6.106. $\left\{\begin{array}{l}u^{2}+v^{2}=u v+13 \\ u+v=\sqrt{u v}+3\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
u^{2}+v^{2}=u v+13 \\
u+v=\sqrt{u v}+3
\end{array}\right.
\]
## Solution.

Domain of definition: $u v \geq 0$.

Let $\left\{\begin{array}{l}\sqrt{u}=x \geq 0, \\ \sqrt{v}=y \geq 0,\end{array}\left\{\begin{array}{l}u=x^{2}, \\ v=y^{2},\end{array}\left\{\begin{array}{l}u^{2}=x^{4}, \\ v^{2}=y^{4} .\end{array}\right.\right.\right.$

With respect to $x$ and $y$, we obtain the system

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x ^ { 4 } + y ^ { 4 } = x ^ { 2 } y ^ { 2 } + 1 3 , } \\
{ x ^ { 2 } + y ^ { 2 } = x y + 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=x^{2} y^{2}+13, \\
x^{2}+y^{2}=x y+3
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
\left(x^{2}+y^{2}\right)^{2}-3 x^{2} y^{2}-13=0 \\
x^{2}+y^{2}=x y+3
\end{array}\right.
\end{aligned}
$$

From the first equation of the system, we have

$$
(x y+3)^{2}-3 x^{2} y^{2}-13=0, x^{2} y^{2}+6 x y+9-3 x^{2} y^{2}-13=0
$$

$x^{2} y^{2}-3 x y+2=0$, from which $x y=2$ or $x y=1$. We obtain two systems of equations:

1) $\left\{\begin{array}{l}x y=2, \\ (x+y)^{2}=3 x y+3\end{array}\right.$ or
2) $\left\{\begin{array}{l}x y=1 \\ (x+y)^{2}=3 x y+3\end{array}\right.$

from which we find

$\left\{\begin{array}{l}x_{1}=1, \\ y_{1}=2,\end{array},\left\{\begin{array}{l}x_{2}=2, \\ y_{2}=1,\end{array}\left\{\begin{array}{l}x_{3}=\frac{\sqrt{6}-\sqrt{2}}{2}, \\ y_{3}=\frac{\sqrt{6}+\sqrt{2}}{2},\end{array}\left\{\begin{array}{l}x_{4}=\frac{\sqrt{6}+\sqrt{2}}{2}, \\ y_{4}=\frac{\sqrt{6}-\sqrt{2}}{2} .\end{array}\right.\right.\right.\right.$

Then $\left\{\begin{array}{l}\sqrt{u}=1, \\ \sqrt{v}=2,\end{array}\left\{\begin{array}{l}u_{1}=1, \\ v_{1}=4 ;\end{array}\left\{\begin{array}{l}\sqrt{u}=2, \\ \sqrt{v}=1,\end{array}\left\{\begin{array}{l}u_{2}=4, \\ v_{2}=1 ;\end{array}\right.\right.\right.\right.$

$$
\left\{\begin{array} { l } 
{ \sqrt { u } = \frac { \sqrt { 6 } - \sqrt { 2 } } { 2 } , } \\
{ \sqrt { v } = \frac { \sqrt { 6 } + \sqrt { 2 } } { 2 } , }
\end{array} \left\{\begin{array} { l } 
{ u _ { 3 } = 2 - \sqrt { 3 } , } \\
{ v _ { 3 } = 2 + \sqrt { 3 } , }
\end{array} \left\{\begin{array} { l } 
{ \sqrt { u } = \frac { \sqrt { 6 } + \sqrt { 2 } } { 2 } , } \\
{ \sqrt { v } = \frac { \sqrt { 6 } - \sqrt { 2 } } { 2 } , }
\end{array} \left\{\begin{array}{l}
u_{4}=2+\sqrt{3} \\
v_{4}=2-\sqrt{3}
\end{array}\right.\right.\right.\right.
$$

Answer: $(1 ; 4),(4 ; 1),(2-\sqrt{3} ; 2+\sqrt{3})(2+\sqrt{3} ; 2-\sqrt{3})$