Can there ever be infinite number of tuples of $(a,b,c,d)$ such that $ac-bd = k$ and $ad+bc = l$ for fixed $k,l$? Suppose, for now, that all numbers are real numbers. Let us fix numbers $k,l$. 
Then can there ever be infinite number of tuples of $(a,b,c,d)$ such that $ac-bd =k$, $ad+bc = l$ for some $k$ and $l$?
What happens if numbers are integers?
Also, what happens if we change the restriction from integers/real numbers to any commutative numbers (field, ring etc.)?
 A: For real numbers, yes. One infinite family of solutions is for instance
$$
(a,b,c,d) = \left(r\frac{l+k}{2}, r\frac{l-k}{2}, \frac{1}{r}, \frac{1}{r}\right),
$$
for $r \neq 0$.
For integers, if $k = l = 0$, then $(a,b,c,d) = (n,n,0,0)$ gives infinite number of solutions. If $k \neq 0$ or $l \neq 0$,
$$
0 <k^2+l^2 = (ac-bd)^2+(ab+cd)^2= (a^2+b^2)(c^2+d^2),
$$
so $1 \leq a^2+b^2,c^2+d^2$, and
$$
k^2+l^2 = (a^2+b^2)(c^2+d^2) \geq a^2+b^2 \geq a^2,
$$
and similarly $a^2,b^2,c^2,d^2 \leq k^2+l^2$, which implies there are only finitely many solutions.
A: Let $R$ be a commutative ring and let $R[i] = R[x]/(x^2 + 1)$. You want to find solutions to
$$(a + bi)(c + di) = k + \ell i$$
The answer depends strongly on $R$. For example, if $R = \mathbb{Z}$ then you are looking for ways to factor $k + \ell i$ in the Gaussian integers, and there are finitely many such factorizations in the same way that there are finitely many factorizations of integers, which can be counted by finding the prime factorization of $k + \ell i$. 
On the other hand, if $R$ has an infinite group of units (for example, if $R$ is an infinite field), then there are infinitely many solutions with $b = d = \ell = 0$. 
A: For the system of equations:
$$\left\{\begin{aligned}&ac-bd=q\\&ad-bc=t\end{aligned}\right.$$
$k$ - choose an integer, so that the bracket was intact.
Then the solutions are.
$$a=q+k^2-\frac{k}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$
$$b=q+k(k+1)-\frac{k}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$
$$c=q+k(k+2)+1-\frac{(k+1)}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$
$$d=q+k(k+1)-\frac{(k+1)}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$
