For how many different values of the parameter $p$ does the system of equations

$$
x^{2}-y^{2}=0 \quad x y+p x-p y=p^{2}
$$

have exactly one solution?
Transform the first equation into a product: $(x+y)(x-y)=0$, hence $x=y$ or $x=-y$. If $x=y$, then from the second equation:

$$
\begin{aligned}
x^{2}+p x-p x & =p^{2}, \\
x^{2} & =p^{2},
\end{aligned}
$$

from which $x_{1}=p, y_{1}=p$ or $x_{2}=-p, y_{2}=-p$. Therefore, for any $p \neq 0$, the system of equations has two solutions.

If $x=-y$, then the second equation can be written as $x^{2}-2 p x+p^{2}=0$. From this, $(x-p)^{2}=0$ so $x=p$ and $y=-p$.

Accordingly, if $p=0$, the system of equations has exactly one solution, and for any other value of $p$, it has three.