$p$-adic numbers $\mathbb{Q}_p$: Why is $(\mathbb{Q}_p,+,\cdot)$ a topological ring? 
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*If we consider the $p$-adic numbers $\mathbb{Q}_p$ (with the metric topology associated to the $p$-adic norm $|\cdot|_p$), how could I proof that both the addition and the multiplication are continuous as maps $\mathbb{Q}_p\times \mathbb{Q}_p\to \mathbb{Q}_p$?

*Morover, why is the subgroup of squares $(\mathbb{Q}_p^*)^2$ an open subset of $\mathbb{Q}_p$? 
The question has to do with Serre's Course in Arithmetic proof of the Hasse-Minkowski Theorem for $n\ge5$.
 A: $\text{}$1. The continuity of the addition follows from the triangle inequality on the norm.
Let $a_n \to a$, $b_n \to b$. Then $(a_n+b_n) \to (a+b)$. How do we prove this? We have $$|(a_n+b_n)-(a+b)|\leq |a_n-a|+|b_n-b|.$$Given $\epsilon >0$, take $\delta=\epsilon/2$. Then if $$|a_n-a|<\delta \text{ and } |b_n-b|<\delta,$$then$$|(a_n+b_n)-(a+b)|<\epsilon.$$The continuity of the multiplication is proved similarly using$$|a_nb_n-ab|=|a_nb_n-ab+a_nb-a_nb|\leq |a_n||b_n-b|+|a-a_n||b|$$and the fact that if $a_n \to a$, then $\{a_n\}$ is bounded: $|a_n|<c$.
Notice that you do not even need the strong triangle inequality, so this is true for any normed fields.
$\text{}$2. Let us prove that the subgroup $(\mathbb{Q}_p^\times)^2$ of the group $\mathbb{Q}_p^\times$ is open in $\mathbb{Q}_p^\times$.
Recall that a $p$-adic number $x \neq 0$ is a square if and only if $x = p^{2n}u^2$ for some $n \in \mathbb{Z}$ and $u \in \mathbb{Z}_p^\times$. Let $x \in (\mathbb{Q}_p^\times)^2$ and $y \in \mathbb{Q}_p$ be such that $|x - y|_p < p^{-2n}$. Then by the isosceles triangle property$\text{}^*$, we have$$|y|_p = |x|_p = p^{-2n},$$and hence $y = p^{2n}v$ for some $v \in \mathbb{Z}_p^\times$. Then we have$$|y - x|_p = p^{-2n}|v - u^2|_p < p^{-2n};$$hence $|v - u^2| < 1$, which means that the two units have the same first digit. It now follows from Hensel's lemma that $v$ is also a square in $\mathbb{Z}_p$, and so, $y \in (\mathbb{Q}_p^\times)^2$.
$\text{}^*$Isosceles triangle property: If the elements $a$, $x$ of  non-Archimidean field $F$ satisfy the inequality $\|x - a\| < \|a\|$, then $\|x\| = \|a\|$.
