Let $A$ be $2×2$ matrix such that their elements are Integer, inverse matrix exits and $A^2=E$. I'm thinking about decomposing $A$ into family by regarding $A$ as a family of $T^{-1}AT$.
I wonder how many families are there? I expect there are only $4$ families , $\begin{bmatrix} ±1 & 0 \\ 0 & ±1 \end{bmatrix}$.
But how to prove that?
By Cayley-Hamilton theorem ,$1+tr(A)+det(A)=0$
By the existence of inverse matrix, $det(A)=±1$
Now I came to a deadlock.
Since $A^2=E$, $A$ is invertible. It is also diagonalizable, because otherwise its Jordan form over $\Bbb{C}$ would be $$ J=\begin{pmatrix} \lambda & 1 \cr 0 & \lambda\end{pmatrix}, $$ with $\lambda=\pm 1$. But then $J$ has infinite order, hence $A$ too, a contradiction to $A^2=E$. So $A$ is diagonalizable, i.e., $$ TAT^{-1}=\begin{pmatrix} \lambda & 0 \cr 0 & \mu\end{pmatrix}. $$ Then we obtain $$ \begin{pmatrix} \lambda^2 & 0 \cr 0 & \mu^2\end{pmatrix}=(TAT^{-1})^2=TA^2T^{-1}=TET^{-1}=E, $$ so that $\lambda^2=\mu^2=\pm 1$. Over the integers there are exactly the four possibilities you have obtained. Two of them are still conjugated.