Over the years, I've been somewhat in the habit of asking questions in this vein to experts in the Langlands programme.
As is well known, given an algebraic number field $K$, the Langlands programme proposesthey propose to replace the reciprocity map
$$A_K^\*/K^*\rightarrow Gal(K^{ab}/K)$$ of abelian class field theory by a correspondence between the $n$-dimensional representations $\rho$ of $Gal(\bar{K}/K)$ and certain automorphic representations $\pi_{\rho}$ of $GL_n(A_K)$. (We'll skip the Weil group business for this discussion.) Substantial arithmetic information is carried on either side by the $L$-functions, which are supposed to be equal.
This involves deep and beautiful mathematics whenever something can be proved, and there are many applications, such as the Sato-Tate conjecture or this recent paper of Chenevier and Clozel:
http://www.math.polytechnique.fr/~chenevier/articles/galoisQautodual2.pdf
(I mention this one because it is in some ways very close to the point of this question.)
However, there are elementary consequences of abelian class field theory that seem not to have an obvious non-abelian analogueanalogues. The one I wish to mention today has to do with the fundamental group. Given a number field $K$ (assume it's totally imaginary to avoid some silly issues), how can we tell if it has non-trivial abelian unramified extensions? Class field theory says we can look at the class group, which is quite computable in principle, and even in practice for small discriminants. But now, suppose we go on to ask the non-abelian question: which number fields have $$\pi_1(Spec(O_K))=0?$$ That is to say, when does $K$ have no unramified extension at all, abelian or not? As far as I know, there is no easy answer to this question. Niranjan Ramachandran has pointed out that there are at least ten examples, $K=\mathbb{Q}$ (oops, that's real) and $K$ an imaginary quadratic field of class number one. I know of no others. Of course I would be happy to collect some more, if someone else has them lying around.
But the question I really wishedwanted to ask today is: Suppose we are in a Langlands paradise where everything reasonably conjectured by the programme is a theorem. Does this give a way to algorithmically resolve this question as in the abelian case (as we run over fields $K$) resolve this question as in the abelian case? Otherwise, what elseis there a sensible refinement of the usual formulation that would havesubsume such applications?
Added:
I'm embarrassed to be addedadmit I hadn't followed the question mentioned by David Hansen (even after commenting on it). Thanks to David for pointing it out. Of course my main question still stands. I changed the usualtitle following Andy Putman's suggestion. The original title evolved from a (humorously) provocative version that I normally use only among friends who already know I'm a Langlands fan: 'What is the Langlands programme to allow such applicationsgood for?'
Regarding jnewton's very natural thought: in addition to other difficulties, one would also need to bound $n$.