A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 -y^2 = 2000^2$ I found this answer here on AoPS. I agree with the answer till it multiplies $49$ by $2$. I think it should be multiplied by $4$ since there are $4$ possible cases:
1) $x+y, x-y$ is positive.
2) $x+y$ is neg and $x-y$ is pos.
3) $x+y$ is pos and $x-y$ is neg.
4) $x+y,x-y$ are neg.
So, I think the answer must be $49 \cdot 4 = 196$.
Consider, the case $x+y=4000$ and $x-y=1000$. There are 4 solutions. $(2500, 1500), (-2500, 1500), (2500, -1500), (-2500, -1500)$.
Am I right? If not, can you highlight my fault and an intuitive understanding to the question?

Take the simpler case $x^2-y^2=(x-y)(x+y)=20^2=2^4\cdot 5^2$  We give a factor $2$ to each of $x,y$ then distribute the factors of $2^2 \cdot 5^2$ to $x+y,x-y$.  There are $9$ of them:  $1,2,4,5,10,20,25,50,100$  The case $x-y=8, x+y=50$ gives $x=29, y=21$.  You cannot separately pick the signs of $x+y$ and $x-y$, because they need to be of the same sign.  There are only two cases, not four.
