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.)?
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 $l \neq 0$,
$$
0 &lt;k^2+l^2 = (ac-bd)^2+(ab+cd)^2= (a^2+b^2)(c^2+d^2),
$$
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.
For the system of equations:

$$\left\{\begin{aligned}&amp;ac-bd=q\\&amp;ad-bc=t\end{aligned}\right.$$

$k$ - choose an integer, so that the bracket was intact.

Then the solutions are.

$$a=q+k^2-\frac{k}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$

$$b=q+k(k+1)-\frac{k}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$

$$c=q+k(k+2)+1-\frac{(k+1)}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$

$$d=q+k(k+1)-\frac{(k+1)}{2}(2k+1+\frac{t+(2k+1)q}{k(k+1)})$$