCoprimePair(N);
    return 0;
}

Java

// Java program to find two coprime
// divisors of a given number
// such that both are greater
// than 1
import java.util.*;

class GFG{

// Function which finds the
// required pair of divisors
// of N
public static void findCoprimePair(int N)
{

    // We iterate upto sqrt(N)
    // as we can find all the
    // divisors of N in this
    // time
    for(int x = 2; x <= Math.sqrt(N); x++)
    {
        if (N % x == 0)
        {

            // If x is a divisor of N
            // keep dividing as long
            // as possible
            while (N % x == 0)
            {
                N /= x;
            }
            if (N > 1)
            {

                // We have found a
                // required pair
                System.out.println(x + " " + N);
                return;
            }
        }
    }

    // No such pair of divisors
    // of N was found, hence
    // print -1
    System.out.println(-1);
}

// Driver code
public static void main(String[] args)
{

    // Sample example 1
    int N = 45;
    findCoprimePair(N);

    // Sample example 2
    N = 25;
    findCoprimePair(N);
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to find two coprime
# divisors of a given number such that
# both are greater than 1
import math

# Function which finds the
# required pair of divisors
# of N
def findCoprimePair(N):

    # We iterate upto sqrt(N)
    # as we can find all the
    # divisors of N in this
    # time
    for x in range(2, int(math.sqrt(N)) + 1):
        if (N % x == 0):

            # If x is a divisor of N
            # keep dividing as long
            # as possible
            while (N % x == 0):
                N //= x

            if (N > 1):

                # We have found a
                # required pair
                print(x, N)
                return;

    # No such pair of divisors
    # of N was found, hence
    # print -1
    print("-1")

# Driver Code

# Sample example 1
N = 45
findCoprimePair(N)

# Sample example 2
N = 25
findCoprimePair(N)

# This code is contributed by Vishal Maurya.

C#

// C# program to find two coprime
// divisors of a given number
// such that both are greater
// than 1
using System;
class GFG{

// Function which finds the
// required pair of divisors
// of N
public static void findCoprimePair(int N)
{
  // We iterate upto sqrt(N)
  // as we can find all the
  // divisors of N in this
  // time
  for(int x = 2;
          x <= Math.Sqrt(N); x++)
  {
    if (N % x == 0)
    {
      // If x is a divisor of N
      // keep dividing as long
      // as possible
      while (N % x == 0)
      {
        N /= x;
      }
      if (N > 1)
      {
        // We have found a
        // required pair
        Console.WriteLine(x +
                          " " + N);
        return;
      }
    }
  }

  // No such pair of divisors
  // of N was found, hence
  // print -1
  Console.WriteLine(-1);
}

// Driver code
public static void Main(String[] args)
{
  // Sample example 1
  int N = 45;
  findCoprimePair(N);

  // Sample example 2
  N = 25;
  findCoprimePair(N);
}
}

// This code is contributed by Chitranayal

Javascript

<script>

// JavaScript program to find two coprime
// divisors of a given number
// such that both are greater
// than 1

// Function which finds the
// required pair of divisors
// of N
function findCoprimePair(N)
{
    // We iterate upto sqrt(N)
    // as we can find all the
    // divisors of N in this
    // time
    for (let x = 2; x <= Math.sqrt(N); x++) {
        if (N % x == 0) {
            // If x is a divisor of N
            // keep dividing as long
            // as possible
            while (N % x == 0) {
                N = Math.floor(N / x);
            }
            if (N > 1) {
                // We have found a
                // required pair
                document.write(x + " "
                    + N + "<br>");
                return;
            }
        }
    }
    // No such pair of divisors
    // of N was found, hence
    // print -1
    document.write(-1 + "<br>");
}

// Driver Code
    // Sample example 1
    let N = 45;
    findCoprimePair(N);

    // Sample example 2
    N = 25;
    findCoprimePair(N);

// This code is contributed by Surbhi Tyagi.

</script>

Output:

3 5
-1

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