Given the equation $x^{2}+p x+q=0$. Let's write down the quadratic equation in $y$ whose roots are

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

where $x_{1}$ and $x_{2}$ are the roots of the given equation.
The desired equation can only be set up if $y_{1}$ and $y_{2}$ are both numbers, that is, their denominators are not 0; in other words, if $x_{1}$ and $x_{2}$ differ from 1; we therefore assume that $x=1$ does not satisfy the given equation, i.e., $1+p+q \neq 0$. The coefficients of the sought equation are essentially given by the sum and product of $y_{1}$ and $y_{2}$. Since the two roots $x_{1}$ and $x_{2}$ can be transformed into each other by swapping, it is hoped that these expressions can be directly converted into expressions involving $x_{1}+x_{2}$ and $x_{1} x_{2}$, and thus can be written in terms of $p$ and $q$ without using the quadratic formula. (Otherwise, the problem - implicitly - clearly wants this.) According to the relationship between the roots and coefficients, if we write the sought equation in the form $a y^{2}+b y+c=0$ and use the fact that $x_{1}+x_{2}=-p, x_{1} x_{2}=q$, then

$$
\begin{aligned}
& \frac{c}{a}=y_{1} y_{2}=\frac{x_{1}\left(1+x_{1}\right) x_{2}\left(1+x_{2}\right)}{\left(1-x_{2}\right)\left(1-x_{1}\right)}=\frac{x_{1} x_{2}\left[1+\left(x_{1}+x_{2}\right)+x_{1} x_{2}\right]}{1-\left(x_{1}+x_{2}\right)+x_{1} x_{2}}=\frac{q(1-p+q)}{1+p+q} \\
& \frac{b}{a}=-\left(y_{1}+y_{2}\right)=-\frac{\left(x_{1}+x_{1}^{2}\right)\left(1-x_{1}\right)+\left(x_{2}+x_{2}^{2}\right)\left(1-x_{2}\right)}{\left(1-x_{2}\right)\left(1-x_{1}\right)}=\frac{x_{1}^{3}+x_{2}^{2}-\left(x_{1}+x_{2}\right)}{1+p+q}= \\
& =\frac{\left(x_{1}+x_{2}\right)^{3}-3 x_{1} x_{2}\left(x_{1}+x_{2}\right)-\left(x_{1}+x_{2}\right)}{1+p+q}=\frac{-p^{3}+3 p q+p}{1+p+q}=\frac{p\left(1+3 q-p^{2}\right)}{1+p+q}
\end{aligned}
$$

thus, for example, with the choice $a=1$, an equation that meets the requirement is:

$$
y^{2}+\frac{p\left(1+3 q-p^{2}\right)}{1+p+q} y+\frac{q(1-p+q)}{1+p+q}=0
$$

or, with the choice $a=1+p+q$ (which ensures $a \neq 0$, so the equation is indeed quadratic):

$$
(1+p+q) y^{2}+p\left(1+3 q-p^{2}\right) y+q(1-p+q)=0
$$

For example, with $p=-1, q=-12$, the roots of the given equation are $x_{1}=-3, x_{2}=4$, and equation (1) is: $-12 y^{2}+36 y+120=0$, whose roots are: $y_{1}=-2, y_{2}=5$, and these, as well as $x_{1}$ and $x_{2}$, indeed satisfy the required relationships.

Kátai Szabolcs (Bp. I., Toldy F. g. I. o. t.)