## Task 20/86

All pairs $(x ; y)$ of real numbers $x$ and $y$ are to be determined for which the two equations are satisfied:

$$
\begin{array}{r}
x^{4}+y^{4}=12\left(x^{2}+y^{2}\right)-6 x^{2} y^{2}+16 \\
x y=3
\end{array}
$$
From (2) it follows that $4 x y=4 \cdot 3=12$. Substituting this into (1), we get

$$
x^{4}+y^{4}=4 x y\left(x^{2}+y^{2}\right)-6 x^{2} y^{2}+16
$$

and after equivalent transformation, $(x-y)^{4}=16$. From this, we obtain the real intermediate solutions

$$
x-y=2 \quad \text { and } \quad x-y=-2 \quad(3 a ; b)
$$

Due to (2), $x ; y \neq 0$; thus, from (2) it follows that $y=\frac{3}{x}$. Substituting this into (3a;b), we get the two equations $x-\frac{3}{x}=2$ and $x-\frac{3}{x}=-2$, i.e.,

$$
x^{2}-2 x-3=0 \quad \text { and } \quad x^{2}+2 x-3=0 \rightarrow x_{11}=-1 ; x_{12}=3 ; x_{21}=-3 ; x_{22}=1
$$

From (2), it then follows that $y_{11}=-3 ; y_{12}=1 ; y_{21}=-1 ; y_{22}=3$.

The solution set is thus $L=\{(-1 ;-3) ;(3 ; 1) ;(-3 ;-1) ;(1 ; 3)\}$.