<s>[INST] List all the pairs $(x,y)$ s.t. $x^2 - y^2 = 2020$ 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 $(x,y)$.
 [/INST] 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 $5$
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