the New Year will be 2017: how many pairs of integer solutions to $x^2 + y^2 = (2017)^3$? We're almost in 2017. I wonder how many pairs of integer solutions has the following diophantine equation:
$$x^2 + y^2 = (2017)^3$$
Thanks in advance.

Borrowing from this answer here
The answer is just $\frac{3+1}{2}=2$. ( or $4$ if the order matters).
I checked it with this code:
#include <bits/stdc++.h>
using namespace std;
typedef long long lli;

lli N=2017;

int isq(lli N){
    lli s=sqrt(N);
    if(s*s==N) return(1);
    return(0);
}

int main(){
    N=N*N*N;
    int res=0;
    for(lli i=0;i*i<N;i++){
        if(isq(N-i*i) ) res++;
    }
    printf("%d\n",res);
}

