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Count pairs (a, b) whose sum of cubes is N (a^3 + b^3 = N)
• Difficulty Level : Easy
• Last Updated : 27 Apr, 2021

Given N, count all ‘a’ and ‘b’ that satisfy the condition a^3 + b^3 = N.
Examples:

```Input : N = 9
Output : 2
1^3 + 2^3 = 9
2^3 + 1^3 = 9

Input : N = 28
Output : 2
1^3 + 3^3 = 28
3^3 + 1^3 = 28```

Note:- (a, b) and (b, a) are to be considered as two different pairs.

```Implementation:
Travers numbers from 1 to cube root of N.
a) Subtract cube of current number from
N and check if their difference is a
perfect cube or not.
i) If perfect cube then increment count.

2- Return count.```

Below is the implementation of above approach:

## C++

 `// C++ program to count pairs whose sum``// cubes is N``#include``using` `namespace` `std;` `// Function to count the pairs satisfying``// a ^ 3 + b ^ 3 = N``int` `countPairs(``int` `N)``{``    ``int` `count = 0;` `    ``// Check for each number 1 to cbrt(N)``    ``for` `(``int` `i = 1; i <= cbrt(N); i++)``    ``{``        ``// Store cube of a number``        ``int` `cb = i*i*i;` `        ``// Subtract the cube from given N``        ``int` `diff = N - cb;` `        ``// Check if the difference is also``        ``// a perfect cube``        ``int` `cbrtDiff = cbrt(diff);` `        ``// If yes, then increment count``        ``if` `(cbrtDiff*cbrtDiff*cbrtDiff == diff)``            ``count++;``    ``}` `    ``// Return count``    ``return` `count;``}` `// Driver program``int` `main()``{``    ``// Loop to Count no. of pairs satisfying``    ``// a ^ 3 + b ^ 3 = i for N = 1 to 10``    ``for` `(``int` `i = 1; i<= 10; i++)``        ``cout << ``"For n = "` `<< i << ``", "``             ``<< countPairs(i) <<``" pair exists\n"``;` `    ``return` `0;``}`

## Java

 `// Java program to count pairs whose sum``// cubes is N` `class` `Test``{``    ``// method to count the pairs satisfying``    ``// a ^ 3 + b ^ 3 = N``    ``static` `int` `countPairs(``int` `N)``    ``{``        ``int` `count = ``0``;``     ` `        ``// Check for each number 1 to cbrt(N)``        ``for` `(``int` `i = ``1``; i <= Math.cbrt(N); i++)``        ``{``            ``// Store cube of a number``            ``int` `cb = i*i*i;``     ` `            ``// Subtract the cube from given N``            ``int` `diff = N - cb;``     ` `            ``// Check if the difference is also``            ``// a perfect cube``            ``int` `cbrtDiff = (``int``) Math.cbrt(diff);``     ` `            ``// If yes, then increment count``            ``if` `(cbrtDiff*cbrtDiff*cbrtDiff == diff)``                ``count++;``        ``}``     ` `        ``// Return count``        ``return` `count;``    ``}``    ` `    ``// Driver method``    ``public` `static` `void` `main(String args[])``    ``{``        ``// Loop to Count no. of pairs satisfying``        ``// a ^ 3 + b ^ 3 = i for N = 1 to 10``        ``for` `(``int` `i = ``1``; i<= ``10``; i++)``            ``System.out.println(``"For n = "` `+ i + ``", "` `+``                     ``+ countPairs(i) + ``" pair exists"``);``    ``}``}`

## Python 3

 `# Python 3 program to count pairs``# whose sum cubes is N``import` `math` `# Function to count the pairs``# satisfying a ^ 3 + b ^ 3 = N``def` `countPairs(N):` `    ``count ``=` `0` `    ``# Check for each number 1``    ``# to cbrt(N)``    ``for` `i ``in` `range``(``1``, ``int``(math.``pow``(N, ``1``/``3``) ``+` `1``)):``    ` `        ``# Store cube of a number``        ``cb ``=` `i ``*` `i ``*` `i` `        ``# Subtract the cube from given N``        ``diff ``=` `N ``-` `cb` `        ``# Check if the difference is also``        ``# a perfect cube``        ``cbrtDiff ``=` `int``(math.``pow``(diff, ``1``/``3``))` `        ``# If yes, then increment count``        ``if` `(cbrtDiff ``*` `cbrtDiff ``*` `cbrtDiff ``=``=` `diff):``            ``count ``+``=` `1``    ` `    ``# Return count``    ``return` `count`  `# Driver program` `# Loop to Count no. of pairs satisfying``# a ^ 3 + b ^ 3 = i for N = 1 to 10``for` `i ``in` `range``(``1``, ``11``):``    ``print``(``'For n = '``, i, ``', '``, countPairs(i),``                                ``' pair exists'``)`  `# This code is contributed by Smitha.`

## C#

 `// C# program to count pairs whose sum``// cubes is N`` ` `using` `System;``class` `Test``{``    ``// method to count the pairs satisfying``    ``// a ^ 3 + b ^ 3 = N``    ``static` `int` `countPairs(``int` `N)``    ``{``        ``int` `count = 0;``      ` `        ``// Check for each number 1 to cbrt(N)``        ``for` `(``int` `i = 1; i <= Math.Pow(N,(1.0/3.0)); i++)``        ``{``            ``// Store cube of a number``            ``int` `cb = i*i*i;``      ` `            ``// Subtract the cube from given N``            ``int` `diff = N - cb;``      ` `            ``// Check if the difference is also``            ``// a perfect cube``            ``int` `cbrtDiff = (``int``) Math.Pow(diff,(1.0/3.0));``      ` `            ``// If yes, then increment count``            ``if` `(cbrtDiff*cbrtDiff*cbrtDiff == diff)``                ``count++;``        ``}``      ` `        ``// Return count``        ``return` `count;``    ``}``     ` `    ``// Driver method``    ``public` `static` `void` `Main()``    ``{``        ``// Loop to Count no. of pairs satisfying``        ``// a ^ 3 + b ^ 3 = i for N = 1 to 10``        ``for` `(``int` `i = 1; i<= 10; i++)``            ``Console.Write(``"For n = "` `+ i + ``", "` `+``                     ``+ countPairs(i) + ``" pair exists"``+``"\n"``);``    ``}``}`

## PHP

 ``

## Javascript

 ``

Output:

```For n= 1, 1 pair exists
For n= 2, 1 pair exists
For n= 3, 0 pair exists
For n= 4, 0 pair exists
For n= 5, 0 pair exists
For n= 6, 0 pair exists
For n= 7, 0 pair exists
For n= 8, 1 pair exists
For n= 9, 2 pair exists
For n= 10, 0 pair exists```

Reference: https://www.careercup.com/question?id=5954491572551680
This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.