elementary number theory - Another approach than Fermat's to find all solutions to $x^2+y^2=2015$

I've used Fermat's theorem on the sum of two squares to solve the following problem:

Find all (if any) solutions to $x^2+y^2=2015$

Let's look at $x^2+y^2=p$ (1)

$p$ is a solution iff $p=2$ or $p \equiv 1 \pmod{4}$

If we divide $2015$ into primes we get:
$5\times13\times31=2015$

And this is where Fermat's theorem kinda comes in..

It can be deduced from Fermat's theorem on sum of two squares that (1) has a solution iff every prime number $p \equiv 3 \pmod{4}$ appears with an even exponent. (2)

We can clearly see that it doesn't and we are finished.

Thus there are no solutions to this problem.

I feel like my attempt at this question falls short since I don't fully comprehend the proof of Fermat's theorem on sum of two square or the special case (2) that can be deduced from the theorem.

If anyone could give me another approach that be much appreciated.

Any comments on my solution would also be nice

One other approach I was thinking about is that $2015$ is odd and thus $(x,y)$ must one be odd and one even.
We know that even squares are even and odd squares are odd and sum of even+odd=odd.

We could thus write the equation on the form $2n+1$ and $2n$ and from here im lost!

Answer

$$2015\equiv3\pmod4$$

For any integer $a, a^2\equiv0,1\pmod4$