4. Find all values of the parameter $a$ for which the system $\left\{\begin{array}{l}y=|a-3| x+1|+x+3|+3|x+1|, \\ 2^{2-y} \log _{\sqrt{3}}\left((x+|a+2 x|)^{2}-6(x+1+|a+2 x|)+16\right)+2^{x+|a+2 x|} \log _{1 / 3}\left(y^{2}+1\right)=0, \text { has a unique } \\ x+|a+2 x| \leq 3,\end{array}\right.$

solution $(x ; y)$, where $x$ and $y$ are integers. Specify this solution for each found $a$. (16 points)
# Solution:

Transform the second equation of the system as follows:

$$
\begin{aligned}
& 2^{3-y} \log _{3}\left((x+|a+2 x|)^{2}-6(x+|a+2 x|)+9+1\right)-2^{x+|a+2 x|} \log _{3}\left(y^{2}+1\right)=0 \\
& 2^{3-y} \log _{3}\left((x+|a+2 x|-3)^{2}+1\right)=2^{x+|a+2 x|} \log _{3}\left(y^{2}+1\right) \\
& 2^{3-x-|a+2 x|} \log _{3}\left((x+|a+2 x|-3)^{2}+1\right)=2^{y} \log _{3}\left(y^{2}+1\right)
\end{aligned}
$$

From the first equation of the system, it follows that \( y \geq 0 \). By the condition \( 3-x-|a+2 x| \geq 0 \). The function \( f(y)=2^{y} \log _{3}\left(y^{2}+1\right) \) is increasing for \( y \geq 0 \). The last equation \( f(3-x-|a+2 x|)=f(y) \), therefore, \( 3-x-|a+2 x|=y \). Substituting into the first equation of the system, we get \( 3-x-|a+2 x|=|a-3| x+1|+x+3|+3|x+1|, \quad|a-3| x+1|+x+3|+|a+2 x|=3-x-3|x+1| \).

Let \( u=a-3|x+1|+x+3 \) and \( v=a+2 x \). Then the original equation will be \( |u|+|v|=u-v \). The solutions to the last equation are all \( u \) and \( v \) such that \( u \geq 0 \) and \( v \leq 0 \), or \( a-3|x+1|+x+3 \geq 0 \) and \( a+2 x \leq 0 \). From this, we have \( 3|x+1|-x-3 \leq a \leq -2 x \).

In the Ocha system, we construct the graphs of the functions \( a=3|x+1|-x-3 \) and \( a=-2 x \).

If \( x \) is an integer, then \( y=|a-3| x+1|+x+3|+3|x+1| \) will be an integer if \( a \) is an integer. For \( a=-2 \), we have the solution \( x=-1, \quad y=0 \). For \( a=-1 \), we have the solution \( x=-1, \quad y=1 \).

For \( a=1 \), we have the solution \( x=-1, \quad y=3 \). For \( a=3 \), we have the solution \( x=-2, y=4 \). For \( a=4 \), we have the solution \( x=-2, \quad y=5 \). For \( a=6 \), we have the solution \( x=-3, \quad y=6 \). Note that all the obtained values satisfy

\( 3-x-|a+2 x| \geq 0 \).

Answer: \( a=-2, x=-1, \quad y=0 \);

\( a=-1, x=-1, \quad y=1 \);

\( a=1, x=-1, \quad y=3 \)

\( a=3, x=-2, \quad y=4 \)

inequality

\( a=4, x=-2, \quad y=5 \)

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-03.jpg?height=1171&width=942&top_left_y=1393&top_left_x=957)

\( a=6, x=-3, \quad y=6 \).