Two doubts about squares in $\Bbb Z_p$ The statement says that for $p \neq 2$ an element $x=p^i u \in \mathbb Q_p^\times$ (with $i \in \mathbb Z$ and $u \in \mathbb Z_p^\times$) is a square if and only if $i$ is even and $u$ is a square in $\mathbb Z_p^\times$. This statement is reduced to finding squares in $Z_p$
$\textbf{Doubt-}$ Now the proof is reduced to this statement- Let $u \in \mathbb Z_p^\times$ and choose a rational integer $a$ such that $u \equiv a \bmod p$. (This I understand).   So if I let $u=(\alpha_0,\alpha_1,.....)$ where $\alpha_i=a_i/p^{i+1}Z$, What rational integer $a$ is he choosing s.t. $u-a $ is divisible by $p$ and moreover shouldn't $a$ be an element of $\mathbb Z_p^\times$ for $u-a$ to be defined even, and why any such $a$ exist? If you can explain this to me the rest of the proof I understand
And for the prime number $p=2$   an element $x=p^i u \in \mathbb Z_p^\times$ is a square if and only if     $u \equiv 1$ (mod 8). It is enough to show this for main result. So here is my doubt-
$\textbf{Doubt-}$ In this one, Let an element $x=p^i u \in \mathbb Z_p^\times$ is a square then $u=v^2$, But then book says that then $v$ is congruent mod $8$ to an odd integer. WHY? Why not to any even integer? In $\Bbb Z_8$ there is $4=2^2$, but $2$ is not odd.
 A: Note that every nonzero element of $\mathbb Q_p^\times$ can be written in a unique way as $p^iu$ with an integer $i$ and a unit $u$. Thus if $x=y^2$ with $x=p^iu$, $y=p^jv$, then by uniqueness $j=2i$ and $u=v^2$, thus the first idea is justified.
Since $\mathbb Z_p$ is the completion of $\mathbb Z$ under $|\cdot|_p$, each element $u\in\mathbb Z_p^\times \subset \mathbb Z_p$ can be approximated arbitrarily well by a rational integer $a\in\mathbb Z\subset\mathbb Z_p$. For our purposes an approximation with $|u-a|_p<1$ is good enough (notationally equivalent: $u\equiv a\pmod p$). Actually, we can use $u=v^2$ and approximate $v$ by $b\in \mathbb Z$ to conclude that $a\approx b^2$, i.e. $a\bmod p$ must be a square. After this, I assume your textbook goes to investigate $a^{-1}u$ ans shows that all numbers sufficiently close to $1$ are squares. 
For $p\ne 2$, this "sufficiently close to $1$" merely means $|u-1|_p<1$.
For $p=2$, however, we note that $u=v^2$ with $v\equiv 1\pmod 2$ implis $u\equiv 1\pmod 8$ (already in $\mathbb Z$, the square of an odd number is $\equiv 1\pmod 8$), whereas $v\equiv 0\pmod 2$ would not allow $u$ to be a unit.
