Find all rational numbers $r$ such that
$$
\frac{r^{3}+5 r^{2}-4 r+1}{r-2 r^{2}} \text{ and } \frac{r^{3}-3 r^{2}+4 r-1}{r-2 r^{2}}
$$

are both integers.
Solution: From $\frac{r^{3}+5 r^{2}-4 r+1}{r-2 r^{2}}-\frac{r^{3}-3 r^{2}+4 r-1}{r-2 r^{2}}$
$$
=\frac{8 r^{2}-8 r+2}{r(1-2 r)}=-4+\frac{2}{r} \in \mathbf{Z} \text {, }
$$

we have $r=\frac{2}{k}, k \in \mathbf{Z}, k \neq 0,4$.
Substituting $r=\frac{2}{k}$ into $\frac{r^{3}+5 r^{2}-4 r+1}{r-2 r^{2}} \in \mathbf{Z}$, we get
$$
\frac{8+20 k-8 k^{2}+k^{3}}{2 k^{2}-8 k} \in \mathbf{Z} \text {. }
$$

Thus, $k=2 n, n \in \mathbf{Z}, n \neq 0,2$,
substituting into equation (1), we get
$$
\frac{1+5 n-4 n^{2}+n^{3}}{n^{2}-2 n} \in \mathbf{Z} \text {. }
$$

Therefore, $n \mid\left(1+5 n-4 n^{2}+n^{3}\right)$.
Hence, $n= \pm 1 \Rightarrow k= \pm 2 \Rightarrow r= \pm 1$.
Upon verification, the rational numbers sought are $r= \pm 1$.
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University,
510405)