Example 2 Given that $a$, $b$, and $c$ are all integers, and for all real numbers $x$, we have
$$
(x-a)(x-2005)-2=(x-b)(x-c)
$$

holds. Find all such ordered triples $(a, b, c)$.
Analysis: $(x-a)(x-2005)-2=(x-b)$. $(x-c)$ always holds, which means $x^{2}-(a+2005) x+$ $2005 a-2=(x-b)(x-c)$ always holds, indicating that
$$
x^{2}-(a+2005) x+2005 a-2=0
$$

has two integer roots $b$ and $c$.
Given by the problem, we have
$$
\begin{array}{l}
x^{2}-(a+2005) x+2005 a-2 \\
=(x-b)(x-c)
\end{array}
$$

always holds, hence
$$
x^{2}-(a+2005) x+2005 a-2=0
$$

has two integer roots $b$ and $c$. Therefore,
$$
\begin{array}{l}
\Delta=(a+2005)^{2}-4(2005 a-2) \\
=(a-2005)^{2}+8
\end{array}
$$

is a perfect square. Let it be $n^{2}$, where $n$ is a positive integer, then
$$
(n-a+2005)(n+a-2005)=8 \text { . }
$$

Since $n-a+2005$ and $n+a-2005$ have the same parity and are both greater than 0, we have
$$
\left\{\begin{array}{l}
n-a+2005=2, \\
n+a-2005=4
\end{array} \text { or } \left\{\begin{array}{l}
n-a+2005=4, \\
n+a-2005=2 .
\end{array}\right.\right.
$$

Solving these, we get $\left\{\begin{array}{l}a=2006, \\ n=3\end{array}\right.$ or $\left\{\begin{array}{l}a=2004, \\ n=3 .\end{array}\right.$
When $a=2004$, the two roots of the equation are $\frac{(a+2005) \pm 3}{2}$, i.e., $(b, c)=(2003,2006)$, (2006,2003);

When $a=2006$, $(b, c)=(2004,2007)$, (2007,2004).
Therefore, the ordered triples $(a, b, c)$ that satisfy the conditions are
$$
\begin{array}{l}
(2004,2003,2006),(2004,2006,2003), \\
(2006,2004,2007),(2006,2007,2004) .
\end{array}
$$