5. If $r$ is a real number, the sequence $\left\{x_{n}\right\}$ is defined by $x_{0}=0$, $x_{1}=1$ and $x_{n+2}=r x_{n+1}-x_{n}, n \geqslant 0$, then for which values of $r$ does $x_{1}+x_{3}+\cdots$ $+x_{2 m-1}=x_{m}^{2}$ hold for $m \geqslant 1$?
5. For all $r$, the characteristic equation of the given recurrence relation is $y^{2}-r y+1=0$, with roots
$$
\frac{r \pm \sqrt{r^{2}-4}}{2}.
$$

If $r=2$, there is a repeated root of 1. Thus, for some $c_{1}, c_{2}$, we have $x_{2}=c_{1}+c_{2} n$. Using $x_{0}=0$ and $x_{1}=1$, we get $c_{1}=0, c_{2}= \pm 1$ and $x_{\mathrm{n}}=n$.
If $r=-2$, then $x_{\mathrm{B}}=(-1)^{\mathrm{n}+1} n$.
In both cases,
$$
\begin{array}{l}
x_{1}+x_{a}+\cdots+x_{2 m-1} \\
=1+3+\cdots+(2 m-1) \\
=( \pm m)^{2}=x_{m 0}^{2}
\end{array}
$$

If $r \neq \pm 2$, then $x_{\mathrm{n}}=c_{1} z^{\mathrm{a}}+c_{2} z^{-\mathrm{n}}$, where $z=\frac{r+\sqrt{r^{2}-4}}{2}$. Using the initial values, we determine $c_{1}=-c_{2}=c$, where $c=\frac{1}{\sqrt{r^{2}-4}}$, so $x_{\mathrm{n}}=c\left(z^{\mathrm{D}}-z^{-\mathrm{n}}\right)$. Therefore,
$$
\begin{aligned}
x_{1} & +x_{3}+\cdots+x_{2 m-1} \\
= & c\left(z+z^{3}+\cdots+z^{2 m-1}-z^{-1}-z^{-8}\right. \\
& \left.\quad-\cdots-z^{2-2 m}\right) \\
= & c \frac{z\left(z^{2 m}-1\right)}{z^{2}-1}-\frac{z^{-1}\left(z^{-2 m}-1\right)}{z^{-2}-1} \\
= & \frac{c z\left(z^{2 m}-2+z^{-2 m}\right)}{z^{2}-1} \\
= & \frac{x_{m}^{2}}{x_{1}}=x_{m}^{2} .
\end{aligned}
$$