One, (20 points) If the two quadratic equations
$$
\begin{array}{l}
a^{2} x^{2}+a x-1=0, \\
x^{2}-a x-a^{2}=0
\end{array}
$$

have a common solution, find all possible values of $a$.
Let $x_{0}$ be the common solution of the two equations, then we have
$$
\begin{array}{l}
a^{2} x_{0}^{2} + a x_{0} - 1 = 0, \\
x_{0}^{2} - a x_{0} - a^{2} = 0 .
\end{array}
$$

Subtracting the two equations gives $\left(a^{2}-1\right)\left(x_{0}^{2}-1\right)=0$.
If $a^{2}-1=0$, then the original equations become
$$
x^{2} + x - 1 = 0, x^{2} - x - 1 = 0
$$

or
$$
x^{2} - x - 1 = 0, x^{2} + x - 1 = 0
$$
neither of which has a common solution. Therefore, we must have $x_{0}^{2}-1=0$. Substituting this into the original equations yields
$$
a^{2} + a - 1 = 0, 1 - a - a^{2} = 0
$$

or
$$
a^{2} - a - 1 = 0, 1 + a - a^{2} = 0.
$$
In both cases, the possible values of $a$ are
$$
\begin{array}{l}
\frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}, \\
\frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2} .
\end{array}
$$