In certain contexts in Diophantine approximation, it is important for the denominators of the convergents to be restricted to a certain residue class. Consider, for instance, this problem where it is shown that $sum_{n geq 1} sin^{n}(n)$ diverges. In their solutions, the answerers utilized the fact that $pi/2$ has infinitely many convergents $p/q$ with odd denominators.

This question concerns whether, in the general case, restricting the denominators to certain residue classes makes no “significant” difference to the theory.

The specific questions I’d ask are as follows. Recall that to each irrational $alpha$, there are two standard quantities we associate to $alpha$ which measure its approximability.

One is the irrationality measure $lambda(alpha) = supleft{mu >0 : |alpha – p/q|<frac{1}{q^{mu}} text{has infinitely many} text{solutions} p,q in mathbb{Z}, q neq 0right}$. The other is the so-called Markov constant $M(alpha) = supleft{C>0: |alpha – p/q|<frac{1}{Cq^{2}} text{has infinitely many} text{solutions} p,q in mathbb{Z}, q neq 0right}$.

Now, let $a,b in mathbb{N}$, and define $lambda_{a,b}(alpha), M_{a,b}(alpha)$ similar to above, but with the additional constraint that $q equiv a mod b$.

Must $lambda(alpha)=lambda_{a,b}(alpha), M(alpha)=M_{a,b}(alpha)$? If not, are there quantitative estimates on how “bad” $lambda_{a,b}(alpha)$ or $M_{a,b}(alpha)$ can get?

A simple solution to problems like this is to simply consider the sequence of continued fraction convergents, and check whether the reduced denominators fall into any given residue class infinitely many times. In the case where $a=1$, $b=2$ (i.e., when we want $q$ odd), it is easy, since consecutive denominators in the convergents are necessarily coprime, hence we cannot have consecutive even denominators, hence we have infinitely many odd denominators. However, I’m not sure how to go about attacking the general problem.