[ Symmetric Systems. Involutory Transformations ] Symmetric Polynomials [ Methods for Solving Problems with Parameters

Solve the system of equations:

$$
\begin{aligned}
& x y=a, \\
& x^{5}+y^{5}=b^{5} .
\end{aligned}
$$

#
Let $a=0$. If $b \neq 0$, there are no solutions; if $b=0$ - an infinite set of solutions of the form $(t,-t)$.

Let $a \neq 0$. Set $x y=t$. Then $x^{5}+y^{5}=(x+y)^{5}-5(x+y)^{3} x y+5(x+y) x^{2} y^{2}=a^{5}-5 a^{3} t+5 a t^{2}$.

Solving the quadratic equation $5 a t^{2}-5 a^{3} t+a^{5}-b^{5}=0$, we find $t=\frac{1}{2}\left(a^{2} \pm \sqrt{\frac{a^{5}+4 b^{5}}{5 a}}\right)$. Further

calculations make sense when

$a\left(a^{5}+4 b^{5}\right) \geq 0$. As a result, we obtain the system of equations $x+y=a, x y=t$.

Thus, $x, y$ are the solutions of the equation $z^{2}-a z+t=0$, that is, they are equal to

$$
\frac{1}{2}\left(a \pm \sqrt{a^{2}-2 a^{2}+2 \sqrt{\frac{a^{5}+4 b^{5}}{5 a}}}\right)=\frac{1}{2}\left(a \pm \sqrt{2 \sqrt{\frac{a^{5}+4 b^{5}}{5 a}}-a^{2}}\right)
$$

We obtain another condition on the parameters: $a\left(16 b^{5}-a^{5}\right) \geq 0$. The final conditions are: $0<a^{5} \leq 16 b^{5}$ or $16 b^{5} \leq$ $a^{5}<0$.

## Answer

For $a=b=0$ - an infinite set of solutions of the form $(t,-t)$;

$$
\text { for } 16 b^{5} \leq a^{5}<0 \text { and } 0<a^{5} \leq 16 b^{5}\{x, y\}=\left\{\frac{a}{2}+\sqrt{\frac{1}{2} \sqrt{\frac{a^{5}+4 b^{5}}{5 a}}-\frac{a^{2}}{4}}, \frac{a}{2}-\sqrt{\frac{1}{2} \sqrt{\frac{a^{5}+4 b^{5}}{5 a}}-\frac{a^{2}}{4}}\right\} \text {. }
$$

[ $\quad \underline{\text { Diophantine equations }}$ ..... ]
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