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. C.

Let $u=2+\sqrt{7}, v=2-\sqrt{7}$. Then $u+v=4, u v=-3$,
knowing that $u, v$ are the two roots of the equation $x^{2}=4 x+3$.
Therefore, $u^{2}=4 u+3, v^{2}=4 v+3$.
Thus, when $n \geqslant 2$,
$$
u^{n}=4 u^{n-1}+3 u^{n-2}, v^{n}=4 v^{n-1}+3 v^{n-2} \text {. }
$$

Let $S_{n}=u^{n}+v^{n}$. Then when $n \geqslant 2$,
$$
S_{n}=S_{n-1}+S_{n-2}, S_{0}=2, S_{1}=4 \text {. }
$$

Hence, all $S_{n}$ are even numbers. Therefore,
$$
\begin{array}{l}
(\sqrt{7}+2)^{2 n+1}-(\sqrt{7}-2)^{2 n+1} \\
=u^{2 n+1}+v^{2 n+1}=S_{2 n+1}=2 k \\
\Rightarrow(\sqrt{7}+2)^{2 n+1}=2 k+(\sqrt{7}-2)^{2 n+1} . \\
\text { Also } 0<(\sqrt{7}-2)^{2 n+1}<1 \text {, so }(\sqrt{7}-2)^{2 n+1} \text { is the }
\end{array}
$$
decimal part of $a_{n}$, i.e., $b_{n}=(\sqrt{7}-2)^{2 n+1}$. Hence
$$
\begin{array}{l}
a_{n} b_{n}=(\sqrt{7}+2)^{2 n+1}(\sqrt{7}-2)^{2 n+1} \\
=3^{2 n+1} \text { (an odd number). }
\end{array}
$$