One, (20 points) Given real numbers $x$, $y$, and $a$ satisfy
$$
x+y=x^{3}+y^{3}=x^{5}+y^{5}=a \text {. }
$$

Find all possible values of $a$.
If $x=-y$, then $a=0$.
Let $x, y$ be the two roots of the quadratic equation $z^{2}-a z+p=0$.
Thus, $x+y=a$.
Then $x^{2}+y^{2}=(x+y)^{2}-2 x y=a^{2}-2 p$, $x^{3}+y^{3}=a^{3}-3 a p$,
$x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}$
$=a^{4}-4 a^{2} p+2 p^{2}$,
$$
x^{5}+y^{5}=a^{5}-5 a^{3} p+5 a p^{2} \text {. }
$$

When $a \neq 0$, from $x^{3}+y^{3}=a$, we get
$a^{3}-3 a p=a \Rightarrow p=\frac{a^{2}-1}{3}$.
Thus, $a^{4}-5 a^{2} \cdot \frac{a^{2}-1}{3}+5\left(\frac{a^{2}-1}{3}\right)^{2}=1$
$\Rightarrow a^{4}-5 a^{2}+4=0$
$\Rightarrow a= \pm 1, a= \pm 2$.
In summary, all possible values of $a$ are $\pm 2, \pm 1, 0$.