Let $x_{1}$ and $x_{2}$ be the two roots of the equation $a x^{2}+b x+c=0$. What relationship must hold between the coefficients $a, b$, and $c$ for the difference of the squares of the two roots to be $\frac{c^{2}}{a^{2}}$?
First solution. According to the problem,

$$
x_{1}^{2}-x_{2}^{2}=\frac{c^{2}}{a^{2}}
$$

$$
x_{1}+x_{2}=-\frac{b}{a}
$$

$$
x_{1} x_{2}=\frac{c}{a}
$$

We eliminate $x_{1}$ and $x_{2}$ from these three equations. Dividing (1) by (2), we get:

$$
x_{1}-x_{2}=-\frac{c^{2}}{a b}
$$

From (1) and (2), we obtain:

$$
x_{1}=-\frac{b^{2}+c^{2}}{2 a b}, \quad x_{2}=\frac{c^{2}-b^{2}}{2 a b}
$$

Substituting the values of $x_{1}$ and $x_{2}$ into (3), we get:

$$
b^{4}-c^{4}-4 a b^{2} c=0
$$

(Jenő Engler, Pécs.)

Second solution. Squaring (1):

$$
\left(x_{1}^{2}-x_{2}^{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{4}-4 x_{1} x_{2}\left(x_{1}+x_{2}\right)^{2}=\frac{c^{4}}{a^{4}}
$$

Considering (2) and (3),

$$
\frac{b^{4}}{a^{4}}-\frac{4 b^{2} c}{a^{3}}=\frac{c^{4}}{a^{4}}
$$

or

$$
b^{4}-c^{4}=4 a b^{2} c
$$

(Nándor Ehrenfeld, Nyitra.)