Let $\alpha, \beta, \gamma$ be three given quantities and $x_{1}, x_{2}$ be the two roots of the following equation:

$$
x^{2}+p x+q=0
$$

What relationship must hold between $p$ and $q$ so that

$$
\alpha x_{1}^{2}+\beta x_{1}+\gamma=\alpha x_{2}^{2}+\beta x_{2}+\gamma
$$

is true?
The proposition can also be written as:

$$
\alpha\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)+\beta\left(x_{1}-x_{2}\right)=0
$$

or

$$
\left(x_{1}-x_{2}\right)\left[\alpha\left(x_{1}+x_{2}\right)+\beta\right]=0
$$

The left side is $=0$ if

$$
x_{1}=x_{2}, \quad \text { or } \quad \alpha\left(x_{1}+x_{2}\right)+\beta=0 \text {. }
$$

In the first case, the discriminant of the given equation is 0, so

$$
p^{2}=4 q
$$

In the second case, $x_{1}+x_{2}=-p$, thus

$$
p=\frac{\beta}{\alpha}
$$

(István Strasser, Budapest.)

Number of solutions: 46.