1. The positive integer pairs $(x, y)$ that satisfy $y=\sqrt{x+1}+\sqrt{x+2005}$ (   ).
(A) There is only one pair
(B) There are exactly two pairs
(C) There are at least three pairs
(D) There are none
-.1.B.
Let $a^{2}=x+1, b^{2}=x+2005$, where $a, b$ are both natural numbers. Then
$$
\begin{array}{l}
y=a+b, \\
b^{2}-a^{2}=(b+a)(b-a)=2004=2^{2} \times 3 \times 167 .
\end{array}
$$

Since $b+a$ and $b-a$ have the same parity, and $b+a$
$$
\begin{array}{l}
>b-a, \text { so, } \\
\left\{\begin{array} { l } 
{ b + a = 1 0 0 2 , } \\
{ b - a = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
b+a=334, \\
b-a=6 .
\end{array}\right.\right. \\
\text { Solving, we get }\left\{\begin{array} { l } 
{ a = 5 0 0 , } \\
{ b = 5 0 2 }
\end{array} \text { or } \left\{\begin{array}{l}
a=164, \\
b=170 .
\end{array}\right.\right.
\end{array}
$$