Let's find two numbers whose sum, product, and the difference of their squares are equal.
In the sense of the task:

$$
x+y=x y
$$

$$
x+y=x^{2}-y^{2}=(x+y)(x-y)
$$

From (2)

$$
x=1+y
$$

substituting into (1):

$$
1+2 y=y+y^{2}
$$

or

$$
y^{2}-y-1=0
$$

from which

$$
y=\frac{1}{2}(1 \pm \sqrt{5})
$$

and thus

$$
x=\frac{1}{2}(3 \pm \sqrt{5}) \text {. }
$$

(2) is also satisfied if

$$
x+y=0
$$

and since in this case $x y$ is also 0, therefore $x=0, y=0$ is also a solution to the system of equations.

(Zoltán Szávay, Györ.)

Number of solutions: 74.