1. For which digits $x, y, z$ is the equality

$$
\overline{x y} \sqrt{y x}+\overline{y z} \sqrt{z y}=\overline{(x+z) x y} ?
$$
Solution. Since $\overline{y x}$ and $\overline{z y}$ must be perfect squares, we conclude that they are among the numbers $16, 25, 36, 49, 64$ or $81$. However, $y$ must simultaneously appear as both the first and second digit of some of the given two-digit numbers, so $y \in\{1,4,6\}$. Now it is easily obtained that $x \in\{6,9,4\}$ and $z \in\{3,6,8\}$. Furthermore, $x+z$ must be a digit, so $x+z \leq 9$, which is only possible if $z=3$ and $x \in\{6,4\}$. Further, from $z=3$ it follows that $y=6$, so we get that $x=4$. Indeed,

$$
46 \sqrt{64}+63 \sqrt{36}=746
$$