8.8. (Austria - PDR, 79). For each value of $n \in \mathbf{N}$, find the greatest value of $k \in \mathbf{Z}^{+}$, for which the number $\left[(3+\sqrt{11})^{2 n-1}\right]$ is divisible by $2^{k}$.
8.8. Let

$$
a_{n}=(3+\sqrt{11})^{n}+(3-\sqrt{11})^{n}
$$

then the following equality holds:

$$
a_{n+2}=6 a_{n+1}+2 a_{n}, \quad n \in \mathbf{Z}^{+}
$$

Indeed, let $\alpha=(3+\sqrt{11})^{n}, \beta=(3-\sqrt{11})^{n}$, then

$$
\begin{gathered}
a_{n}=\alpha+\beta \\
a_{n+1}=(3+\sqrt{11}) \alpha+(3-\sqrt{11}) \beta \\
a_{n+2}=(3+\sqrt{11})^{2} \alpha+(3-\sqrt{11})^{2} \beta=(20+6 \sqrt{11}) \alpha+(20-6 \sqrt{11}) \beta= \\
=(18+6 \sqrt{11}) \alpha+(18-6 \sqrt{11}) \beta+(2 \alpha+2 \beta)=6 a_{n+1}+2 a_{n}
\end{gathered}
$$

From this and the equalities $a_{0}=2, a_{1}=6$ it follows that the number $a_{n}$ is an integer for any $n \in Z^{+}$. Since $-1<3-\sqrt{11}<0$, then for $n \in N$ we have

$a_{2,-1}=(3+\sqrt{11})^{2 n-1}+(3-\sqrt{11})^{2 n-1}<(3+\sqrt{11})^{2 n-1}<a_{2 n-1}+1$, i.e.

$$
a_{2 n-1}=\left[(3+\sqrt{11})^{2 n-1}\right]
$$

Now we will prove by induction on $n \in \mathbf{N}$ the following statement: the number $a_{2,-2}$ is divisible by $2^{n}$, and the number $a_{2,2-1}$ is divisible by $2^{n}$, but not by $2^{n+1}$. For $n=1$ the statement is true, since $a_{0}=2, a_{1}=6$. Suppose it is true for some value $n \in N$. Then the number $a_{2 l}=$ $=6 a_{2 n-1}+2 a_{2 n-2}$ is divisible by $2^{n+1}$, since both numbers $a_{2 n-1}$ and $a_{2 n}$ are divisible by $2^{n}$. Further, the number $a_{2 n+1}=6 a_{2 n}+2 a_{2 n-1}$ is divisible by $2^{n+1}$ (since both numbers $a_{2 n}$ and $a_{2 n-1}$ are divisible by $2^{n}$), but does not divide $2^{2 n+2}$ (since the number $a_{22}$ is divisible by $2^{n+1}$, and the number $a_{2 n-1}$ is not divisible). The statement is proved. Thus, the highest power of two by which the number $a_{2 n-1}$ is divisible is $2^{n}$, i.e. $k=n$.