A extending the p-adic valuation to a quadratic extension of $\mathbb{Q}_p$ I'm trying to solve the following problem. Prove that, if $d \in \mathbb{Z}_p$ is non-square, then
$|a + b \sqrt{d}|p = |a^2 − b^2d|^{1/2}_p$ , for any $a, b \in \mathbb{Q}p$,
defines a non-Archimedean valuation on $\mathbb{Q}_p(\sqrt{d})$ which extends the usual $|. |_p$ on $\mathbb{Q}_p$.
The hint we are given is to show that $|\alpha|_p\le 1 \Rightarrow |\alpha +1|_p \le 1$. I've managed to reduce this to different cases, and the only case where I'm still having trouble $|a|_p=|b|_p$. Apparently I need to use Hensel's Lemma somehow, but I don't know how. Can anyone give me a hint? 
 A: Sure, here’s a hint. Use the good form of Hensel to show this: let $\alpha$ be algebraic over $\Bbb Q_p$, with minimal polynomial $f(X)$ whose constant term is $c$. If $c\in\Bbb Z_p$ (the $p$-adic integers), then all the coefficients of $f$ are in $\Bbb Z_p$. You definitely do not need to divide the problem into cases.
A: The hint of @Lubin works for all finite extensions of a complete field, not only for quadratic ones. I will try to elaborate the hint in the case of degree $2$ extensions of a complete field $F$ with a non-archimedean valuation $|\cdot |$. 
Let $\alpha = a + b \sqrt{d}$ with $b \ne 0$. $\, \alpha$ and $\bar \alpha = a - b \sqrt{d}$ are the roots of the polynomial $x^2 - 2 a x + (a^2 - d b^2)$. Assume that $|a^2 - d b^2| \le 1$ but $|2 a| > 1$. We will show that the equation $x^2 - 2 a x + (a^2 - d b^2)= 0$ has a root in $F$. The equation is equivalent to 
$$x = \frac{1}{2a}x^2 + \frac{a^2 - d b^2}{2a}= u x^2 + v$$
The function $x \mapsto u x^2 + v$ maps the complete space (a ring) $R\colon = \{x \ \colon \ |x|\le 1\}$ to itself and is a contraction, hence it has a (unique) fixed point, contradiction.
Or: assume that we have $|a^2 - d b^2 | \le 1 $.  We have the easy to check equality
$$\left(\frac{a}{b}\right)^2 \cdot \left( 1 - \frac{ 4( a^2 - d b^2)}{(2a)^2}\right)= d$$
If we had $|2 a | >1 $ then the norm of $\delta\, \colon = \frac{ ( a^2 - d b^2)}{(2a)^2}$ would be $<1$ and so $(1-4 \delta)$ and thus $d$ would be a square, contradiction. Therefore, $|2a| \le 1$ and this implies $|(a+1)^2 - d b^2|\le 1$.
