$8 \cdot 24$ Let $r$ be a real number, and the sequence $\left\{x_{n}\right\}$ satisfies: $x_{0}=0, x_{1}=1$, and $x_{n+2}=$ $r x_{n+1}-x_{n}, n \geqslant 0$. For what value of $r$ does it hold that
$x_{1}+x_{3}+\cdots+x_{2 m-1}=x_{m}^{2}$, for any $m \geqslant 1$.
The characteristic equation of the given recurrence relation is
$$
\lambda^{2}-r \lambda+1=0,
$$

its roots are $\frac{1}{2}\left(r \pm \sqrt{r^{2}-4}\right)$.
If $r=2$, then the characteristic roots are the double root 1, in this case
$$
x_{n}=n, n=0,1,2, \cdots
$$

If $r=-2$, then $\lambda=-1$ is a double characteristic root, in this case
$$
x_{n}=(-1)^{n+1} n, n=0,1,2, \cdots \text {. }
$$

In both cases, we have
$$
x_{1}+x_{3}+\cdots+x_{2 m-1}=1+3+\cdots+2 m-1=m^{2}=x_{m}^{2} .
$$

If $r^{2} \neq 4$, then there are two distinct characteristic roots $\lambda_{1}, \lambda_{2}\left(\lambda_{1}, \lambda_{2}\right.$ are not necessarily real numbers $)$.
Since $\lambda_{1} \lambda_{2}=1$, we denote
$$
z \cdot=\lambda_{1}, \frac{1}{z}=\lambda_{2} \text {. }
$$

At this time, $z \neq 1$, and
$$
x_{n}=c_{1} z^{n}+c_{2} z^{-n}, n=0,1,2, \cdots
$$

where $c_{1}, c_{2}$ are constants (not necessarily real numbers) determined by $x_{0}=0$ and $x_{1}=1$. From this, we get
$$
\begin{aligned}
\sum_{k=1}^{m} x_{2 k-1} & =c_{1} \sum_{k=1}^{m} z^{2 k-1}+c_{2} \sum_{k=1}^{m} z^{1-2 k} \\
& =c_{1} \frac{z\left(z^{2 m}-1\right)}{z^{2}-1}+c_{2} \frac{z^{-1}\left(z^{-2 m}-1\right)}{z^{-2}-1}
\end{aligned}
$$

From $x_{0}=c_{1}+c_{2}=0$, we know that $c_{1}=-c_{2}=c$, then
$$
\begin{aligned}
\sum_{k=1}^{m} x_{2 k-1} & =c \frac{z\left(z^{2 m}-1\right)}{z^{2}-1}-c \frac{z\left(z^{-2 m}-1\right)}{1-z^{2}} \\
& =\frac{c z}{z^{2}-1}\left(z^{2 m}-2+z^{-2 m}\right) \\
& =\frac{c z}{z^{2}-1}\left(z^{m}-z^{-m}\right)^{2}
\end{aligned}
$$

From $x_{1}=c\left(z-z^{-1}\right)=\frac{c\left(z^{2}-1\right)}{z}=1$, we immediately get
$$
\sum_{k=1}^{m} x_{2 k-1}=c^{2}\left(z^{m}-z^{-m}\right)^{2}=x_{m}^{2} .
$$

In summary, when $r$ is any real number, we have
$$
x_{1}+x_{3}+\cdots+x_{2 m-1}=x_{m}^{2}
$$