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?

An even number that's a square is always a multiple of $4$. An odd number that's a square is always one larger than a multiple of $4$. So the sum of two perfect squares is always either 


*

*A multiple of four (if they are both even)

*One larger than a multiple of four (if one of them is odd)

*Two larger than a multiple of four (if they're both odd)


$2015$ is none of these, since it's three larger than a multiple of four ($2012$).
Or, said more concisely, consider the equation modulo $4$.
