1. the following equations apply to the real numbers $x, y, a$:

$$
\begin{aligned}
x+y & =a \\
x^{3}+y^{3} & =a \\
x^{5}+y^{5} & =a
\end{aligned}
$$

Determine all possible values of $a$.
## 1st solution:

The polynomials on the left-hand side of the three equations are symmetrical in $x$ and $y$, we can therefore express them by the elementary symmetrical polynomials $u=x+y$ and $v=x y$ and obtain

$$
\begin{aligned}
u & =a \\
u^{3}-3 u v & =a \\
u^{5}-5 u^{3} v+5 u v^{2} & =a
\end{aligned}
$$

If you insert $x=y=0$ into the equations, you can see that $a$ can take the value 0. We therefore assume $a \neq 0$ in the following. Inserting $u=a$ into the second equation gives $a^{3}-3 a v=a$, because of $a \neq 0$ we can divide by $a$ and get $v=\left(a^{2}-1\right) / 3$. If we insert this and $u=a$ into the third equation, we get $a^{5}-5 a^{3}\left(a^{2}-1\right) / 3+5 a\left(a^{2}-1\right)^{2} / 9=a$, after multiplying by $-9 / a$ and simplifying, we get

$$
a^{4}-5 a^{2}+4=0 \quad \Longleftrightarrow \quad\left(a^{2}-1\right)\left(a^{2}-4\right)=0
$$

We obtain the possible values $a^{2}=1$ or $a^{2}=4$, i.e. for $a$ the possible values $-2,-1,1,2$. We also show that these four values can actually occur. To do this, we insert the following values for $x$ and $y$ into the three equations in sequence: $x=y=-1, x=0, y=-1, x=0, y=1$ and $x=y=1$. In total, $a$ can therefore have exactly five values $-2,-1,0,1,2$.