A system of equations with degree 2 Let $a,b,c,d \in \mathbb R$. Suppose the following holds
\begin{align} 
a^2-c^2 &=1 \\
b^2-d^2 &=-1 \\
ad-bc &= \pm1 \\
ab-cd &=0
\end{align}
How can I find
$a,b,c$ and $d$. I'm trying to show something regarding differential geometry
and arrive with these equations. I tried to solved it but my algebra sucks.  
Any help or hints on how to solved this system of equations would be much appreciated!
 A: From the first two, we can set $$a=\cosh u,c=\sinh u,b=\sinh v,d=\cosh v.$$
The fourth implies
$$\cosh u\sinh v-\sinh u\cosh v=\sinh(v-u)=0\to u=v,$$
and by the third
$$\cosh u\cosh u-\sinh u\sinh u=\cosh(v-u)=\pm1.$$
Only the plus sign is possible, but any $u$ yields a solution.

Update:
Ooops, I forgot possible negative signs.
$$a=\pm\cosh u,c=\pm\sinh u,b=\pm\sinh v,d=\pm\cosh v.$$
The fourth implies
$$\pm\cosh u\sinh v\pm\sinh u\cosh v=\pm\sinh(v\pm u)=0\to u=\pm v,$$
and by the third
$$\pm\cosh u\cosh u\pm\sinh u\sinh u=\pm\cosh(v\pm u)=\pm1.$$
Hence, any $u=\pm v$.
A: Just two by two matrices,
$$
\left(
\begin{array}{cc}
a&c \\
b&d \\
\end{array}
\right)
\left(
\begin{array}{cc}
a&-b \\
-c&d \\
\end{array}
\right) = 
\left(
\begin{array}{cc}
1&0 \\
0&1 \\
\end{array}
\right)
$$
so that the right hand factor must be the inverse of $ M= 
\left(
\begin{array}{cc}
a&c \\
b&d \\
\end{array}
\right)$
When $M$ has determinant $1,$  we are demanding
$$
\left(
\begin{array}{cc}
a&-b \\
-c&d \\
\end{array}
\right) =
\left(
\begin{array}{cc}
d&-c \\
-b&a \\
\end{array}
\right) = M^{-1}
$$
so that $a=d$ and $b=c,$
or
$$
M=
\left(
\begin{array}{cc}
a&c \\
c&a \\
\end{array}
\right)
$$
When $M$ has determinant $-1,$  we are demanding
$$
\left(
\begin{array}{cc}
a&-b \\
-c&d \\
\end{array}
\right) =
\left(
\begin{array}{cc}
-d&c \\
b&-a \\
\end{array}
\right) = M^{-1}
$$
so that $a=-d$ and $b=-c,$
or
$$
M=
\left(
\begin{array}{cc}
a&c \\
-c&-a \\
\end{array}
\right)
$$
One may fill in both versions using $\cosh t$ and $\sinh t$
A: If we add the third equation to the fourth, we get
$$\pm 1 = ab+ad-cb-cd = (a-c)(b+d)$$
If we subtract the third equation from the fourth, we get
$$\mp 1 = ab-ad+cb-cd= (a+c)(b-d)$$
The first two equations are
$$(a+c)(a-c) = 1 \hspace{1cm} (b+d)(b-d)=-1$$
So if we set $w=a-c, x=a+c, y=b-d, z=b+d$, the system becomes
$$wx=1 \hspace{1cm} yz=-1 \hspace{1cm} wz= \pm 1 \hspace{1cm} xy= \mp 1$$
This allows us to write
$$x = \frac1w \hspace{1cm} y = \mp w \hspace{1cm} z = \pm \frac1w$$
and hence
$$a = \frac{w+x}2 = \frac{w+\frac1w}2 \hspace{1cm} b = \frac{y+z}2 = \mp \frac{w-\frac1w}2$$
$$c = \frac{w-x}2 = \frac{w-\frac1w}2 \hspace{1cm} d = \frac{y-z}2 = \mp \frac{w+\frac1w}2$$
Note the resemblance to hyperbolic trig functions, as Richard suggested. If we set $w = e^t$, we get
$$a = \cosh t \hspace{1cm} b = \mp \sinh t \hspace{1cm} c = \sinh t \hspace{1cm} d = \mp\cosh t$$
(we could instead set $w = -e^t$, which negates all four variables and hence still solves the original system).
