# Thread: the number of solutions in N²

1. ## the number of solutions in N²

Determine the number of solutions in N² of the equation:
x² + y² + 2xy-2019x-2019y-2020 = 0

2. ## Re: the number of solutions in N²

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.

3. ## Re: the number of solutions in N²

Originally Posted by SlipEternal
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.
Thanks ...if zero is a natural number then you have" n" solutions whether zero is not a natural number "n-1" solutions..

4. ## Re: the number of solutions in N²

\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.

5. ## Re: the number of solutions in N²

Originally Posted by Archie
\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.
It is one solution for $u$ (over natural numbers) but 2019 or 2020 over $x,y$. Example: (1,2019), (2,2018), (3,2017), etc. are all distinct solutions over the natural numbers.