5. (5 points) Solve the system of equations:

$$
\left\{\begin{array}{c}
x^{3}+4 y=y^{3}+16 x \\
\frac{y^{2}+1}{x^{2}+1}=5
\end{array}\right.
$$
# Solution:

The following equivalent transformations are valid:

$$
\left\{\begin{array} { c } 
{ x ^ { 3 } + 4 y = y ^ { 3 } + 1 6 x } \\
{ \frac { y ^ { 2 } + 1 } { x ^ { 2 } + 1 } = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
y^{2}=5 x^{2}+4 \\
x^{3}+4 y=y^{3}+16 x
\end{array} \Leftrightarrow\right.\right.
$$

$$
\begin{aligned}
& {\left[\begin{array}{c}
y=\sqrt{5 x^{2}+4} \\
\left\{\begin{array}{c}
y \\
x^{3}+4 \sqrt{5 x^{2}+4}=\left(5 x^{2}+4\right) \sqrt{5 x^{2}+4}+16 x
\end{array}\right. \\
\left\{\begin{array}{c}
y=-\sqrt{5 x^{2}+4} \\
x^{3}-4 \sqrt{5 x^{2}+4}=-\left(5 x^{2}+4\right) \sqrt{5 x^{2}+4}+16 x
\end{array}\right.
\end{array} \Leftrightarrow\right.}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f859f267290c335bb97ag-04.jpg?height=562&width=1153&top_left_y=593&top_left_x=523)

$$
\begin{aligned}
& {\left[\begin{array}{c}
\left\{\begin{array}{c}
x=0 \\
y= \pm 2
\end{array}\right. \\
\left\{\begin{array}{c}
y=\sqrt{5 x^{2}+4} \\
\sqrt{5 x^{2}+4}=\frac{x^{2}-16}{5 x}
\end{array} \Leftrightarrow\right. \\
\left\{\begin{array}{c}
y=-\sqrt{5 x^{2}+4} \\
\sqrt{5 x^{2}+4}=-\frac{x^{2}-16}{5 x}
\end{array}\right.
\end{array}\right.}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f859f267290c335bb97ag-05.jpg?height=1077&width=456&top_left_y=184&top_left_x=880)

Solve the equation \(5 x^{2}+4=\frac{\left(x^{2}-16\right)^{2}}{25 x^{2}}\) for \(x \neq 0\). This equation is equivalent to the biquadratic equation \(31 x^{4}+33 x^{2}-64=0\), whose solutions are \(x=1\) and \(x=-1\).

Substituting the obtained values into the constraints of the system (1), we get the answer.

Answer: \((0 ; \pm 2), (\pm 1 ; \mp 3)\).