I am trying to construct a sequence in $\mathbb Q_2$ that is formed of rational numbers and converges to $\sqrt{-7}$, to prove that $(\mathbb Q, |\cdot|_2)$ is not complete. My lecturer stated that this was an easy example but I am getting confused by my working. (Also I have just been introduced to $p$-adic numbers, so I can't use Hansel's Lemma, for example.)
So far, I said that if $\{x_n\}_{n\in \mathbb N} \to \sqrt{-7}$ then $\{x_n^2\}_{n\in \mathbb N} \to -7$ and so I must have $x_n^2 + 7$ divisible by a large power of $2$ for large $n$. Suppose I want that $2^n|x_n^2+7$ for all $n.$ I would then, for every $n$, need a root of $x^2+7$ in $\mathbb Z/2^n\mathbb Z$. But I'm unsure how to approach this kind of problem; finding quadratic residues modulo $p^n$ (here $p=2$) rather than modulo primes or products of distinct primes (where I can use the Chinese Remainder Theorem).
I would appreciate any help with my proof or an explanation of how to find quadratic residues modulo $p^n$, or both.
Suppose you have $x_n$ such that $2^n\mid x_n^2+7$. If $2^{n+1}\mid x_n^2+7$, you can let $x_{n+1}=x_n$. If not, then $$(x_n+2^{n-1})^2+7=(x_n^2+7)+2^n x_n+ 2^{2n-2}$$ is divisible by $2^{n+1}$ (as long as $n\geq 3$) since $x_n$ must be odd. So in this case, you can define $x_{n+1}=x_n+2^{n-1}$. By induction, you can thus define a sequence $(x_n)$ such that $2^n\mid x_n^2+7$. Moreover, this sequence is Cauchy with respect to the $2$-adic norm, since by construction $|x_n-x_m|_2\leq 2^{-n+1}$ for $m\geq n$.