Number of non-integer solutions in an diophantine equation of order 2 
Consider the equation $x^2 + y^2 = 2015$ where $x\geq 0$ and $y\geq 0$. hoe many solutions $(x, y)$ exist such that both $x$ and $y$ are non-negative integers?

*

*Greater than two

*Exactly two

*Exactly one

*None


I tried all the combinations of x and y values and found that there are no non-negative integers. Is there a better method to solve it?

$2015=5\cdot13\cdot31$ then $2015$ is not a sum of two squares because the factor $31$ is not congruent to $1$ modulo $4$ and its exponent is not even. For example, if the exponent of $31$ is $2$ instead of $1$ we have $62465$ instead of $2015$ and in this case one has $31^2+248^2=5\cdot13\cdot31^2=62465$.
