Example 4 Let $a$ be a positive number. If the geometric sequence $\left\{a_{n}\right\}$ with $a$ as the first term satisfies that $a_{1}+1, a_{2}+2, a_{3}+3$ also form a geometric sequence, then $\left\{a_{n}\right\}$ is called a $P$ sequence corresponding to $a$. If $a_{n}=a^{2 n-1}(n=1,2, \cdots)$ is a $P$ sequence, where, $\frac{1}{2}<a<1), x_{n}^{2}-x_{n-1} x_{n+1}$ are all constants. 

 

Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Prove that if $a, a^{3}, a^{5}, \cdots$ is a sequence $P$, i.e., $a+1, a^{3}+2, a^{5}+3$ also form a geometric sequence, then
$$
\begin{array}{l}
\left(a^{3}+2\right)^{2}=(a+1)\left(a^{5}+3\right) \\
\Rightarrow a^{5}-4 a^{3}+3 a-1=0 \\
\Rightarrow\left(a^{2}+a-1\right)\left(a^{3}-a^{2}-2 a+1\right)=0 .
\end{array}
$$

It is easy to see that the roots of the equation $a^{2}+a-1=0$ are $a=\frac{-1 \pm \sqrt{5}}{2}$, and its positive root $a=\frac{\sqrt{5}-1}{2}$ satisfies $\frac{1}{2}<a<1$. For the equation $a^{3}-a^{2}-2 a+1=0$, we have:
$$
\begin{array}{l}
f(-2)=-1<0, \\
f(0)=1>0, \\
f\left(\frac{1}{2}\right)=-\frac{1}{8}<0, \\
f(1)=-1<0, \\
f(2)=1>0,
\end{array}
$$

which indicates that the equation $a^{3}-a^{2}-2 a+1=0$ has one root in each of the intervals $(-2,0)$, $\left(0, \frac{1}{2}\right)$, and $(1,2)$.

Thus, the equation (1) has only one root $\frac{1}{2}<a<1$, which is $a=\frac{\sqrt{5}-1}{2}$. This root satisfies $a^{2}+a-1=0$, i.e., $a-\frac{1}{a}=-1$, or $a_{1}-\frac{1}{a_{1}}=-1$, which means $x_{1}=-1$.
$$
\begin{array}{l}
\text { By }\left(a-\frac{1}{a}\right)^{3}=-1 \Rightarrow a^{3}-\frac{1}{a^{3}}=3\left(a-\frac{1}{a}\right)-1 \\
\Rightarrow x_{2}=3 x_{1}-1 \Rightarrow x_{2}=-4, \\
\text { By }\left(a-\frac{1}{a}\right)^{2}=1 \Rightarrow a^{2}+\frac{1}{a^{2}}=3 .
\end{array}
$$

Therefore, for $n \geqslant 2$,
$$
\begin{array}{l}
3 x_{n}=\left(a^{2}+\frac{1}{a^{2}}\right)\left(a^{2 n-1}-\frac{1}{a^{2 n-1}}\right) \\
=a^{2 n+1}-\frac{1}{a^{2 n+1}}+a^{2 n-3}-\frac{1}{a^{2 n-3}}=x_{n+1}+x_{n-1},
\end{array}
$$

which implies
$$
x_{n+1}=3 x_{n}-x_{n-1} \text {. }
$$

Given $x_{1}=-1, x_{2}=-4$, we have
$$
x_{3}=-11, \cdots
$$

Thus, all terms of the sequence $\left\{x_{n}\right\}$ are integers.
Let $\varphi(n)=x_{n}^{2}-x_{n+1} x_{n-1}$. Then
$$
\begin{array}{l}
\varphi(n-1)=x_{n-1}^{2}-x_{n} x_{n-2}, \\
\varphi(2)=x_{2}^{2}-x_{3} x_{1}=5, \\
\varphi(n)-\varphi(n-1) \\
=x_{n}^{2}-x_{n+1} x_{n-1}-x_{n-1}^{2}+x_{n} x_{n-2} \\
=x_{n}\left(3 x_{n-1}-x_{n-2}\right)-\left(3 x_{n}-x_{n-1}\right) x_{n-1}- \\
\quad x_{n-1}^{2}+x_{n} x_{n-2} \\
=0 .
\end{array}
$$

Hence, $\varphi(n)=\varphi(n-1)=\cdots=\varphi(2)=5$, i.e., for each $n \geqslant 2$, we have $x_{n}^{2}-x_{n-1} x_{n+1}=5$.