Solve the following system of equations in natural numbers:

$$
\begin{aligned}
& x+y=z t \\
& z+t=x y .
\end{aligned}
$$
The given system of equations is symmetric with respect to the pairs of unknowns $x, y$ and $z, t$, so it suffices to look for solutions where $x \leq y, z \leq t$; furthermore, the two pairs can be interchanged, so we can also impose the constraint $x \leq z$.

First, let's look for a solution where $x=1$. From the simplified system

$$
1+y=z t, \quad z+t=y
$$

by eliminating $y$ and further transforming, we get

$$
\begin{gathered}
1+z+t=z t \\
t=\frac{z+1}{z-1}=1+\frac{2}{z-1}
\end{gathered}
$$

(we can divide by $z-1$ because the case $z=1$ contradicts equation (2)). Therefore, $z>1$ and $z-1$ is a natural number divisor of 2:

$$
z-1=1 \quad \text { or } \quad 2
$$

In the first case, from (3) and then (1) we get

$$
z=2, \quad t=3, \quad y=5, \quad x=1
$$

In the latter case, however, $z=3$ and from (3) $t=2<z$, which, according to the preliminary remark, is a solution that is equivalent to the one found.

When looking for a solution where each unknown is a natural number greater than 1, the following holds:

$$
\begin{gathered}
x-1 \geq 1, \quad y-1 \geq 1, \quad(x-1)(y-1) \geq 1, \quad \text { that is } \\
x y \geq x+y
\end{gathered}
$$

Similarly,

$$
z t \geq z+t
$$

and considering our equations,

$$
x y \geq x+y=z t \geq z+t=x y
$$

The last expression is equal to the first, so the equality holds at each $\geq$ sign, hence also above, so

$$
x-1=y-1=1, \quad x=y=2 \quad \text { and } \quad z=t=2
$$

This completes the solution. From the first solution found, we can write down 7 more solutions that are essentially different (not just in order).