Proving Arnold's cat map has only rational periodic points. I was trying to prove that all periodic points of the map are rational points. I started with $\begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}= {\begin{bmatrix}
           1 & 1 \\
           1 & 2 \\
         \end{bmatrix}}^n \begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}$ and manipulated it using diagnolization and arrived at $\begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}=
         \begin{bmatrix}
           a & b \\
           b & c \\
         \end{bmatrix} \begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}$ where a,b,c are fibonacci numbers. This seems to me to imply the only periodic point is the zero vector. But I know that isn't correct. I think it might have to do with me no including mod 1, but how do I account for that throughout the problem?
 A: Denote
$$
x = (x_1,x_2)^\top, \quad A = \pmatrix{1&1\\1&2}.
$$
Suppose that $x$ is periodic. That is, there exists an integer $n$ such that $A^nx - x \in \Bbb Z^2$. In other words, $(A^n - I)x \in \Bbb Z^2$. Let $y = (A^n - I)x$, and let $M = A^n - I$. We see that $M$ is a matrix with integer entries and, importantly, $M$ is invertible (since neither of the eigenvalues of $A$ satisfy $\lambda^n = 1$). It follows that $M^{-1}$ is a matrix with integer entries, and
$$
Mx = y \implies x = M^{-1}y.
$$
Because $M^{-1}$ and $y$ have rational entries, we conclude that $x$ must have rational entries.

The converse of this statement (that every rational point is periodic) is a bit more interesting. Suppose that $x$ has rational entries; we want to show that there exists a positive integer $n$ for which $(A^n - I)x$ has integer entries. We can write $x$ in the form $x = \frac 1d z$ for some common denominator $d \in \Bbb Z$ and vector $z \in \Bbb Z^2$. With that in mind, it suffices to show that there exists an $n$ for which the entries of $A^n - I$ are a multiple of $d$.
In other words, it suffices to show that there necessarily exists an $n$ for which
$$
A^n = I \pmod d.
$$
We can prove this as follows: by the pigeonhole principle, there exist positive integers $n_1<n_2$ for which
$$
A^{n_1} \equiv A^{n_2} \pmod d.
$$
However, $A$ has an inverse with integer entries. So, we can deduce that
$$
[A^{-1}]^{n_1} A^{n_1} \equiv [A^{-1}]^{n_1} A^{n_2} \pmod d \implies\\
I \equiv A^{n_2 - n_1} \pmod d.
$$
Thus, the desired conclusion holds with $n = n_2 - n_1$.
A: Of course you can just account for mod 1 at the end and have Our equation tells us that $(a-1)x_1+bx_2=0$ mod 1, that is to say that for some integer k, $(a-1)x_1+bx_2=k$. In fact if we let the first equation equal a $k_1$ and the second, $k_2$, in general we see $x_1=\frac{k_1}{a-1}-\frac{b(k_2+\frac{bk_1}{1-a})}{\frac{b}{1-a}+c-1}$ and $x_2=\frac{k_2+\frac{bk_1}{1-a}}{\frac{b}{1-a}+c-1}$ both of which must be rational numbers.
