2A. If each root of the equation $x^{2}+p x+q=0$ is increased by 1, the roots of the equation $x^{2}-p^{2} x+p q=0$ are obtained. Determine $p$ and $q$.
Solution. From Vieta's formulas for the first equation, we have $\alpha+\beta=-p$ and $\alpha \beta=q$, and for the second equation, $\alpha+\beta+2=p^{2}$ and $(\alpha+1)(\beta+1)=p q$. Therefore, $p^{2}+p-2=0$ and $q+p^{2}-1=p q$. From the first equation, we get $p_{1}=1, p_{2}=-2$, and the second equation is equivalent to $(p-1) q=p^{2}-1$. If $p=1$, then $q$ is any real number. If $p=-2$, then $q=-1$.