Convergence in $\mathbb{Q}_2$ and $\mathbb{Q}_3$

Question

Prove that for any $a,b\in \mathbb{Z}$ there is a sequence $c_n \in \mathbb{Z}$ such that $c_n \to a$ in $\mathbb{Q}_2$ and $c_n \to b$ in $\mathbb{Q}_3$ (here $\mathbb{Q}_p$ denotes the $p$-adic rationals). I am given as a hint that $3^{2^n} \to 1$ in $\mathbb{Q}_2$ (easy to show) and I am trying to make up an example from here. So it follows that $a\cdot 3^{2^n} \to a$ in $\mathbb{Q}_2$ and we also have $3^{2^n} \to 0$ in $\mathbb{Q}_3$ and I tried some combinations of these with no success.

Any help appreciated!

Answer 1

$ gcd(2^n,3^n)=1$, so there exist integers $\lambda $ and $\mu $ such that $$\lambda 2^n + \mu 3^n =1 $$For definiteness, take the absolutely least such $\lambda $. Multiply the equation by $a-b.$Then $$(a-b)\lambda 2^n+ (a-b)\mu 3^n=a-b $$ so $$a-(a-b)\lambda 2^n=b-(b-a)\mu 3^n $$. Let $$c_n=a-(a-b)\lambda 2^n $$. Then $2^n $divides $c_n-a$and $3^n$divides $c_n-b$ Thus the 2-adic limit of $c_n$ as $n \to \infty$ is $a$ and the 3-adic limit of $c_n$ as $n \to \infty$ is b.

Answer 2

We also have, by induction on $n$, $2^{3^n}\equiv-1\bmod 3^{n+1}$ therefore $2^{3^n}\to-1\in\mathbb Q_3$. So

$a(3^{2^n})-b(2^{3^n})\to a\in\mathbb Q_2$

$a(3^{2^n})-b(2^{3^n})\to b\in\mathbb Q_3$