10.2. Find all solutions of the system of equations in real numbers:

$$
\left(\begin{array}{l}
x^{3}-x+1=y^{2} \\
y^{3}-y+1=x^{2}
\end{array}\right.
$$
Answer.. ( $\pm 1, \pm 1)$ - a total of 4 solutions.

Solution. Move 1 to the right in each equation and multiply the equations, obtaining the equality $\left(x^{2}-1\right)\left(y^{2}-1\right) x y=\left(x^{2}-1\right)\left(y^{2}-1\right)$, from which either $x^{2}-1=0$, or $y^{2}-1=0$, or $x y=1$. In the first two cases, we immediately get 4 solutions $(x, y)=( \pm 1, \pm 1)$. In the third case, substituting $y$ with $\frac{1}{x}$ in the first equation and bringing to a common denominator, we get the equation $x^{5}-x^{3}+x^{2}-1=0$. We factorize it: $\left(x^{2}-1\right)\left(x^{3}+1\right)=\left(x^{2}-1\right)(x+1)\left(x^{2}-x+1\right)=0$. The first two brackets give the already found solutions, the third bracket has no real roots, as its discriminant is less than 0. Therefore, the solutions to the system are 4 pairs $(x, y)=( \pm 1, \pm 1)$.

Grading criteria. Correct answers provided without proof that there are no others: 1 point. Loss of part of the solutions: minus $1-3$ points.