## Task A-2.1. (10 points)

Depending on the real parameter $a$, solve the equation

$$
(a-1)\left(1+x+x^{2}\right)^{2}=(a+1)\left(1+x^{2}+x^{4}\right)
$$

in the set of real numbers.
## Solution.

We have

$$
\begin{aligned}
1+x^{2}+x^{4} & =\left(1+x+x^{2}\right)^{2}-\left(2 x+2 x^{2}+2 x^{3}\right) \\
& =\left(1+x+x^{2}\right)^{2}-2 x\left(1+x+x^{2}\right) \\
& =\left(1+x+x^{2}\right)\left(\left(1+x+x^{2}\right)-2 x\right)
\end{aligned}
$$

Therefore, the given equation can be written in the form

$$
\begin{array}{cc}
(a-1)\left(1+x+x^{2}\right)^{2}=(a+1)\left(1+x+x^{2}\right)\left(\left(1+x+x^{2}\right)-2 x\right) & \\
\left(1+x+x^{2}\right)\left(((a-1)-(a+1))\left(1+x+x^{2}\right)+2 x(a+1)\right)=0 & 1 \text { point } \\
\left(1+x+x^{2}\right)\left(-2\left(1+x+x^{2}\right)+2 x(a+1)\right)=0 & \\
\left(1+x+x^{2}\right)\left(-2-2 x-2 x^{2}+2 a x+2 x\right)=0 & 1 \text { point }
\end{array}
$$

The equation $x^{2}+x+1=0$ has no real solutions.

The discriminant of the equation $x^{2}-a x+1=0$ is $D=a^{2}-4$, so it has real solutions if and only if $|a| \geqslant 2$.

For $a \in(-2,2)$, the given equation has no real solutions,

for $a=2$ the equation has one solution, $x=1$, and for $a=-2$ the solution is $x=-1$.

For $a>2$ the equation has two distinct real solutions,

$$
x_{1,2}=\frac{a \pm \sqrt{a^{2}-4}}{2}
$$