$\overline{x y z}$ is a three-digit number written in the decimal system, its square is a five-digit number: $\overline{x y z^{2}}=\overline{a b c d e}$ and $\overline{z y x^{2}}=\overline{e d c b a}$. How many such three-digit numbers are there and which are they?
A three-digit number squared is a five-digit number if the square of its first digit is a one-digit number. According to this, $(x=0, z=0$ values are excluded)

$$
x=1,2,3 \text { and } z=1,2,3 \text {. }
$$

Thus,

$$
\overline{x y z}^{2}=\overline{a b c d e} \text { implies } e=z^{2}, \ldots
$$

and

$$
\overline{z y x}^{2}=\overline{e d c b a} \text { implies } a=x^{2}, \ldots
$$

The first digits of the square numbers are themselves square numbers; therefore, the products $2 x y$ and $2 y z$ do not increase the first digit of the square. Hence, $2 x y$ and $2 y z$ are one-digit numbers, that is,

$$
x y<5 \text { and } z y<5 \ldots
$$

Since $z^{2}=e$ and $x^{2}=a$, $z^{2}$ and $x^{2}$ do not contribute to the increase of the tens in the square numbers. The number of tens can only be the units of the

$$
2 z(10 x+y) \quad \text { or } \quad 2 x(10 z+y)
$$

products, that is, $2 z y$ and $2 x y$. However, these are one-digit numbers, and thus

$$
d=2 z y, \quad b=2 x y \ldots
$$

The product $2 z x$ increases the hundreds of the square numbers in both cases. On the other hand, according to (4), the thousands of the square numbers ($b$ and $d$) do not increase $y^{2}$, so $y^{2}$ is also a one-digit number.

Now, the hundreds of the square numbers are

$$
\begin{array}{r}
c=y^{2}+2 z x<10 \ldots \\
x \geqq 1 \text { and } z \geqq 1 \text { so } y^{2}<8 ; \text { thus } \\
y=0,1,2, \text { and } z x<5
\end{array}
$$

Considering the possible values for $x, y, z$, the following meet the requirement:

$$
\begin{array}{lcl}
y=0 & \text { case } & 101,102,103,201,202,301 \\
y=1 & " & 111,112,113,211,212,311 \\
y=2 & " & 121,122,221
\end{array}
$$