Solve $ab + cd = -1; ac + bd = -1; ad + bc = -1$ over the integers I am trying to solve this problem:

Solve the system of equations
  \begin{align}
\begin{cases}
ab + cd = -1 \\
ac + bd = -1 \\
ad + bc = -1
\end{cases}
\end{align}
  for the integers $a$, $b$, $c$ and $d$.

I have found that the first equation gives $d = \dfrac{-1-cd}{a}$, which gives $a\neq0$. Other than that, I don't know where to start.
Tips, help or solution is very appreciated. Thanks!

Hint:
Squaring the equations gives
\begin{align}
\begin{cases}
a^2b^2 +2abcd+ c^2d^2 = 1 \\
a^2c^2 +2abcd +b^2d^2 = 1 \\
a^2d^2 + 2abcd+b^2c^2 = 1
\end{cases}
\end{align}
Subtracting them yields
\begin{align}
\begin{cases}
(a^2-b^2)(c^2-d^2)=0 \\
(a^2-c^2)(b^2-d^2)=0 \\
(a^2-d^2)(b^2-c^2)=0
\end{cases}
\end{align}
Now consider some cases.
