Problem 969

## Solution.

By the formula for the sum of cubes, we get

$$\begin{aligned} & \left\{\begin{array} { l } { ( x ^ { 2 } + y ^ { 2 } ) ( x ^ { 4 } - x ^ { 2 } y ^ { 2 } + y ^ { 4 } ) = 6 5 , } \\ { x ^ { 4 } - x ^ { 2 } y ^ { 2 } + y ^ { 4 } = 1 3 } \end{array} \Leftrightarrow \left\{\begin{array}{l} \left(x^{2}+y^{2}\right) 13=65, \\ x^{4}-x^{2} y^{2}+y^{4}=13 \end{array}\right.\right. \\ & \Leftrightarrow\left\{\begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 5 , } \\ { x ^ { 4 } - x ^ { 2 } y ^ { 2 } + y ^ { 4 } = 1 3 } \end{array} \Leftrightarrow \left\{\begin{array}{l} x^{2}+y^{2}=5, \\ \left(x^{2}+y^{2}\right)^{2}-3 x^{2} y^{2}=13 \end{array} \Leftrightarrow\right.\right. \\ & \Leftrightarrow\left\{\begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 5 , } \\ { 2 5 - 3 x ^ { 2 } y ^ { 2 } = 1 3 } \end{array} \Leftrightarrow \left\{\begin{array}{l} x^{2}+y^{2}=5, \\ x^{2} y^{2}=4 . \end{array}\right.\right. \end{aligned}$$

The system is equivalent to the combination of two systems:

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

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