TITLE: Proving $\sqrt{2}\in\mathbb{Q_7}$?
QUESTION [12 upvotes]: Why does Hensel's lemma imply that $\sqrt{2}\in\mathbb{Q_7}$?
I understand Hensel's lemma, namely: 

Let $f(x)$ be a polynomial with integer coefficients, and let $m$, $k$ be positive integers such that $m \leq k$. If $r$ is an integer such that $f(r) \equiv 0 \pmod{p^k}$ and $f'(r) \not\equiv 0 \pmod{p}$ then there exists an integer $s$ such that $f(s) \equiv 0 \pmod{p^{k+m}}$ and $r \equiv s \pmod{p^{k}}$.

But I don't see how this has anything to do with $\sqrt{2}\in\mathbb{Q_7}$?
I know a $7$-adic number $\alpha$ is a $7$-adically Cauchy sequence $a_n$ of rational numbers. We write $\mathbb{Q}_7$ for the set of $7$-adic numbers. 
A sequence $a_n$ of rational numbers is $p$-adically Cauchy if $|a_{m}-a_n|_p \to 0$ as $n \to \infty$.
How do we show $\sqrt{2}\in\mathbb{Q_7}$?

REPLY [6 votes]: There are so many different ways to see that $\sqrt2\in\mathbb Z_7\subset\mathbb Q_7$ that one can get dizzy at the vista. My favorite is the correct statement of Hensel given by @Cocopuffs, but you may like this better:
Show rather that $\sqrt{2/9}\in\mathbb Z_7$, by using the expansion of $(1+t)^{1/2}$ from the Binomial Theorem. Since $t=-7/9$ is $7$-adically small, its powers go to zero, and if the coefficients of the series are $7$-adically bounded, you automatically have a Cauchy series (series whose partial sums form a Cauchy sequence). But when you look at the series for $(1+t)^{1/2}$, you see that the only denominators are powers of $2$, in other words $|c_n|\le1$, for every coefficient $c_n$.