Roth's theorem states that for every real algebraic $\alpha$ and $\epsilon>0$, there is a $c>0$ such that
$$|\alpha -\frac{p}{q}| > \frac{c}{q^{2+\epsilon}}.$$


Lang's conjecture strengthened this to 
$$|\alpha -\frac{p}{q}| > \frac{c}{q^2 (\log q)^{1+\epsilon}}.$$ 


A naive further strengthening would be to ask for 
$$|\alpha -\frac{p}{q}| > \frac{c}{q^2},$$
and this is satisfied by all $\alpha$ with bounded
partial quotients $a\_n$ (Khinchin Thm 23). 


Life would be nice and simple if it were true that all algebraic $\alpha$ has bounded $a\_n$. This seems too much to be asking for but I am ignorant of any counterexample. Does anyone knows a counter example or even some heuristic reasoning why this is unlikely to be true ? Thanks in advance.



Is it possile for all real algebraic numbers to have continued fractions with bounded partial quotients ?
There is no example of an algebraic number of degree $> 2$ for which the boundedness or not of the entries of the continued fraction has been determined. In a 1976 Annals of Math paper, Baum & Sweet treat the analogous problem with the ring of rational integers replaced by $\mathbb{F}\_ 2[x]$ (and its infinite place). They found algebraic quantities over $\mathbb{F}\_ 2(x)$ contained in $\mathbb{F}\_ 2((x))$ of degree $> 2$ whose "continued fraction" entries have bounded degree, and others with unbounded degree. So it all looks a bit puzzling.


