Consider a generalised form of the two dimensional Cat map, $S(v)$, where $$ S(v) = Av\,\, (\text{mod 1}) = \begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix} (\text{mod 1}) $$ where it is assumed that all entries of matrix $A$ are integers. I would like to demonstrate that the number of solutions to the equation $$ S(v) = v_0 $$ for a fixed vector $v_0$, is given by $|\text{det}\, A|$. I am stuck on this.
Here is what I have done so far: suppose that $S(v_1) = S(v_2)$. Therefore we would have $$ \begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}x_1 \\ y_1\end{pmatrix} (\text{mod 1}) = \begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}x_2 \\ y_y\end{pmatrix} (\text{mod 1}) $$ The linear equations coming from this system correspond to $$ ax_1 + by_1 \,\,(\text{mod 1}) = ax_2 + by_2\,\,(\text{mod 1}) $$ and $$ cx_1 + dy_1 \,\,(\text{mod 1}) = cx_2 + dy_2\,\,(\text{mod 1}) $$ These can be combined and solved, say, for $x$, getting us to the following: $$ (ad - bc)x_1 \,\,(\text{mod 1}) = (ad-bc)x_2\,\,(\text{mod 1}) $$ Now, this of course can be written like $$ (\text{det}\, A)x_1 \,\,(\text{mod 1}) = (\text{det}\, A)x_2 \,\,(\text{mod 1}) $$ Getting rid of the modulo by finding the unique integers $m$ and $n$ then we get $$ (\text{det}\, A)x_1 + m = (\text{det}\, A)x_2 + n $$ So that both the LHS and RHS $\in [0, 1)$. This can be written like $$ x_2 = x_1 + \frac{p}{\text{det}\,A} $$ where $p = m-n \in \mathbb{Z}$.
Now, we know that both $x_1$ and $x_2$ belong to $[0, 1)$. I would like the result that, given a value of $x_2 \in [0, 1)$, there are precisely $\text{det}\,A$ different combinations of $x_1$ and integer $p$ that make this equation true. However, I'm not sure I see that this is true.
Hints and comments much appreciated.
Presumably you mean the number of solutions in $X = [0,1)^2$.
$S(v) = v_0$ means $Av - v_0 \in \mathbb Z^2$. Now $A$ maps the square $X$ to the parallelogram $P$ with vertices $\pmatrix{0\cr0\cr}$, $\pmatrix{a\cr c\cr}$, $\pmatrix{a+b\cr c+d\cr}$, $\pmatrix{b\cr d\cr}$, which has area $|\det(A)|$. If you cut this parallelogram along the grid lines and translate the pieces back to $X$ (by an integer in each coordinate), each point of $X$ will be covered an average of $|\det(A)|$ times. Now you just need to show that each point will be covered the same number of times.