1. For which $p$ is one of the roots of the equation $x^{2}-p x+p=0$ the square of the other? (We assume that the roots of the equation are distinct)
Answer: $2 \pm \sqrt{5}$.

Solution. Let the roots be $a$ and $a^{2}$. By Vieta's theorem, $a + a^{2} = p$, and $a \cdot a^{2} = p$. Therefore, $a + a^{2} = a^{3}$. The root $a = 0$ is not valid because the roots of the equation $x^{2} = 0$ are identical. Considering $a \neq 0$, we get $a^{2} - a - 1 = 0$, the roots of which are $\frac{1 \pm \sqrt{5}}{2}$. In this case, $p = a^{2} + a = (a + 1) + a = 2a + 1$.

Criteria. If the extraneous root is not discarded, 2 points are deducted. For a fully justified solution, the score is 7 points.