2. Given that $n^{2}+5 n+13$ is: a perfect square. Then the natural number $n$ ( )
(A) does not exist
(B) has only one
(C) has more than one, but a finite number
(D) has infinitely many
2. (B).
$$
\begin{array}{l}
\because(n+2)^{2}<n^{2}+5 n+13<(n+4)^{2}, \\
\therefore n^{2}+5 n+13=(n+3)^{2} .
\end{array}
$$

Solving gives $n=4$.