23. Let $x$, $y$, and $a$ be real numbers. And satisfy $x+y=x^{3}+y^{3}=$ $x^{5}+y^{5}=a$. Find all possible values of $a$.
(Greece for the 43rd IMO selection exam)
If $x=-y$, then $a=0$.
Let $x, y$ be the two roots of the quadratic equation $z^{2}-a z+p=0$, then
$$
\begin{array}{l}
x+y=a, \\
x^{2}+y^{2}=(x+y)^{2}-2 x y=a^{2}-2 p, \\
x^{3}+y^{3}=a^{3}-3 a p, \\
x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=a^{4}-4 a^{2} p+2 p^{2}, \\
x^{5}+y^{5}=a^{5}-5 a^{3} p+5 a p^{2} .
\end{array}
$$

When $a \neq 0$, from $x^{3}+y^{3}=a$, we get
$$
a^{3}-3 a p=a, p=\frac{a^{2}-1}{3} \text {. }
$$

From $x^{5}+y^{5}=a$, we get $a^{5}-5 a^{3} p+5 a p^{2}=a$. Thus,
$$
a^{4}-5 a^{2} \cdot \frac{a^{2}-1}{3}+5\left(\frac{a^{2}-1}{3}\right)^{2}=1 \text {, }
$$

which simplifies to $a^{4}-5 a^{2}+4=0$.
Therefore, $a= \pm 1, a= \pm 2$.
In summary, the possible values of $a$ are $-2,-1,0,1,2$.