List all the pairs $(x,y)$ s.t. $x^2 - y^2 = 2020$

Question

List all of the pairs $(x,y) \in \mathbb{Z}^2$ s.t. $^2 - ^2 = 2020$.

The prime factorization of $2020$ is $2^2 \cdot 5 \cdot 101$. I used the fact that there exists a solution $(x,y) \in \mathbb{Z}^2$ to the Diophantine equation $x^2 - y^2 = n$ if and only if $n$ is odd or $n$ is a multiple of $4$. Since $n$ is a multiple of $4$, there exist a solution.

I know there is a solution but I am neither able to obtain the number of solution nor able to get the list all the pairs $(x,y)$.

Answer 1

Hint

$x+y,x-y$ have the same parity as $x+y+x-y$ or $x+y-(x-y)$ is even

Now as $2020$ is even, both must be even

$$\implies\dfrac{x+y}2\cdot\dfrac{x-y}2=\dfrac{2020}4$$

Now if $x,y>0$ $$x+y>x-y$$

As $505=1\cdot505=5\cdot101,$

$\dfrac{x-y}2=1$ or $5$

Answer 2

$x^2 - y^2 = (x+y)(x-y) = 2020 = 2^2 \cdot 5 \cdot 101$

So $$(x+y) = 202$$

(think about it)

and $$(x-y) = 10$$

Yielding $x=106$ and $y = 96$ as one pair. Follow the logic to find the other(s) pair(s), i.e., other solutions to $x+y=...$.

Answer 3

Another possibility, besides $(x,y)=(106,96)$ listed in another answer,

is $x+y=1010$ and $x-y=2$, yielding $x=506$ and $y=504$.

Of course, if $(x,y)$ is a solution, then so is $(-x,y)$, $(x,-y)$, and $(-x,-y)$.