6.210. $\left\{\begin{array}{l}x^{4}+y^{4}=17 \\ x^{2}+y^{2}=5\end{array}\right.$
## Solution.

Let's rewrite the system of equations as follows:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ ( x ^ { 2 } + y ^ { 2 } ) ^ { 2 } - 2 x ^ { 2 } y ^ { 2 } = 1 7 , } \\
{ x ^ { 2 } + y ^ { 2 } = 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 2 5 - 2 x ^ { 2 } y ^ { 2 } = 1 7 , } \\
{ x ^ { 2 } + y ^ { 2 } = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2} y^{2}=4, \\
x^{2}+y^{2}=5
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x y= \pm 2 \\
(x+y)^{2}-2 x y=5
\end{array}\right. \\
& (x+2)
\end{aligned} \Leftrightarrow
$$

This system is equivalent to two systems of equations:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x y = - 2 , } \\
{ ( x + y ) ^ { 2 } - 2 x y = 5 }
\end{array} \text { or } \left\{\begin{array} { l } 
{ x y = 2 , } \\
{ ( x + y ) ^ { 2 } - 2 x y = 5 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ x _ { 1 } = 2 , } \\
{ y _ { 1 } = 1 ; }
\end{array} \left\{\begin{array}{l}
x_{2}=1 \\
y_{2}=2
\end{array}\right.\right.\right.\right. \\
& \left\{\begin{array} { l } 
{ x _ { 3 } = - 2 , } \\
{ y _ { 3 } = 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 4 } = 1 , } \\
{ y _ { 4 } = - 2 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 5 } = 2 , } \\
{ y _ { 5 } = - 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 6 } = - 1 , } \\
{ y _ { 6 } = 2 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 7 } = - 2 , } \\
{ y _ { 7 } = - 1 ; }
\end{array} \left\{\begin{array}{l}
x_{8}=-1 \\
y_{8}=-2
\end{array}\right.\right.\right.\right.\right.\right.
\end{aligned}
$$

Answer: $(2 ; 1),(1 ; 2),(-2 ; 1),(1 ;-2),(2 ;-1),(-1 ; 2),(-2 ;-1),(-1 ;-2)$.