1. Let the sum of three real numbers be 0, and the sum of their squares be 4. Calculate the sum of the fourth powers of the three numbers.
Solution. Let's denote the three numbers by $x, y$, and $z$. According to the conditions of the problem, we have

$$
\begin{aligned}
& x+y+z=0 \\
& x^{2}+y^{2}+z^{2}=4
\end{aligned}
$$

If we square the first equation, we get:

$$
x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=0
$$

from which we obtain

$$
x y+y z+z x=-2
$$

If we square equation (3), we get:

$$
\begin{aligned}
& x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+2\left(x^{2} y z+x y^{2} z+x y z^{2}\right)=4 \\
& x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+2 x y z(x+y+z)=4
\end{aligned}
$$

from which, due to (1), it follows that:

$$
x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}=4
$$

If we square equation (2), we get:

$$
x^{4}+y^{4}+z^{4}+2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)=16
$$

from which, due to (4), it follows that $x^{4}+y^{4}+z^{4}=8$.