Are there any non-trivial rational integers in the $p$-adic closure of $\{1,q,q^2,q^3,...\}$? If $p$ is prime and not a divisor of $q$, are there any non-trivial rational integers in the $p$-adic closure of the set or powers of $q$?
Edit: $q$ is also a (rational) integer, not a $p$-adic.
 A: You never said what $q$ is (or what a nontrivial rational integer is :)). I presume you meant that $q$ is some $p$-adic integer and you want an integer outside the sequence of nonnegative powers of $q$ which is a $p$-adic limit of a subsequence of the nonnegative integral powers of $q$. 
In that case, there are lots of examples: if $p > 2$ and $q = -1 \bmod p$ then the subsequence $q^{p^k}$ tends to -1 as $k$ tends to infinity. More generally, for any $p$-adic integer $q$, the subsequence $q^{p^k}$ tends to the unique solution of $t^p = t$ in the $p$-adic integers satisfying $t = q \bmod p$. So when $p$ is odd and $q = -1 \bmod p$, $t = -1$.
A: If $p>2$ then there is an integer $q$ not divisible by $p$, with the properties: $q^k\not\equiv 1$ mod $p$ for $1\leq k\leq p-2$ and $q^{p-1}\not\equiv 1$ mod $p^2$. Under these conditions the $p$-adic closure of $\{1,q,q^2,\dots\}$ is the whole $\mathbb{Z}_p^\times$ - in particular it contains all the rational integers not divisible by $p$.
For $p=2$ and $q\equiv 5$ mod $8$ then the closure is $1+4\mathbb{Z}_2$ - i.e. it contains all the integers which are $1$ mod $4$.
