Problem 5. Solve the system of equations in real numbers

$$
\left\{\begin{array}{l}
x+y+2+4xy=0 \\
y+z+2+4yz=0 \\
z+x+2+4zx=0
\end{array}\right.
$$
Answer: No solutions.

Solution. Note that $x \neq-\frac{1}{4}$ : otherwise, substituting into the first equation, we get $-\frac{1}{4}+2=0$, which is false. Therefore, from the first and third equations, we obtain $y=-\frac{x+2}{1+4 x}, z=-\frac{x+2}{1+4 x}$, from which $y=z$. Similarly, $x=y$. Therefore, $x=y=z$.

Substituting the obtained equality into the first equation, we get $2 x+2+4 x^{2}=0$, or $2 x^{2}+x+1=0$. The discriminant of this equation is $1-8<0$, i.e., the system has no solutions.