Problem 1. If $x+y=0$ and $x^{2}+y^{2}=\frac{1}{4}$, calculate the sum $x^{4}+y^{4}$.
Solution. If $x+y=0$, then $(x+y)^{2}=0$, i.e., $2 x y=-(x^{2}+y^{2})$, so $2 x y=-\frac{1}{4}$, which means $x y=-\frac{1}{8}$. According to this,

$$
x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=\left(\frac{1}{4}\right)^{2}-2\left(-\frac{1}{8}\right)^{2}=\frac{1}{16}-\frac{2}{64}=\frac{1}{32}
$$