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

The roots of the equation are 1 greater than the roots of the

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

equation. Calculate $p$ and $q$! What are the roots?
If the roots of 2) are $x_{1}$ and $x_{2}$, then

$$
\begin{gathered}
x_{1}+x_{2}=-p \ldots \\
x_{1} x_{2}=q \ldots
\end{gathered}
$$

The roots of 1) are $x_{1}+1$ and $x_{2}+1$. For these, we have

$$
\begin{aligned}
& x_{1}+1+x_{2}+1=p^{2} \ldots \\
& \left(x_{1}+1\right)\left(x_{2}+1\right)=p q \ldots
\end{aligned}
$$

Thus, we have 4 equations for calculating the 4 unknowns $[3)-6)]$ \((x_{1}, x_{2}, p, q)\). Based on 3) and 4), we eliminate \(x_{1}\) and \(x_{2}\) from 5) and 6):

$$
\begin{array}{rcc}
-p+2=p^{2}, & \text { or } & p^{2}+p-2=(p+2)(p-1)=0 \ldots \\
-p+q+1=p q & \text { or } \quad p q+p-q-1=(p-1)(q+1)=0 \ldots
\end{array}
$$

The solutions are therefore:
I. \(p=1\); in this case, \(q\) can be any number. The resulting

$$
x^{2}+x+q=0 \quad \text { roots are } \quad x=-\frac{1}{2} \pm \sqrt{\frac{1}{4}-q}
$$

\(x_{1}+1\) and \(x_{2}+1\) satisfy the equation \(x^{2}-x+q=0\).

II. \(p=-2\). Then \(q=-1\) and the roots of 2) are: \(x=1 \pm \sqrt{2}\), while the roots of 1) are: \(x=2 \pm \sqrt{2}\).