Number of non-integer solutions in an diophantine equation of order 2

Question

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?

Answer 1

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 $4$.

Answer 2

$2015=5\cdot13\cdot31$ 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 $31^2+248^2=5\cdot13\cdot31^2=62465$.