(2) Express 2008 as the difference of squares of two integers, then the number of different ways to do this is ( .
(A) 4
(B) 6
(C) 8
(D) 16
(2) Let $x^{2}-y^{2}=2008$, which means $(x+y)(x-y)=2008.2008$ has 8 positive divisors, which are $1, 2, 4, 8, 251, 502, 1004, 2008$. Moreover, $(x+y)$ and $(x-y)$ can only be even numbers at the same time, so the corresponding system of equations is
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y=-2, \\
x-y=-1004,
\end{array}\right. \\
\left\{\begin{array}{l}
x+y=-4, \\
x-y=-502,
\end{array}\right. \\
\left\{\begin{array}{l}
x+y=-502, \\
x-y=-4,
\end{array}\right. \\
\left\{\begin{array}{l}
x+y=-1004, \\
x-y=-2,
\end{array}\right. \\
\left\{\begin{array}{l}
x+y=2, \\
x-y=1004,
\end{array}\right. \\
\left\{\begin{array}{l}
x+y=4, \\
x-y=502,
\end{array}\right. \\
\left\{\begin{array}{l}
x+y=502, \\
x-y=4,
\end{array}\right. \\
\left\{\begin{array}{l}
x+y=1004 \\
x-y=2 .
\end{array}\right. \\
\end{array}
$$

Therefore, $(x, y)$ has 8 different sets of values: $(503,501),(-503,-501),(-503, 501),(503,-501),(253,249),(-253,-249),(-253,249),(253, - 249)$. Hence, the answer is C.