6. The integer part and the fractional part of $(\sqrt{5}+2)^{2 n+1}(n \in \mathbf{N})$ are denoted as $P$ and $Q$, respectively, then $Q(P+Q)=$
$\qquad$
6. 1 Detailed Explanation: Construct $(\sqrt{5}-2)^{2 n+1}$. Since $(\sqrt{5}-2)^{2 n+1} \cdot(\sqrt{5}+2)^{2 n+1}=1$, then $(\sqrt{5}-2)^{2 n+1}=\frac{1}{P+Q}$. By the binomial theorem, $(\sqrt{5}+2)^{2 n+1}-(\sqrt{5}-2)^{2 n+1}$ is an integer, i.e., $P+Q-\frac{1}{P+Q}$ is an integer, thus $Q-\frac{1}{P+Q}$ is an integer. Since $0<Q<1,0<\frac{1}{P+Q}<1$, it follows that $Q-\frac{1}{P+Q}=0$, hence $Q(P+Q)=1$.