p-adic series converging to two different values for two different p-adic norms

Question

I am aware of the definition of p-adic numbers, and the notion of convergence that applies.

The task given to me is as follows:

Find an explicit sequence of natural numbers that converges to $3$ in $Q_3$ and $5$ in $Q_5$.

This is what I know : the norms above are not equivalent, and the proof is by considering the sequence $3^m$, for example, which converges in one norm and not the other.

I am still unable to even think of a sequence that converges in both norms. I was wondering if $15^k$ does the job, but I'm not sure.

My last sub-approach is, to find something that converges to zero in $Q_3$ but not in $Q_5$ (any non-zero value will do, e.g. $1$), and vice-versa, and then scale both by weights and add them to get the desired series.

Answer 1

This is just Chinese Remainder Theorem. You want, for each $n$, an integer $z_n$ satisfying the pair of simultaneous congruences $z_n\equiv3\pmod{3^n}$ and $z_n\equiv5\pmod{5^n}$. For $n=1$, of course $0$ works, but for $n=2$, you are in effect finding a number satisfying a congruence modulo $9\cdot25=225$.

Answer 2

How about getting a sequence $(a_n)$ converging to $0$ $3$-adically, and to $1$ $5$-adically? Then $3+2a_n$ does the job. To do the first, how about $a_n=3^nb_n$ with $b_n\in \Bbb Z$? If we could ensure that $a_n\equiv1\pmod{5^n}$ that might be good...