Question: Determine the positive rational number $m$ that makes $\frac{1}{\pi} \operatorname{arctg} \sqrt{m}$ always a rational number.
If $\sqrt{ } \bar{m}=\operatorname{tg} r \pi$, where $r$ and $m$ are rational numbers, then $2 \cos 2 r \pi=2\left(\frac{2}{\sec ^{2} r \pi}-1\right)$
$$
=\frac{4}{m+1}-2 \text {. }
$$

Since $2 \cos 2 r \pi=e^{2 \mathrm{r} \pi i}+e^{-2 \mathrm{r} \pi i}$ is both an algebraic integer and a rational number, it must be an integer with an absolute value no greater than 2. Therefore, $\frac{4}{m+1}$ must be an integer between 0 and 4. However, the values 0 and 4 are impossible. When the values are 1, 2, and 3, we get $m=3,1$, and $\frac{1}{3}$, respectively.
(215002, Translated by Yin Yigong from Suzhou Education College from "The American Mathematical Monthly", Nov, 1989)