Two (real) numbers twice their sum is equal to the difference of their squares, but 56 less than a quarter of their product. Which are these numbers?
Let the two numbers be $x$ and $y$. According to the problem,

$$
\begin{aligned}
& 2(x+y)=x^{2}-y^{2} \ldots \\
& 2(x+y)=\frac{x y}{4}-56 \ldots
\end{aligned}
$$

$x+y=0$ is not possible; otherwise, from (2),

$x y=4 \cdot 56$, which means $x$ and $y$ would have the same sign, whereas if $x+y=0$, they would have opposite signs. Therefore, we can divide both sides of (1) by the factor $(x+y)$, resulting in

$$
x-y=2 \ldots
$$

Now, substituting $x=y+2$ into (2): $2(2y+2)=\frac{y(y+2)}{4}-56$, and after rearranging:

$$
y^{2}-14 y-240=0
$$

From this,

$$
\begin{aligned}
y=\frac{14 \pm \sqrt{196+960}}{2} & =\frac{14 \pm 34}{2} ; \quad y_{1}=24 ; \quad y_{2}=-10 \\
\text { and } x_{1} & =26 ; \quad x_{2}=-8
\end{aligned}
$$

For the value pair $x_{1}=26 ; y_{1}=24$:

$$
\begin{aligned}
2(26+24) & =100=26^{2}-24^{2}=676-576 \\
\text { and } 2(26+24) & =\frac{624}{4}-56=156-56=100
\end{aligned}
$$

For the value pair $x_{2}=-8, y_{2}=-10$:

$$
\begin{array}{r}
2(-8-10)=-36=(-8)^{2}-(-10)^{2}= \\
=64-100=-36 \text { and } 2(-8-10)=\frac{-8-10}{4}-56=20-56=-36
\end{array}
$$

Lipsitz Imre (Izr. g. VI. o. Debrecen)

Note. A common mistake among a large portion of our readers is to simply divide the equation by an indeterminate expression, such as $(x+y)$ in this problem. This step overlooks a part of the solutions to the equation.