Determinant value of $2 \times 2$ matrices Let $a,b,c,d$ be integers such that $\dfrac ac \in \mathbb Q^+$\ $\mathbb Z^+ $ and $\dfrac bd \in \mathbb Q^- $ \ $ \mathbb Z^-$ ; then how many solutions does $|ad-bc|=1$ have ?
 A: The requirements on $a/c$ and $b/d$ forbid any of $a, b, c, d$ being zero.
Assuming $ad - bc = \pm 1$  
i) Assuming further $a > 0$ and $b > 0$ then $c > 0$ to keep $a/c$ positive then 
$$
d = (\pm 1 + bc)/a
$$ 
and because $bc \ge 1$ there is no way to have $d < 0$ as required by negative $b/d$.
ii) Assuming further $a > 0$ and $b < 0$ then $c > 0$ to keep $a/c$ positive and we have
$$
d = (\pm 1 - |b| c) /a
$$ 
which can not be positive for integer $b$, $c$ (we have $|b|c \ge 1$)
iii) Assuming further $a < 0$ and $b > 0$ then $c < 0$ to keep $a/c$ positive then 
$$
d = (\pm 1 - b|c|)/a
$$ 
and we need $\pm 1 - b|c| > 0$ to have a negative $d$. As $b|c| \ge 1$ this will not be possible.
iv) Assuming further $a < 0$ and $b < 0$ then $c < 0$ to keep $a/c$ positive and we have
$$
d = (\pm 1 + |b| |c|) /a
$$
which we need positive to keep $b/d$ negative, which is not possible with negative $a$.
So in case I missed no case there is no case for a solution.
A: The determinant $|ad-bc|$ represents the area of the parallelogram with vertices:
$$
(0,0),\  (a,c),\ (a+b,c+d),\ (b,d).
$$
Without loss of generality you can supppose $a>0$ (otherwise change sign to all numbers) and hence $c>0$. Also you can suppose $b<0$ (otherwise swap $b$ with $d$ and $a$ with $c$) and hence $d>0$. So $a\ge 1$, $c\ge 1$, $b\le 1$ and $d\ge 1$ mean that the point with coordinates $(0,1)$ is an internal point to the parallelogram. So you have at least one internal point and 4 vertex points with integer coordinates, hence by Pick's theorem the area is at least $1+4/2-1=2$. This means: no solution.
edit. Actually it seems to me that the determinant is $\ge 4$.
