least prime in a arithmetic progression  Hello
Here I want to consider the simplest arithmetic progression $n\equiv 1\pmod{q}$ where $q$ is a prime. Is it true that we can find a prime $p\leq q^2$ in this arithmetic progression? 
This question is usually asked for any $q$ instead of only prime value $q$, but I think if one restrict to only prime number $q$ Linnik bound would be easier to compute. Am I right?  
Thank you.
 A: One can clearly improve the L if one restricted the modulus q to prime numbers, however not anywhere near L=2 or so. E.g. in Heath-Brown's article 1992, it's mentioned how you get better results if q has bounded cube-part (i.e. if every prime number p, such that p^3 | q, is less then c1, then there is a c2=c2(1), such that p <= c2 * q^L. Here, the L will be a slightly better constant than if you don't use the assumption of the bounded cube-part, maybe something like 4.5 instead of 5.5, compare a note of "Meng").
The reason for this improvement is, that you have better bounds for character sums (to modulo q) due to Burgess, if you restrict q e.g. to primes (compare §2 of Heath-Brown's 1992 article). Improvement of these bounds then directly translate into improvements of zero-free regions for Dirichlet L-functions (modulo q).
As regards the thought if it would be easier if one restricted oneself only to one residue class, e.g. a mod q where the number a is fixed and q runs, I don't think you can gain something since everything you prove for a fixed a might likely be proven for any other a as well. That thought stems from looking at the proof of Linnik's theorem (meaning the classical proof, there are also some others out there, originating more from sieve theory; I don't know if they would give a different answer but I would be surprised): The whole reasoning is very "symmetric" for whatever a you take. It's not like your average proof in elementary prime number theory (or sieve theory), where you work with a particular sequence a+kq and have sums over different values etc.
Hope this helps.
