Lower bounds (or less) for the period of sqrt(D) and related sequences This is a continuation of Lower bounds for period length of continued fraction of square root which is a continuation of Upper bound of period length of continued fraction representation of very composite number square root.
Yesterday, I came across a nice article with many problems on continued fractions of Michel Mendes France (from 1993):
http://retro.seals.ch/digbib/fr/view?rid=ensmat-001:1993:39::498&id=&id2=&id3= (broken link)
In problem 6 he asks: Let $x$ be quadratic irrational and $\pi(x)$ the length of its period. Is $sup_n(\pi(x^n))=\infty$?
Is this problem still open? I saw many articles about upper bounds for $\pi(\sqrt{D})$ but none on lower bounds (expect the Golubeva one that is mentioned in the first mentioned question, but due to lack of background, I fail to understand the results there).
As Will Jagy asked: Do we even know whether  $[\pi(\sqrt{n})):n\in \mathbb N]$ is bounded or not?
 A: Edit: The following argument has an error in the second paragraph. I claimed that
the set $\{ D^n\sqrt{D} (\mod 1)\}$ has uncountably many accumulation points, but as
Sam Nead points out in the comments, this doesn't follow from $\sqrt{D}$ being
irrational. On the other hand, it is true that the set $\{ \sqrt{n}(\mod 1)\}$ is
uniformly distributed, and therefore that $\sup \pi(\sqrt{n}) =\infty$. 
This is because $\sqrt{n+1}-\sqrt{n} \to 0$ as $n\to \infty$. See a paper of Elkies
and McMullen for related information on the distribution of this set.
The issue is here that the set of accumulation points of $\{ x^n \mod 1\}$ for
$x$ irrational may be a countable set. This question is actually discussed in Mendes'
paper referred to in the question (I should have looked at it more closely before!). 
As pointed out in that paper, this can occur for Pisot numbers. Also, the
statement is proved by Grisel for many quadratic numbers (as mentioned in a note to Mendes paper). * See also another paper of Grisel.
 I think I can show that $sup_n(\pi(x^n))=\infty$ when $x=\sqrt{D}$. 

Suppose that $x$ is an irrational, non-quadratic number in the interval $(0,1)$, 
and $n_i$ is a sequence of integers such that $\underset{i\to\infty}{\lim} \sqrt{n_i} (\mod 1) = x$. Then the claim is that $\pi(\sqrt{n_i})$
is unbounded. Basically, since $x$ has non-repeating continued fraction expansion, one can show that the continued fraction expansions of $\sqrt{n_i}$ have unbounded period. One can see this geometrically, by noting that there is an element of $PGL_2(\mathbb{Z})$ fixing $\pm \sqrt{n_i}$, and fixing the hyperbolic geodesic in $\mathbb{H}^2$ joining these two points. If $a^2-n_i c^2=\pm 1$ is the minimal solution to the Pell equation, then the matrix is:
$$\begin{bmatrix} a & cn_i \\\ c & a\end{bmatrix}$$

Moreover, the pattern of horoballs which the geodesic penetrates in the Farey packing 
 (source)
tells you the periodic continued fraction. When we take $\sqrt{n_i} (\mod 1)$, these geodesics shifted by $-\left[ \sqrt{n_i} \right]$ approach the vertical geodesic through $x$, whose Farey pattern gives the continued fraction expansion for $x$. Therefore $\pi(\sqrt{n_i})$ is unbounded. 
Now we have to understand the continued fractions of $\sqrt{D^{2n+1}} = D^n \sqrt{D}$, where $D$ is squarefree. The accumulation points of $D^n \sqrt{D} (\mod 1)$ correspond to the truncations of the base $D$ representation of $\sqrt{D}$.  Since $\sqrt{D}$ is irrational, the limit points of these numbers in $[0,1]$ will be uncountable, and therefore contains a non-quadratic irrational $x$ which is a limit of $\sqrt{D^{2n_i+1}} (\mod 1)$, as $n_i\to \infty$. So we see that $\pi(\sqrt{D^{2n_i+1}}) \to \infty$. 
A: I just came across an interesting Ph.D. thesis by Roger Patterson which is very relevant to this topic: http://arxiv.org/pdf/math/0703519v1
There are some $D$ which have unusually short continued fraction periods:
If $D = n^2 + 1$, and the CF of $\sqrt{D}$ is $[n,\overline{2n}]$
If $D = n^2 + 4$ and $n$ odd then the CF of $\sqrt{D}$ is $[n,\overline{\frac{n-1}{2},1,1,\frac{n-1}{2},2n}]$
There are others given in the thesis that I refer to.  He treats various families with unusually short periods and discusses a guess of Kaplansky that these are the only ones with unusually short periods (more precise definitions are given in the paper).
A: Not sure this is relevant: following Mendes-France's paper and on his advice,
I wrote a paper in Acta Arithmetica (1975?) showing in particular that if you
multiply any periodic CF by a rational number $r$ the new period length is
always between $1/f(r)$ and $f(r)$ times the old period length, for an explicit function $f$. For instance $f(2)=5$, so when you multiply by $2$ the period
length is at most multiplied by $5$ (and of course at least $1/5$ of the
initial length).
