Task A-2.3.

Find all pairs of complex numbers $(w, z), w \neq z$, that satisfy the system of equations

$$
\begin{aligned}
w^{5}+w & =z^{5}+z \\
w^{5}+w^{2} & =z^{5}+z^{2}
\end{aligned}
$$
## Solution.

By subtracting the given equations, we obtain the equation

$$
w-w^{2}=z-z^{2}
$$

which is equivalent to the equation

$$
(w-z)(w+z-1)=0
$$

Since $w-z \neq 0$, it follows that

$$
w+z=1
$$

Squaring this equality, we get

$$
w^{2}+z^{2}=1-2 w z
$$

and squaring again,

$$
w^{4}+z^{4}=1-4 w z+2(w z)^{2}
$$

The first equation of the given system is equivalent to the equation

$$
\frac{w^{5}-z^{5}}{w-z}=-1
$$

It can be written in the form

$$
\begin{aligned}
\frac{(w-z)\left(w^{4}+w^{3} z+w^{2} z^{2}+w z^{3}+z^{4}\right)}{w-z} & =-1 \\
w^{4}+w^{3} z+w^{2} z^{2}+w z^{3}+z^{4}+1 & =0 \\
w^{4}+z^{4}+w z\left(w^{2}+z^{2}+w z\right)+1 & =0
\end{aligned}
$$

Substituting (2) and (3) and simplifying, we obtain the equation

$$
(w z)^{2}-3 w z+2=0
$$

Solving this, we find that $w z=1$ or $w z=2$.

Substituting $z=\frac{1}{w}$ and $z=\frac{2}{w}$ into (1), we get the quadratic equations $w^{2}-w+1=0$ and $w^{2}-w+2=0$, from which we finally obtain the four solutions

$$
(w, z)=\left(\frac{1 \pm \sqrt{3} i}{2}, \frac{1 \mp \sqrt{3} i}{2}\right) \quad \text { and } \quad(w, z)=\left(\frac{1 \pm \sqrt{7} i}{2}, \frac{1 \mp \sqrt{7} i}{2}\right)
$$