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:
1^{5} - (-2)^{5} = 1 - (-32) = 33 == N

Input: N = 211
Output: -2 -3
Explanation:
(-2)^{5} - (-3)^{5} = -32 - (-243) = 211 == N

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)yp <= (y + 1)p+1 – yp+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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

1 -2

Time Complexity: O(240*240)

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