## Task 6 - 191236

Given a positive real number $a$.

Determine (for each value of this given number $a$) all real solutions $(x ; y)$ of the system of equations

$$
x^{5}+y^{5}=1 \quad, \quad x+y=a
$$
We utilize the symmetry by substituting $x=\frac{a}{2}+z$ and $y=\frac{a}{2}-z$. The second condition is automatically satisfied. From the first condition, we get:

$$
\begin{gathered}
\left(\frac{a}{2}+z\right)^{5}+\left(\frac{a}{2}-z\right)^{5}=1 \\
2\left(\frac{a}{2}\right)^{5}+2 \cdot 10\left(\frac{a}{2}\right)^{3} z^{2}+2 \cdot 5\left(\frac{a}{2}\right) z^{4}=1
\end{gathered}
$$

This is a biquadratic equation, which is easy to solve:

$$
\begin{gathered}
\frac{a^{5}}{16}+\frac{5 a^{3}}{2} z^{2}+5 a z^{4}=1 \\
z^{4}+\frac{1}{2} a^{2} z^{2}+\frac{a^{4}}{80}-\frac{1}{5 a}=0 \\
z^{2}=-\frac{1}{4} a^{2}+\sqrt{\frac{1}{16} a^{4}-\frac{1}{80} a^{4}+\frac{1}{5 a}}
\end{gathered}
$$

(Since $z^{2} \geq 0$ must hold, only this solution is valid).

$$
z^{2}=-\frac{1}{4} a^{2}+\sqrt{\frac{1}{20} a^{4}+\frac{1}{5 a}}=-\frac{1}{4} a^{2}+\frac{1}{2} \sqrt{\frac{a^{5}+4}{5 a}}
$$

Thus, we have:

$$
\begin{aligned}
z & = \pm \sqrt{-\frac{1}{4} a^{2}+\frac{1}{2} \sqrt{\frac{a^{5}+4}{5 a}}} \\
x_{1,2} & =\frac{1}{2} a \pm \frac{1}{2} \sqrt{2 \sqrt{\frac{a^{5}+4}{5 a}}-a^{2}}
\end{aligned}
$$

The solutions are then, due to symmetry, $(x, y)=\left(x_{1}, x_{2}\right)$ and $(x, y)=\left(x_{2}, x_{1}\right)$.