Quadratic Irrationality of the Periodic points of the Gauss map

Question

If $G:[0,1] \rightarrow [0,1]$ is the Gauss map which is defined as $$G(x) = \left\{\frac{1}{x}\right\} = \frac{1}{x} - \left\lfloor\frac{1}{x}\right\rfloor,$$ show that if $x$ is periodic of order $n$ for $G$ then it must be an quadratic irrational.

I am looking for a quadratic equation I believe. One can solve $G(x)=x$ for $n=1$ quite easily and with some more work can solve $G^2(x)=x$. However, this method does not generalise well to general $n$, due to the number of terms. Moreover, I could not see a formula that induction could be used upon that may make the problem easier.

Can anyone solve this problem?

Answer 1

Ok, I'll assume you are aware that the Gauss map is the same, after a change of 'encoding', as the left shift map on the continued fraction expansion of a real number. Stated in these terms, the (slightly stronger version of the) theorem we want to prove is

Theorem: If $x$ is a real number with eventually periodic continued fraction expansion $$x=[q_0,q_1,\ldots q_{k-1}, \overline{q_k,\ldots,q_{k+n}}]$$ then $x$ is a quadratic irrational.

First let's prove a Lemma. Let $x=[q_0,q_1,\ldots]$ be a real number with infinite continued fraction expansion and let $$\begin{array}{ccc} a_0=q_0 & a_1=q_0q_1+1 & a_{n+2}=a_{n+1}q_{n+2}+a_n\\ b_0=1 & b_1=q_1 & b_{n+2}=b_{n+1}q_{n+2}+b_n. \end{array}$$ The rational $a_n/b_n$ is the usual $n$th convergent to $x$.

Lemma: Let $x_n=G^n(x)=[q_n,q_{n+1},\ldots]$. If $x>1$ then $$x=\frac{x_na_n + a_{n-1}}{x_nb_n + b_{n-1}}.$$

Proof: We'll prove this by induction. The $n=1$ case is some simple algebra to see that $$x=[q_0,q_1,x_1]=\frac{x_1q_0q_1+x_1+q_0}{x_1q_1+1}=\frac{x_1a_1 + a_0}{x_1b_1 + b_0}.$$ Suppose the Lemma is true for some $n\geq 1$, then $$\begin{array}{rcl} x=[q_0,\ldots, q_{n}, q_{n+1}, x_{n+1}] & = & [q_0,\ldots,q_{n},q_{n+1}+\frac{1}{x_{n+1}}]\\ & = & \dfrac{(q_{n+1} + \frac{1}{x_{n+1}})a_n + a_{n-1}}{(q_{n+1} + \frac{1}{x_{n+1}})b_n + b_{n-1}}\\ & = & \dfrac{x_{n+1}a_{n+1} + a_n}{x_{n+1} b_{n+1} + b_n} \end{array}$$ where the inductive hypothesis was used to get from the first to second line because $x_n = q_{n+1} + \frac{1}{x_{n+1}}$. We used the recursive definition of the $a_i$ and $b_i$ to get to the last line, along with some simple algebra (multiply through by $x_{n+1}$).

Now we can prove the theorem.

Proof: Let $x=[q_0,q_1,\ldots, q_{k-1}, \overline{q_{k},\ldots, q_{k+n}}]$ so that by the lemma we have $$x=\frac{x_k a_k + a_{k-1}}{x_k b_k + b_{k-1}}$$ and $$x_k=[q_k,\ldots, q_{k+n-1},x_k]=\frac{x_k a'_n + a'_{n-1}}{x_k b'_n + b'_{n-1}}$$ where $a'_n/b'_n$ is the $n$th convergent to $x_k$. So we can write $x_k(x_k b'_{n} + b'_{n-1})=x_k a'_n + a'_{n-1}$ which is a quadratic in $x_k$ with rational coefficients so $x_k$ is a quadratic irrational. If $x_k$ is the root of a quadratic, then it follows quickly from the first equation above that $x$ must also be a quadratic irrational and so we are done.