8・32 Let $a_{1}=a_{2}=\frac{1}{3}$, and for $n=3,4,5, \cdots$
$$
a_{n}=\frac{\left(1-2 a_{n-2}\right) a_{n-1}^{2}}{2 a_{n-1}^{2}-4 a_{n-2} a_{n-1}^{2}+a_{n-2}} .
$$
(i) Find the general term formula for the sequence $\left\{a_{n}\right\}$.
(ii) Prove: $\frac{1}{a_{n}}-2$ is a perfect square of an integer.
［Solution］It is easy to prove by induction that
$$
00$. From the recursive formula, we get
$$
\frac{1}{a_{n}}=\frac{2 a_{n-1}^{2}-4 a_{n-2} a_{n-1}^{2}+a_{n-2}}{\left(1-2 a_{n-2}\right) a_{n-1}^{2}}=2+\frac{a_{n-2}}{\left(1-2 a_{n-2}\right) a_{n-1}^{2}},
$$

which means $b_{n}=\frac{1}{b_{n-2} a_{n-1}^{2}}=\frac{\left(b_{n-1}+2\right)^{2}}{b_{n-2}}$,
from which we can derive $\sqrt{b_{n} b_{n-2}}=b_{n-1}+2$.
By substituting $n+1$ for $n$, we get $\sqrt{b_{n+1} b_{n-1}}=b_{n}+2$.
Thus, $\sqrt{b_{n+1} b_{n-1}}-\sqrt{b_{n} b_{n-2}}=b_{n}-b_{n-1}$,
and therefore $\sqrt{b_{n-1}}\left(\sqrt{b_{n+1}}+\sqrt{b_{n-1}}\right)=\sqrt{b_{n}}\left(\sqrt{b_{n}}+\sqrt{b_{n-2}}\right)$.
Let $c_{n}=\frac{\sqrt{b_{n}}+\sqrt{b_{n-2}}}{\sqrt{b_{n-1}}}, n=3,4, \cdots$ then
$$
c_{n+1}=c_{n}=\cdots=c_{3}=\frac{\sqrt{b_{3}}+\sqrt{b_{1}}}{\sqrt{b_{2}}} \text {. }
$$

Since $a_{1}=a_{2}=\frac{1}{3}$, we have $b_{1}=b_{2}=1, b_{3}=\frac{\left(b_{2}+2\right)^{2}}{b_{1}}=9$.
Therefore, $c_{n}=4, n=3,4,5, \cdots$ which means
$$
\sqrt{b_{n}}=4 \sqrt{b_{n-1}}-\sqrt{b_{n-2}}, n=3,4,5, \cdots
$$

Since $\sqrt{b_{1}}=\sqrt{b_{2}}=1$, it is easy to prove by induction that $\sqrt{b_{n}}$ is a positive integer, i.e., $b_{n}=\frac{1}{a_{n}}-2$ is a perfect square, proving the conclusion of (ii).

The characteristic equation of the recursive relation (1) is $\lambda^{2}-4 \lambda+1=0$, with characteristic roots $2 \pm \sqrt{3}$, so
$$
\sqrt{b_{n}}=\alpha(2+\sqrt{3})^{n}+\beta(2-\sqrt{3})^{n}, n=1,2, \cdots
$$

where $\alpha, \beta$ are undetermined constants. From $\sqrt{b_{1}}=\sqrt{b_{2}}=1$, we can calculate
$$
\alpha=\frac{3}{2}-\frac{5}{6} \sqrt{3}, \beta=\frac{3}{2}+\frac{5}{6} \sqrt{3} \text {. }
$$

From this, we can derive
$$
\begin{aligned}
b_{n} & =\left[\alpha(2+\sqrt{3})^{n}+\beta(2-\sqrt{3})^{n}\right]^{2} \\
& =\alpha^{2}(2+\sqrt{3})^{2 n}+\beta^{2}(2-\sqrt{3})^{2 n}+2 \alpha \beta(2+\sqrt{3})^{n}(2-\sqrt{3})^{n} \\
& =\alpha^{2}(7+4 \sqrt{3})^{n}+\beta^{2}(7-4 \sqrt{3})^{n}+2 \alpha \beta \\
& =\left(\frac{13}{3}-\frac{5}{2} \sqrt{3}\right)(7+4 \sqrt{3})^{n}+\left(\frac{13}{3}+\frac{5}{2} \sqrt{3}\right)(7-4 \sqrt{3})^{n}+\frac{1}{3} .
\end{aligned}
$$

Thus, the general term formula is
$$
\begin{aligned}
a_{n} & =\left(b_{n}+2\right)^{-1} \\
& = \\
& \left(\left(\frac{13}{3}-\frac{5}{2} \sqrt{3}\right)(7+4 \sqrt{3})^{n}+\left(\frac{13}{3}+\frac{5}{2} \sqrt{3}\right)(7-4 \sqrt{3})^{n}+\frac{7}{3}\right)^{-1},
\end{aligned}
$$

where $n=1,2, \cdots$.