[ $\left.\begin{array}{c}\text { Theorem of Bezout. Factorization }\end{array}\right]$

Find the solution to the system

$$
x^{4}+y^{4}=17
$$

$$
x+y=3 \text {. } \quad .
$$

#
Let $x y=u$. Then $x^{2}+y^{2}=9-2 u,(9-2 u)^{2}-2 u x^{2}=x^{4}+y^{4}=17$. From this, $u^{2}-18 u+32=0$, that is, $u=2$ or 16.

The system $x+y=3, x y=2$ has the obvious solution $\{1,2\}$, while the system $x+y=3, x y=16$ has no solutions.

## Answer

$\{1,2\}$.