Let's calculate the coefficients of the equation

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

if we know that the equation whose roots are one greater than those of the previous one, is of the following form

$$
x^{2}-p^{2} x+p q=0
$$
Let's calculate the coefficients of the equation

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

if we know that the equation whose roots are one unit larger than the previous ones, has the following form:

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

Solution. If the roots of the first equation are $x^{\prime}$ and $x^{\prime \prime}$, then it is known that:

$$
x^{\prime}+x^{\prime \prime}=-p, \quad x^{\prime} x^{\prime \prime}=q
$$

According to the problem,

$$
\left(x^{\prime}+1\right)+\left(x^{\prime \prime}+1\right)=p^{2}, \quad\left(x^{\prime}+1\right)\left(x^{\prime \prime}+1\right)=p q
$$

or, rearranging and substituting the values from (3):

$$
-p+2=p^{2}
$$

$$
q-p+1=p q
$$

These equations can be further simplified to:

$$
(p-1)(p+2)=0
$$

$$
(p-1)(q+1)=0
$$

If $p=1$, then, as can be seen directly from (6), the value of $q$ can be arbitrary, while if $p=-2$, then $q=-1$. Therefore, the given equation can have the following forms:

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

or

$$
x^{2}-2 x+1=0
$$

(János Galter, secondary school, VIII. class, Székely-Udvarhely).

The problem was also solved by: Bernát Friedmann, S.-A.-Ujhely; Béla Grünhut, Pécs; János Imre, Jr., Nyíregyháza; György Jankovich, Fülek; Elemér Meitner, Budapest; Aladár Visnya and Lipót Weisz, Pécs.