4. Find all values of the parameter $a$ for which the system

$$
\left\{\begin{array}{c}
\left(a\left|\log _{2} y\right|+a\left|\log _{2} x\right|-2\right)\left(\left(\log _{2} x\right)^{2}+\left(\log _{2} y\right)^{2}-48 a^{2}\right)=0 \\
\left(\log _{a} x\right)^{2}\left(\log _{a} y\right)^{2}\left(\log _{2} a^{2}\right)^{4}=256 a^{2}
\end{array}\right.
$$

has eight distinct solutions. (16 points)
Solution. Let's make a change of variables: $\log _{2} x=u, \log _{2} y=v$. Variables $u$ and $v$ can take any values, $x$ and $y$ are uniquely determined. Simplify the second equation of the system: $\left(\log _{a} x\right)^{2}\left(\log _{a} y\right)^{2}\left(\log _{2} a^{2}\right)^{4}=256 a^{2} \Leftrightarrow a>0, a \neq 1,\left(\log _{2} x\right)^{2}\left(\log _{2} y\right)^{2}=16 a^{2}$, $|u v|=4 a$. Then we get

$$
\left\{\begin{array} { c } 
{ ( a | v | + a | u | - 2 ) ( u ^ { 2 } + v ^ { 2 } - 4 8 a ^ { 2 } ) = 0 , } \\
{ | u v | = 4 a , a > 0 , a \neq 1 , }
\end{array} \Leftrightarrow \left[\begin{array}{c}
|v|+|u|=2 / a \\
|| u v \mid=4 a, a>0, a \neq 1 \\
\left\{\begin{array}{c}
u^{2}+v^{2}=48 a^{2}
\end{array}\right. \\
|u v|=4 a, \quad a>0, a \neq 1
\end{array}\right.\right.
$$

I. $\{\quad|v|+|u|=2 / a$, $\{|u v|=4 a, a>0, a \neq 1$.

1) The system has 4 different solutions if

$$
\frac{2}{a}=4 \sqrt{a}, a=\frac{1}{\sqrt[3]{4}}
$$

2) The system has 8 different solutions if

$$
\frac{2}{a}>4 \sqrt{a}, 00, a \neq 1 \text {. }\end{array}\right.$

1) The system has 4 different solutions if

$$
4 \sqrt{3} a=2 \sqrt{2 a}, \quad a=1 / 6
$$

2) The system has 8 different solutions if $4 \sqrt{3} a>2 \sqrt{2 a}, a>1 / 6, a \neq 1$.

Possible coincidences of solutions in cases I and II:

$$
\left\{\begin{array}{c}
|v|+|u|=2 / a \\
u^{2}+v^{2}=48 a^{2} \\
|u v|=4 a, \quad a>0, a \neq 1
\end{array}\right.
$$

By symmetry, we can consider only the case where $u>0, v>0$. Then

$$
\left\{\begin{array} { c } 
{ v + u = 2 / a , } \\
{ u ^ { 2 } + v ^ { 2 } = 4 8 a ^ { 2 } , } \\
{ u v = 4 a , a > 0 , a \neq 1 . }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
v+u=2 / a \\
(u+v)^{2}=48 a^{2}+8 a \\
u v=4 a, a>0, a \neq 1
\end{array}\right.\right.
$$

We have $4 / a^{2}=48 a^{2}+8 a, \quad 12 a^{4}+2 a^{3}-1=0$, $a=1 / 2, \quad 12 a^{4}+2 a^{3}-1=(a-1 / 2)\left(12 a^{3}+8 a^{2}+\right.$

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-13.jpg?height=756&width=774&top_left_y=256&top_left_x=1189)
$4 a+2)$. The polynomial $12 a^{3}+8 a^{2}+4 a+2$ has no positive roots. When $a=1 / 2$, the solutions in cases I and II coincide, and the system will have 8 solutions.

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-13.jpg?height=471&width=1607&top_left_y=2015&top_left_x=196)

Answer: $a \in(0 ; 1 / 6) \cup\{1 / 2\} \cup(1 / \sqrt[3]{4} ; 1) \cup(1 ;+\infty)$.