6.197. $\left\{\begin{array}{l}x^{2}+y^{4}=20, \\ x^{4}+y^{2}=20 .\end{array}\right.$
Solution.

From the condition we have $x^{2}+y^{4}=x^{4}+y^{2}, y^{4}-x^{4}+x^{2}-y^{2}=0$,

$$
\left(y^{2}-x^{2}\right)\left(y^{2}+x^{2}\right)-\left(y^{2}-x^{2}\right)=0 \Leftrightarrow\left(y^{2}-x^{2}\right)\left(y^{2}+x^{2}-1\right)=0
$$

equivalent to two equations: $y^{2}-x^{2}=0$ or $y^{2}+x^{2}-1=0, y^{2}=x^{2}$ or $y^{2}=1-x^{2} \Rightarrow x^{4}+x^{2}=20$ or $x^{4}+\left(1-x^{2}\right)=20 ; x^{4}+x^{2}-20=0$ or $x^{4}-x^{2}-19=0$, from which $x_{1}^{2}=4, x_{2}^{2}=-5, x_{3}^{2}=\frac{1-\sqrt{77}}{2}, x_{4}^{2}=\frac{1+\sqrt{77}}{2}$; $x_{2}^{2}$ and $x_{3}^{2}$ are not suitable. Then $x= \pm 2$ or $x= \pm \sqrt{\frac{1+\sqrt{77}}{2}} ; y_{1}^{2}=4$, $y_{2}^{2}=\frac{1-\sqrt{77}}{2}$, from which $y= \pm 2$ (the value of $y_{2}^{2}$ is not suitable). Answer: $(2 ; 2),(-2 ;-2),(2 ;-2),(-2 ; 2)$.