1. Let $x_{1}$ and $x_{2}$ be the roots of the equation $x^{2} + p x - \frac{1}{2 p^{2}} = 0$, where $p$ is an arbitrary real parameter. For which value of the parameter $p$ does the following hold:

$$
x_{1}^{4} + x_{2}^{4} = 2 + \sqrt{2}
$$
Solution. According to Vieta's formulas, we have

$$
x_{1}+x_{2}=-p, \quad x_{1} x_{2}=-\frac{1}{2 p^{2}}
$$

On the other hand,

$$
\left(x_{1}+x_{2}\right)^{4}=x_{1}^{4}+4 x_{1}^{3} x_{2}+6 x_{2}^{2} x_{2}^{2}+4 x_{1} x_{2}^{3}+x_{2}^{3}
$$

from which

$$
x_{2}^{4}+x_{1}^{4}=\left(x_{1}+x_{2}\right)^{4}-4 x_{1}^{2} x_{2}^{2}\left(x_{1}+x_{2}\right)+6\left(x_{1} x_{2}\right)^{2}=2+p^{4}+\frac{1}{2 p^{4}}
$$

From the equation

$$
2+p^{4}+\frac{1}{2 p^{4}}=2+\sqrt{2}
$$

we get $\left(p^{2}-\frac{1}{\sqrt{2} p^{2}}\right)^{2}=0$, i.e., $p= \pm \frac{1}{\sqrt[8]{2}}$.