(5) Let $a_{n}=(2+\sqrt{7})^{2 n+1}$, $b_{n}$ be the fractional part of $a_{n}$, then when $n \in \mathbf{N}^{*}$, the value of $a_{n} b_{n}$ ( ).
(A) must be an irrational number
(B) must be an even number
(C) must be an odd number
(D) can be either an irrational number or a rational number
5 Let $u=2+\sqrt{7}, v=2-\sqrt{7}$, then $u+v=4, u v=-3$, which means $u, v$ are the roots of the equation $x^{2}=4 x+3$, thus
$$
u^{2}=4 u+3, v^{2}=4 v+3,
$$

Therefore, when $n \geqslant 2$,
$$
u^{n}=4 u^{n-1}+3 u^{n-2}, v^{n}=4 v^{n-1}+3 v^{n-2} .
$$

Let $S_{n}=u^{n}+v^{n}$, then when $n \geqslant 2$, $S_{n}=4 S_{n-1}+3 S_{n-2}, S_{0}=2, S_{1}=4$, hence all $S_{n}$ are even numbers.
$$
\begin{array}{c}
(\sqrt{7}+2)^{2 n+1}-(\sqrt{7}-2)^{2 n+1}=u^{2 n+1}+v^{2 n+1}=S_{2 n+1}=2 k, \\
(\sqrt{7}+2)^{2 n+1}=2 k+(\sqrt{7}-2)^{2 n+1},
\end{array}
$$

Since
$$
0<(\sqrt{7}-2)^{2 n+1}<1,
$$

$(\sqrt{7}-2)^{2 n+1}$ is the fractional part of $a_{n}$, that is
$$
\begin{array}{c}
b_{n}=(\sqrt{7}-2)^{2 n+1}, \\
a_{n} b_{n}=(\sqrt{7}+2)^{2 n+1} \cdot(\sqrt{7}-2)^{2 n+1}=3^{2 n+1}
\end{array}
$$

is an odd number. Therefore, the answer is $\mathrm{C}$.