Consider the equation

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

which has roots $x_{1}$ and $x_{2}$. Determine the coefficients $r$ and $s$ as functions of $p$ and $q$ such that the equation

$$
y^{2}+r y+s=0
$$

has roots $y_{1}$ and $y_{2}$ which are related to the roots $x_{1}$ and $x_{2}$ by

$$
y_{1}=\frac{x_{1}}{x_{1}-1} \text { and } y_{2}=\frac{x_{2}}{x_{2}-1}
$$

What relationship must hold between $p$ and $q$ for equations (1) and (2) to be identical?
If we apply the familiar

$$
x_{1}+x_{2}=-p \quad \text { and } \quad x_{1} x_{2}=q
$$

relationships, then we obtain that

$$
-r=y_{1}+y_{2}=\frac{x_{1}}{x_{1}-1}+\frac{x_{2}}{x_{2}-1}=\frac{2 x_{1} x_{2}-\left(x_{1}+x_{2}\right)}{x_{1} x_{2}-\left(x_{1}+x_{2}\right)+1}=\frac{2 q+p}{q+p+1}
$$

and

$$
s=y_{1} \cdot y_{2}=\frac{x_{1}}{x_{1}-1} \cdot \frac{x_{2}}{x_{2}-1}=\frac{x_{1} x_{2}}{x_{1} x_{2}-\left(x_{1}+x_{2}\right)+1}=\frac{q}{q+p+1}
$$

The two given equations are identical if $p=r$ and $q=s$, that is, if:

$$
p=\frac{2 q+p}{q+p+1}
$$

$$
q=\frac{q}{q+p+1}
$$

From (4), either $q=0$, in which case (3) gives $p=0$, or

$$
p+q=0
$$

which condition includes the first one. Therefore, $p+q=0$, since this also satisfies the equation under (3), which is the necessary and sufficient condition for the two equations to be identical.

(Alfréd Haar, Budapest.)

Number of solutions: 33.