TITLE: For a sufficiently large $a$, are there distinct (mod $a$) integers such that all powers up to the $n$-th are "close" modulo $a$?
QUESTION [5 upvotes]: Given $n\in\Bbb N$ is there an $a_n\in\Bbb N$ such that for every $a>a_n$ there are two distinct integers $0<b<c<a$ such that $b^i\bmod a,c^i\bmod a\in(\sqrt a,\sqrt a\log a)$ for every $i\in\{1,\dots,n\}$? 
If so what is minimum such $a$?

REPLY [2 votes]: This is only a heuristic, but perhaps it can be made rigorous with some
effort (using Weil's bound etc).
I write $m$ instead of your $a$. Let $I:=(\sqrt m,\sqrt m\log m)\pmod m$. You
want to show that there are at least two distinct values of $b\in{\mathbb
Z}_m$ such that $b,\dotsc,b^n\in I$. The number of such $b$ is
  $$ \frac1{m^n} \sum_{x_1,\dotsc,x_n\in I} \sum_{u_1,\dotsc,u_n\in{\mathbb Z}_m}
      \sum_{b\in{\mathbb Z}_m} e^{ \frac{(b-x_1)u_1+\dotsb+(b^n-x_n)u_n}{m} }. $$
The main term obtained for $u_1=\dotsb=u_n=0$ is $|I|^n/m^{n-1}$, which is of
the order of magnitude $m^{1-n/2}(\log m)^n$. This is a small number for
$n\ge 3$ and $m$ large, and it is thus plausible to expect that in this case you cannot (in general) find even one single $b\in{\mathbb Z}_m$ with the property in question.