## Solution.

From the condition $\left\{\begin{array}{l}x^{2}+y=20, \\ x+y^{2}=20,\end{array}\right.$ we get $x^{2}+y=x+y^{2}, x^{2}-y^{2}-x+y=0$, $(x-y)(x+y)-(x-y)=0,(x-y)(x+y-1)=0$.

Therefore, the given system is equivalent to two systems of equations:

$$
\left\{\begin{array}{l}
x-y=0 \\
x+y^{2}-20=0
\end{array}\right.
$$

or

$$
\left\{\begin{array} { l } 
{ x + y - 1 = 0 , } \\
{ x + y ^ { 2 } - 2 0 = 0 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ x _ { 1 } = - 5 , } \\
{ y _ { 1 } = - 5 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 2 } = 4 , } \\
{ y _ { 2 } = 4 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 3 } = \frac { 1 + \sqrt { 7 7 } } { 2 } , } \\
{ y _ { 3 } = \frac { 1 - \sqrt { 7 7 } } { 2 } ; }
\end{array} \left\{\begin{array}{l}
x_{4}=\frac{1-\sqrt{77}}{2} \\
y_{4}=\frac{1+\sqrt{77}}{2}
\end{array}\right.\right.\right.\right.
$$

Answer: $(-5 ;-5),(4 ; 4),\left(\frac{1+\sqrt{77}}{2} ; \frac{1-\sqrt{77}}{2}\right),\left(\frac{1-\sqrt{77}}{2} ; \frac{1+\sqrt{77}}{2}\right)$.
6.216. $\left\{\begin{array}{l}x^{2}+y-20=0 \\ x+y^{2}-20=0 .\end{array}\right.$

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.

6.216. $\left\{\begin{array}{l}x^{2}+y-20=0 \\ x+y^{2}-20=0 .\end{array}\right.$