TITLE: $2$-adic sequence converging to $\sqrt{-7}$.
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: Just use the Binomial expansion, for $(1-8)^{1/2}$. It’s a fact, not too terribly hard to prove, that $(1+4t)^{1/2}$ is a series in $t$ with all integer coefficients. Plug in $-2$ for $t$, voilà.