Determine the number of solutions in N² of the equation:
x² + y² + 2xy-2019x-2019y-2020 = 0
Factor it. You should find either 2019 or 2020 solutions in $\mathbb{N}^2$ depending on whether zero is a natural number or not. Some books say it is, some say it is not.
\begin{align}(x+y)^2 - 2019(x+y) - 2020 &= 0 &\text{now set $u=x+y$ for clarity} \\ u^2 - 2019u -2020 &= 0 \\ (u-2020)(u+1) &= 0 &\text{and replace $u=x+y$} \\ (x+y-2020)(x+y+1) &= 0\end{align}
That's two solutions to be interpreted.