# Find x, y, z that satisfy 2/n = 1/x + 1/y + 1/z

Given n, find x, y, z such that x, y, z satisfy the equation “2/n = 1/x + 1/y + 1/z”

There are multiple x, y and z that satisfy the equation print anyone of them, if not possible then print -1.

Examples:

```Input : 3
Output : 3 4 12
Explanation: here 3 4 and 12 satisfy
the given equation

Input : 7
Output : 7 8 56
```

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

Note that for n = 1 there is no solution, and for n > 1 there is solution x = n, y = n+1, z = n·(n+1).

To come to this solution, represent 2/n as 1/n+1/n and reduce the problem to represent 1/n as a sum of two fractions. Let’s find the difference between 1/n and 1/(n+1) and get a fraction 1/(n*(n+1)), so the solution is

2/n = 1/n + 1/(n+1) + 1/(n*(n+1))

## C++

 `// CPP program to find x y z that ` `// satisfies 2/n = 1/x + 1/y + 1/z... ` `#include ` `using` `namespace` `std; ` ` `  `// function to find x y and z that ` `// satisfy given equation. ` `void` `printXYZ(``int` `n) ` `{ ` `    ``if` `(n == 1) ` `        ``cout << -1; ` ` `  `    ``else` `        ``cout << ``"x is "` `<< n << ``"\ny is "` `              ``<< n + 1 << ``"\nz is "` `              ``<< n * (n + 1); ` `} ` ` `  `// driver program to test the above function ` `int` `main() ` `{ ` `    ``int` `n = 7; ` `    ``printXYZ(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find x y z that ` `// satisfies 2/n = 1/x + 1/y + 1/z... ` `import` `java.io.*; ` ` `  `class` `Sums { ` `    ``// function to find x y and z that ` `    ``// satisfy given equation. ` `    ``static` `void` `printXYZ(``int` `n){ ` `        ``if` `(n == ``1``) ` `            ``System.out.println(-``1``); ` `        ``else``{ ` `        ``System.out.println(``"x is "``+ n); ` `        ``System.out.println(``"y is "``+ (n+``1``)); ` `        ``System.out.println(``"z is "``+ (n * (n + ``1``))); ` `        ``} ` `    ``} ` `    ``// Driver program to test the above function ` `    ``public` `static` `void` `main (String[] args) { ` `        ``int` `n = ``7``; ` `        ``printXYZ(n); ` `    ``} ` `} ` ` `  `// This code is contributed by Chinmoy Lenka `

## Python3

 `# Python3 code to find x y z that ` `# satisfies 2/n = 1/x + 1/y + 1/z... ` ` `  `# function to find x y and z that ` `# satisfy given equation. ` `def` `printXYZ( n ): ` `    ``if` `n ``=``=` `1``: ` `        ``print``(``-``1``) ` `    ``else``: ` `        ``print``(``"x is "` `, n ) ` `        ``print``(``"y is "` `,n ``+` `1``) ` `        ``print``(``"z is "` `,n ``*` `(n ``+` `1``)) ` ` `  `# driver code to test the above function ` `n ``=` `7` `printXYZ(n) ` ` `  `# This code is contributed by "Sharad_Bhardwaj". `

## C#

 `// C# program to find x y z that ` `// satisfies 2/n = 1/x + 1/y + 1/z... ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `    ``// function to find x y and z that ` `    ``// satisfy given equation. ` `    ``static` `void` `printXYZ(``int` `n) ` `    ``{ ` `        ``if` `(n == 1) ` `            ``Console.WriteLine(-1); ` `        ``else` `        ``{ ` `            ``Console.WriteLine(``"x is "``+ n); ` `            ``Console.WriteLine(``"y is "``+ (n+1)); ` `            ``Console.WriteLine(``"z is "``+ (n * (n + 1))); ` `        ``} ` `    ``} ` `     `  `    ``// Driver program  ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `n = 7; ` `        ``printXYZ(n); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m `

## PHP

 ` `

Output:

```x is 7
y is 8
z is 56
```

Time complexity: O(1)

Alternate Solution
We can write 2/n = 1/n + 1/n. And further as 2/n = 1/n + 1/2n + 1/2n.

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Improved By : vt_m