Find two numbers whose difference of fourth power is equal to N

Given an integer N, the task is to find two non-negative integers X and Y such that X⁴ – Y⁴ = N. If no such pair exists, print -1.

Examples:

Input: N = 15 
Output: X = 2, Y = 1 
Explanation:
X⁴ – Y⁴ = (2)⁴ – (1)⁴ = (16) – (1) = 15

Input: N = 10 
Output: -1 
Explanation :
No such value of X and Y are there which satisfy the condition. 

Approach:
To solve the problem mentioned above, we have to observe that we need to find the minimum and the maximum values of x and y that is possible to satisfy the equation. 

• The minimum value for the two integers can be 0 since X & Y are non-negative.
• The maximum value of X and Y can be ceil(N^(1/4)).
• Hence, iterate over the range [0, ceil(N^(1/4))] and find any suitable pair of X and Y that satisfies the condition.

Below is the implementation of the above approach:

C++

// C++ implementation to find the
// values of x and y for the given
// equation with integer N

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

// Function which find required x & y
void solve(int n)
{
    // Upper limit of x & y,
    // if such x & y exists
    int upper_limit = ceil(pow(
        n, 1.0 / 4));

    for (int x = 0; x <= upper_limit; x++) {

        for (int y = 0; y <= upper_limit; y++) {

            // num1 stores x^4
            int num1 = x * x * x * x;

            // num2 stores y^4
            int num2 = y * y * y * y;

            // If condition is satisfied
            // the print and return
            if (num1 - num2 == n) {
                cout << "x = " << x
                     << ", y = " << y;
                return;
            }
        }
    }

    // If no such pair exists
    cout << -1 << endl;
}

// Driver code
int main()
{
    int n = 15;

    solve(n);

    return 0;
}

Java

// Java implementation to find the
// values of x and y for the given
// equation with integer N
import java.util.*;

class GFG{

// Function which find required x & y
static void solve(int n)
{

    // Upper limit of x & y,
    // if such x & y exists
    int upper_limit = (int) (Math.ceil
                            (Math.pow(n, 1.0 / 4)));

    for(int x = 0; x <= upper_limit; x++)
    {
       for(int y = 0; y <= upper_limit; y++)
       {

          // num1 stores x^4
          int num1 = x * x * x * x;

          // num2 stores y^4
          int num2 = y * y * y * y;

          // If condition is satisfied
          // the print and return
          if (num1 - num2 == n)
          {
              System.out.print("x = " + x +
                             ", y = " + y);
              return;
          }
       }
    }

    // If no such pair exists
    System.out.print(-1);
}

// Driver code
public static void main(String[] args)
{
    int n = 15;

    solve(n);
}
}

// This code is contributed by shivanisinghss2110

Python3

# Python3 implementation to find the
# values of x and y for the given
# equation with integer N
from math import pow, ceil

# Function which find required x & y
def solve(n) :

    # Upper limit of x & y,
    # if such x & y exists
    upper_limit = ceil(pow(n, 1.0 / 4));

    for x in range(upper_limit + 1) :

        for y in range(upper_limit + 1) :

            # num1 stores x^4
            num1 = x * x * x * x;

            # num2 stores y^4
            num2 = y * y * y * y;

            # If condition is satisfied
            # the print and return
            if (num1 - num2 == n) :
                print("x =", x, ", y =" , y);
                return;

    # If no such pair exists
    print(-1) ;

# Driver code
if __name__ == "__main__" :

    n = 15;

    solve(n);

# This code is contributed by AnkitRai01

C#

// C# implementation to find the
// values of x and y for the given
// equation with integer N
using System;

class GFG{

// Function which find required x & y
static void solve(int n)
{

    // Upper limit of x & y,
    // if such x & y exists
    int upper_limit = (int) (Math.Ceiling
                            (Math.Pow(n, 1.0 / 4)));

    for(int x = 0; x <= upper_limit; x++)
    {
       for(int y = 0; y <= upper_limit; y++)
       {

          // num1 stores x^4
          int num1 = x * x * x * x;

          // num2 stores y^4
          int num2 = y * y * y * y;

          // If condition is satisfied
          // the print and return
          if (num1 - num2 == n)
          {
              Console.Write("x = " + x +
                          ", y = " + y);
              return;
          }
       }
    }

    // If no such pair exists
    Console.Write(-1);
}

// Driver code
public static void Main(String[] args)
{
    int n = 15;

    solve(n);
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

    // Javascript implementation to find the
    // values of x and y for the given
    // equation with integer N

    // Function which find required x & y
    function solve(n)
    {
        // Upper limit of x & y,
        // if such x & y exists
        let upper_limit = Math.ceil(Math.pow(n, 1.0 / 4));

        for (let x = 0; x <= upper_limit; x++) {

            for (let y = 0; y <= upper_limit; y++) {

                // num1 stores x^4
                let num1 = x * x * x * x;

                // num2 stores y^4
                let num2 = y * y * y * y;

                // If condition is satisfied
                // the print and return
                if (num1 - num2 == n) {
                    document.write("x = " + x + ", y = " + y);
                    return;
                }
            }
        }

        // If no such pair exists
        document.write(-1);
    }

    let n = 15;

    solve(n);


</script>

Output

x = 2, y = 1

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

Another Approach:

In this implementation, we create an unordered map (hash table) to store the values of x^4 – n for each x from 0 to the upper limit. Then, we iterate through the possible values of y and check if y^4 – n is already in the hash table. If it is, we retrieve the corresponding value of x and print the solution. If no such pair exists, we print -1.

C++

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

void solve(int n) {
    unordered_map<int, int> hashTable;

    // Calculate upper limit of x & y
    int upper_limit = ceil(pow(n, 1.0/4));

    // Store x^4 - n in hash table
    for (int x = 0; x <= upper_limit; x++) {
        int x4 = x * x * x * x;
        hashTable[x4 - n] = x;
    }

    // Check if there is a y such that y^4 - x^4 = n
    for (int y = 0; y <= upper_limit; y++) {
        int y4 = y * y * y * y;
        if (hashTable.find(y4) != hashTable.end()) {
            int x = hashTable[y4];
            cout << "x = " << x << ", y = " << y << endl;
            return;
        }
    }

    // If no such pair exists
    cout << -1 << endl;
}

int main() {
    int n = 15;

    solve(n);

    return 0;
}

Output

x = 2, y = 1

Time Complexity: O(N^(1/4))
Auxiliary space: O(N^(1/4))