A Derived Pythagorean Triple

Question

I again stumbled upon an interesting number theory question: $x^2+y^2 = 2019$ with $x, y \in \mathbb{N}$. This one is derived from another Diophantine equation that I am trying to solve. Is there a way to solve this without trying out all the possible cases for $x = 1$ to $31$ ? I mean any sufficiently advanced or unusual technique or quick insight that could be used? Thank you for your time.

Answer 1

For any $a\in\Bbb{Z}$, if you divide $a^2$ by $4$, then the remainder is either $0$ or $1$. Hence, for any $x,y\in\Bbb{Z}$, if you divide $x^2+y^2$ by $4$, then the remainder must be either $0$, $1$, or $2$. If you divide $2019$ by $4$, then you get a remainder of $3$. It follows that there is no $x,y\in\Bbb{Z}$ with $x^2+y^2=2019$.