Three, (20 points) Given that $a, b, c, d$ are different real numbers, and $a, c$ are the roots of the equation $x^{2} + a x - b = 0$, $b, d$ are the roots of the equation $x^{2} + c x + d = 0$. Find the values of $a, b, c, d$.

---

Please note that the translation retains the original formatting and structure of the text.
$$
\begin{array}{l}
a+c=-a, a c=-b ; \\
b+d=-c, b d=d .
\end{array}
$$
(1) If $d \neq 0$, from $b d=d$, we know $b=1$.

From $a+c=-a$, we know $c=-2 a$.
From $a c=-b$, we know $-2 a^{2}=-1$, thus $a= \pm \frac{\sqrt{2}}{2}$.
When $a= \pm \frac{\sqrt{2}}{2}$, $c=\mp \sqrt{2}$, we get
$$
\begin{array}{l}
d=-c-b=\sqrt{2}-1 ; \\
d=-c-b=-\sqrt{2}-1 .
\end{array}
$$

Upon verification,
$$
\begin{array}{l}
(a, b, c, d) \\
=\left(\frac{\sqrt{2}}{2}, 1,-\sqrt{2}, \sqrt{2}-1\right),\left(-\frac{\sqrt{2}}{2}, 1, \sqrt{2},-\sqrt{2}-1\right)
\end{array}
$$

are two sets of solutions that meet the conditions.
(2) If $d=0$, then $b=-c$.
From $a+c=-a$, we know $c=-2 a$.
From $a c=-b$, we know $a c=c$.
If $c=0$, then $a=0$, which contradicts the condition that $a, b, c, d$ are distinct real numbers.

If $c \neq 0$, then $a=1$. From $c=-2 a$, we know $c=-2$, thus, $b=-c=2$.

Upon verification, $(a, b, c, d)=(1,2,-2,0)$ is also a solution that meets the conditions.