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.
 A: Without loss of generality, let's say $x < y$. Thus, $2x^2 < x^2+y^2=2017^3$, meaning $x < \sqrt{\frac{2017^3}{2}} < 64054$. Thus, just test all of the integers $x$ from $x=0$ to $x=64053$. Here's the Python code for this problem and the program's output.
>>> from math import sqrt
... for x in range(64054):
...     y = int(sqrt(2017**3-x**2))
...     if x*x*+y*y == 2017**3: print(x, y)

18153 88748
51543 74492

Thus, we have:
$$(x, y) \in \{(\pm 18153,\pm 88748), (\pm 51543,\pm 74492), (\pm 74492,\pm 51543), (\pm 88748,\pm 18153)\}$$
A: Noble Mushtak's solution is a quick way to solve this problem completely using a computer. What I will try to present here is a method for finding a solution that can be done a little more simply.
We know that $2017$ is a prime number. Thus, the factorization of $2017^3$ is just that: $2017^3$.
By Fermat's theorem on two squares, $2017$ is expressible as the sum of two squares, since $2017\equiv 1 (\operatorname{mod} 4)$. This can be done by hand relatively easily: the only solution to $a^2+b^2=2017$, excluding trivial rearrangments, is:
$9^2+44^2=2017$.
Consider $(2017\cdot 9)^2+(2017\cdot 44)^2$. We can factor out $2017^2$, so this gives
$2017^2(9^2+44^2)$,
and knowing the value of the sum of squares, this is just
$2017^3$.
Thus, we have quickly found a solution
$(2017\cdot 9)^2+(2017\cdot 44)^2=2017^3$,
matching up with one of the solutions the programs gave.
There are other primitive solutions to this equation; this method won't give them, but it is a quick way to solve the problem by hand. A consequence of this is that given any integer $n$ such that there exists a Diophantine solution to
$a^2+b^2=n$,
$n^3$ can be decomposed as well (and in an easily constructible way).
A: 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);
}

A: For future reference, if you just enter

m^2 + n^2 = (2017)^3

into Mathematica (or WolframAlpha) then it returns all integer solutions:

A: I want to share with everyone that I just found an algebraic method to calculate primitive triples in my specific case of raising to the third power:
$$[x(x^2-3y^2)]^2+[y(3x^2-y^2)]^2=n^3$$
I do not know why but on specialist internet sites this is not often found.
Happy new year for all.
