Count pairs (a, b) whose sum of cubes is N (a^3 + b^3 = N) in C++

We are given a number N. The goal is to find ordered pairs of positive numbers such that the sum of their cubes is N.

We will do this by finding solutions to the equation a³ + b³ = N. Where a is not more than cube root of N and b can be calculated as cube root of (N-a³).

Let’s understand with examples.

Input 

N=35

Output 

Count of pairs of (a,b) where a^3+b^3=N: 2

Explanation 

Pairs will be (2,3) and (3,2). 23+33=8+27=35

Input 

N=100

Output 

Count of pairs of (a,b) where a^3+b^3=N: 0

Explanation 

No such pairs possible.

Approach used in the below program is as follows

• We take integer N.

• Function cubeSum(int n) takes n and returns the count of ordered pairs with sum of cubes as n.

• Take the initial variable count as 0 for pairs.

• Traverse using for loop to find a.

• Start from a=1 to a<=cbrt(n) which is cube root of n.

• Calculate cube of b as n-pow(a,3).

• Calculate b as cbrt(bcube)

• If pow(b,3)==bcube. Increment count by 1.

• At the end of all loops count will have a total number of such pairs.

• Return the count as result.

Example

 Live Demo

#include <bits/stdc++.h>
#include <math.h>
using namespace std;
int cubeSum(int n){
   int count = 0;
   for (int a = 1; a < cbrt(n); a++){
      int bcube=n - (pow(a,3));
      int b = cbrt(bcube);
      if(pow(b,3)==bcube)
         { count++; }
   }
   return count;
}
int main(){
   int N = 35;
   cout <<"Count of pairs of (a,b) where a^3+b^3=N: "<<cubeSum(N);
   return 0;
}

Output

If we run the above code it will generate the following output −

Count of pairs of (a,b) where a^3+b^3=N: 2