# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 `

Output:

```x = 2, y = 1
```

Time Complexity: O(sqrt(N))

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