A Diophantine Equation Revisited

Question

This question relates to the one I just asked several hours ago: Find $x+y$ if $x,y \in \mathbb{N}$ and $x^2+y^2 = 2019^2$. There is a short cut to this equation but I haven’t found it yet. I tried estimate $x+y$ but it has lots of possible outcomes. Any one sees the trick here? Thank you in advance.

Answer 1

We have $2019=3\cdot 673$ and $0^2+3^2=3^2$, $$ 385^2+552^2=673^2 $$ by using the formulas here. Now we can use two square theorem by Brahmagupta to solve $2019^2=x^2+y^2$.