What automorphisms of $\mathbb{Z}^2$ have order $3$? Phrased differently, what integer matrices
$A = \begin{bmatrix}
a & b
\\
c & d
\end{bmatrix}$
satisfy $A^3 = I$, where $I$ is the identity matrix? I have encountered this problem studying the Fermat curve $F_3$ defined in complex projective 2-space by the equation $x_1^3 + x_2^3 + x_3^3 = 0$. This curve has fundamental group isomorphic to $\mathbb{Z}^2$ and is equipped with an automorphism of order $3$. I want to show that this automorphism must be the identity on the fundamental group, or something nice like that.

The greatest common divisor of $x^3-1 = (x-1)(x^2+x+1)$ and the characteristic polynomial of $A$ must make the matrix vanish.
If the greatest common divisor is $(x-1)$ then $A=I$.
If the greatest common divisor is $(x^2+x+1)$ then the characteristic polynomial of $A$ must be $x^2+x+1$.
We can rewrite the characteristic polynomial as $x^2-(a+d)x +ad - bc$
Hence we must have $d=-a-1$
We now get $-(a)(a+1) - cb = 1$, so $cb = -(a^2+a+1)$
Hence the solutions can be parametrized as:
$\begin{pmatrix}
a & -(a^2+a+1)/k\\
k & -(a+1)
\end{pmatrix}$
where $a$ is an integer and $k$ a divisor of $-(a^2+a+1)$, along with $I$.
