Consider the quadratic equation $a x^{2}+b x+c=0$. If we substitute $y+k$ for $x$, we obtain a quadratic equation in terms of $y$. Determine $k$ such that the product of the roots of the two equations is equal. Furthermore, calculate the sum of the four roots of the two equations in this case.
If in the given equation we substitute $(y+k)$ for $x$, then

$$
a y^{2}+(2 a k+b) y+a k^{2}+b k+c=0
$$

Thus, according to the problem,

$$
\frac{c}{a}=\frac{a k^{2}+b k+c}{a}
$$

from which, since $a \neq 0$, it follows:

$$
(a k+b) k=0
$$

and thus

$$
k=0, \quad \text { or } \quad k=-\frac{b}{a}
$$

The sum of the four roots:

$$
\begin{aligned}
\left(1^{\circ}\right) \quad x_{1}+x_{2}+y_{1}+y_{2} & =-\frac{b}{a}-\frac{b}{a}=-\frac{2 b}{a} \\
\left(2^{\circ}\right) & x_{1}+x_{2}+y_{1}+y_{2}=-\frac{b}{a}-\frac{-2 b+b}{a}=0
\end{aligned}
$$

(Endre Mellinger, Budapest.)

Number of solutions: 52.