TITLE: How to construct a small coprime?
QUESTION [18 upvotes]: Given an integer $n$, is there a deterministic algorithm to find in poly$(\log n)$ time an integer $q$, $n < q< n^{c}$, such that $gcd(q,n!)=1$?  Here $c>1$ is some fixed constant. 
Obviously, a small prime $n< q < 2n$ works.  But those are really hard to find deterministically.  One can of course, deterministically test primality of $n+1, n+2, \ldots, n+C(\log n)^2,$ but that's not yet proved to work.  Taking $q=n!+1$ works for the $gcd$ condition, but such $q$ is too large for my applications.  If $q< n^c$ is open, but say for the case $q< n^{O((\log n)^k)}$ this is known, what are the best known bounds?

REPLY [2 votes]: If we could find a number co-prime to $2^n!$ in $n^{O(1)}$ time, or even just a number divisible by at least one prime greater than $2^n$, we could factor it to find such a prime.  This would constitute a solution to the strong conjecture with factoring, so it is an open problem.
As far as I know, it is open whether or not it is possible to find a prime larger than $2^n$ in time $2^{\frac{n}{2}+o(1)}$ with or without a factoring oracle.  The finding primes page doesn't say this explicitly but I don't see any obvious method to beat this using a factoring oracle from the information presented.  Perhaps someone can confirm that this weaker problem is open.