Return two prime numbers

Given an even number ( greater than 2 ), return two prime numbers whose sum will be equal to given number.

NOTE A solution will always exist. read Goldbach’s conjecture

Example:

Input : 4
Output: 2 + 2 = 4

If there are more than one solutions possible, return the lexicographically smaller solution.

If [a, b] is one solution with a <= b,
and [c,d] is another solution with c <= d, then

[a, b] < [c, d] 

If a < c OR a==c AND b < d. 

Approach : Use Sieve of Eratosthenes to find all the prime numbers and then check if i and n - i is set, return i & n - i.
*/

 #include <iostream>  
 #include <cmath>  
 using namespace std;  
 int *allPrimes(int N){  
   int *arr = new int[N + 1];  
   arr[0] = 0;  
   arr[1] = 0;  
   for(int i = 2; i <= N; i++){  
     arr[i] = 1;  
   }  
   for(int i = 2; i <= sqrt(N); i++){  
     if(arr[i]){  
       for(int j = i * i; j <= N; j = j + i){  
         arr[j] = 0;  
       }  
     }  
   }  
   return arr;  
 }  
 int main(){  
   //should be prime   
   int N = 20;  
   int *arr = new int[N + 1];  
   arr = allPrimes(N);  
   for(int i = 2; i <= N; i++){  
     if(arr[i] && arr[N - i]){  
       cout << i << " " << N - i << endl;  
     }  
   }  
   return 0;  
 }