Pair of integers having difference of their fifth power as X

Given an integer X, the task is to find a pair A and B such that their difference of fifth power is X, i.e., A⁵ – B⁵ = X. If there is no such pair print “Not Possible”.

Input: X = 33 
Output: 1 -2 
Explanation:

Input: N = 211 
Output: -2 -3 
Explanation:

Naive Approach: A simple solution is to use two for loops, one for A and one for B, ranging from -10⁹ to 10⁹. 
Efficient Approach: The idea is to narrow down the range of A and B using mathematical techniques. 

Since A⁵ – B⁵ = X => A⁵ = X + B⁵. For A to be as high as possible, B also has to be as high as possible, as it is evident from the inequality. 

Consider A = N and B = N – 1 
=> N⁵ – (N – 1)⁵ = X. 

 By binomial expansion, we know

(p + 1)y^p <= (y + 1)^p+1 – y^p+1 <= (p+1)(y+1)^p

So we can say that the maximum value of LHS is 4N⁴.

Hence 4N⁵ <= X 
=> N <= (X/5)^1/5. 
=> This gives us N ~ 120.

Since A and B can also be negative, we simply extrapolate the range and the final range we get is [-120, 120].

Below is the implementation of the above approach: 

C++

// C++ implementation to find a pair
// of integers A & B such that
// difference of fifth power is
// equal to the given number X

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

// Function to find a pair
// of integers A & B such that
// difference of fifth power is
// equal to the given number X
void findPair(int x)
{
    int lim = 120;

    // Loop to choose every possible
    // pair with in the range
    for (int i = -lim; i <= lim; i++) {
        for (int j = -lim; j <= lim; j++) {

            // Check if equation holds
            if (pow(i, 5) - pow(j, 5) == x) {
                cout << i << ' ' << j << endl;
                return;
            }
        }
    }
    cout << "-1";
}

// Driver Code
signed main()
{
    int X = 33;

    // Function Call
    findPair(X);
    return 0;
}

Java

// Java implementation to find a
// pair of integers A & B such
// that difference of fifth power
// is equal to the given number X
class GFG{

// Function to find a pair
// of integers A & B such that
// difference of fifth power is
// equal to the given number X
static void findPair(int x)
{
    int lim = 120;

    // Loop to choose every possible
    // pair with in the range
    for(int i = -lim; i <= lim; i++)
    {
       for(int j = -lim; j <= lim; j++)
       {

          // Check if equation holds
          if (Math.pow(i, 5) -
              Math.pow(j, 5) == x)
          {
              System.out.print(i + " " +
                               j + "\n");
              return;
          }
       }
    }
    System.out.print("-1");
}

// Driver Code
public static void main(String[] args)
{
    int X = 33;

    // Function Call
    findPair(X);
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation to find
# a pair of integers A & B such
# that difference of fifth power
# is equal to the given number X
import math

# Function to find a pair
# of integers A & B such that
# difference of fifth power is
# equal to the given number X
def findPair(x):

    lim = 120

    # Loop to choose every possible
    # pair with in the range
    for i in range(-lim, lim + 1):
        for j in range(-lim, lim + 1):

            # Check if equation holds
            if (math.pow(i, 5) -
                math.pow(j, 5) == x):
                print (i, end = ' ')
                print (j, end = '\n')
                return

    print ("-1")

# Driver Code
X = 33

# Function Call
findPair(X)

# This code is contributed by PratikBasu

C#

// C# implementation to find a
// pair of integers A & B such
// that difference of fifth power
// is equal to the given number X
using System;

class GFG{

// Function to find a pair of
// integers A & B such that
// difference of fifth power is
// equal to the given number X
static void findPair(int x)
{
    int lim = 120;

    // Loop to choose every possible
    // pair with in the range
    for(int i = -lim; i <= lim; i++)
    {
       for(int j = -lim; j <= lim; j++)
       {

          // Check if equation holds
          if (Math.Pow(i, 5) -
              Math.Pow(j, 5) == x)
          {
              Console.Write(i + " " +
                            j + "\n");
              return;
          }
       }
    }
    Console.Write("-1");
}

// Driver code
public static void Main(String[] args)
{
    int X = 33;

    // Function call
    findPair(X);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript implementation to find a
// pair of integers A & B such
// that difference of fifth power
// is equal to the given number X


// Function to find a pair
// of integers A & B such that
// difference of fifth power is
// equal to the given number X
function findPair(x)
{
    let lim = 120;

    // Loop to choose every possible
    // pair with in the range
    for(let i = -lim; i <= lim; i++)

    for(let j = -lim; j <= lim; j++)


        // Check if equation holds
        if (Math.pow(i, 5) -Math.pow(j, 5) == x)
        {
            document.write(i + " "+ j);
            return;
        }

    document.write("-1");
}

// Driver Code

    let X = 33;

    // Function Call
    findPair(X);

// This code is contributed by mohan

</script>

Output:

1 -2

Time Complexity: O(240*240)