Solve the following system of equations:

$$
\begin{aligned}
& x^{x}=y \ldots \\
& x^{y}=y^{x} \ldots
\end{aligned}
$$
A 2) equation is satisfied by every $(x, y)$ value pair where $y=x$. We exclude the $x=y=0$ value pair, since expressions of the form $0^{0}$ are indeterminate. Taking this into account, from 1) we get $x^{2}=x$, hence $x= \pm 1$.

Accordingly, the system of equations is satisfied by the value pairs $x=1, y=1$ and $x=-1, y=-1$.

If now $x \neq y$, then

from 1) $\quad x=y^{\frac{1}{x}}, \quad$ from 2) $\quad x=y^{\frac{x}{y}}$,

thus $\frac{1}{x}=\frac{x}{y}, \quad$ that is, $\quad y=x^{2}$.

Substituting into 1): $x^{x}=x^{2}$, hence $x=2, y=4$.

The system of equations has three solutions.