Behind this provocative title is a precise question. I have been working on some books and papers claiming the following.
Let $F$ be a number field, and denote by $p$ its finite places. Here is the claim I would like to understand:
For all place $p$ above 2, there is $n_p$ sufficiently large so that for all $x \in F_p$, if $\mathrm{ord}_p(a-1) \geqslant n_p$ implies $x \in F_p^2$.
That means that for even places, there are neighbourhood of 1 made only of squares. Why is that true? If I think of the rational dyadics $\mathbb{Q}_2$, that would mean with usual notations $$x = 1 + \sum_{n \geqslant n_p} a_n 2^n,$$
but I am far from seing a square here. What am I missing? And how can I adapt this to more general number fields?
Over $\Bbb Q_2$, the neighbourhood $1+8\Bbb Z_2$ of $1$ is made up of squares. An argument using Hensel's lemma proves that $x^2=a$ is soluble over $\Bbb Z_2$ whenever $a\equiv1\pmod8$ in $\Bbb Z_2$.
The same is true in any finite extension of $\Bbb Q_p$. It's fairly obvious when $p$ is odd, but one has to use a stronger form on Hensel when $p=2$.
Alternatively it drops out from the theory of the $p$-adic exponential and logarithm. (They are continuous bijections near the identity.)