$a^2-b^2 = k$, $ab = l$ for fixed integer $k,l$ when $a,b$ are both integers

Question

Let us fix integers $k,l$. Let all numbers be integers.

Now we want integer $a,b$ to satisfy: $$a^2-b^2 = k, \,\,\,2ab = l.$$

We want to maximize the number of possible $(a,b)$. In order to do this, how should I select $k,l$?

In other words, is it possible to select $k,l$ such that the number of possible $(a,b)$ is not limited to some finite number?

Answer 1

Write the first as $4a^2k=4a^4-l^2$ This has at most four solutions.

Answer 2

You can also do it by first determining $a^2, b^2$ using a straightforward quadratic identity.

$$a^2-b^2=k$$

$$(a^2+b^2)^2=(a^2-b^2)^2-4a^2b^2=k^2-l^2$$So that $$a^2+b^2=\sqrt{k^2-l^2}$$[Have to take the positive square root if you are in $\mathbb R$]

This system is easy to solve and you should see that $a^2$ and $b^2$ are determined by $k$ and $l$. Then the only choice is the sign of $a$ and $b$. Remembering that $l$ is fixed, you should find two solutions except when $(k,l)=(0,0)$.