How to show $x^2=20$ has no solution in $2$-adic ring of integers $\mathbb{Z}_2$? How to show $x^2=20$ has no solution in $2$-adic ring of integers $\mathbb{Z}_2$ ?
What is the general criterion fora solution of $x^2=a$ in  $\mathbb{Z}_2$ ?
I know for odd prime $x^2=a$ has solution in  $\mathbb{Z}_p$ or in other word $a$ is quadratic residue modulo $p$ if $a_0$ is quadratic residue modulo $p$, where $a=a_0+a_1p+a_2p^2+\cdots \in  \mathbb{Z}_p$.
But what about for $p$ even i.e., $p=2$ ?
We have two following results as well.

Result $1$: For $p \neq 2$, an $\epsilon \in \mathbb{Z}_p^{\times}$ is square in $\mathbb{Z}_p$ iff it is square in the residue field of $\mathbb{Z}_p$.
Result $2$: An unit $\epsilon \in \mathbb{Z}_2^{\times}$ is square iff $\epsilon \equiv 1 (\mod 8)$.

But as $20$ is not a unit in $\mathbb{Z}_2$, the above Result $2$ is not applicable here.
So we have to use other way.
I am trying as follows:
For a general $2$-adic integer $a$, we have the following form $$ a=2^r(1+a_1 \cdot 2+a_2 \cdot 2^2+\cdots).$$
Thus one necessary criterion for $a \in \mathbb{Q}_2$ to be square in $\mathbb{Q}_2$ is that $r$ must be even integer.
Are there any condition on $a_1$ and $a_2$ ?
For, let $\sqrt{20}=a_0+a_1 2+a_22^2+a_32^3+\cdots$.
Then squaring and taking modulo $2$, we get $$ a_0^2 \equiv 0 (\mod 2) \Rightarrow a_0 \equiv 0 (\mod 2).$$ Thus, $\sqrt {20}=a_12+a_22^2+a_32^3+\cdots$.
Again squaring and taking modulo ($2^2)$, we get
$$a_1 \equiv 1 (\mod 2^2).$$ Thus we have $\sqrt{20}=2+a_22^2+a_32^3+\cdots$.
Again squaring and taking modulo $(2^3)$, we get no solution for $a_i,\ i \geq 2$.
What do I conclude here from ?
How do I conclude that $x^2=20$ has no solution in $\mathbb{Z}_2$ ?
Help me
 A: The other answers do a good job of answering your more general question, but I want to address the specific case of $x^2 = 20$ in the simplest way possible.
Remember how Hensel's lemma works: we perform an iterative procedure to find a solution to $f(x) \equiv 0\pmod {2^n}$ for all $n$. We then apply completeness to show that $f$ has a solution in $\mathbb Z_2$.
The problem here is that, since $x^2 -20$ has a repeated root modulo $2$, the iterative procedure fails.
The simplest way to prove that $x^2 -20$ has no solution in $\mathbb Z_2$ is, therefore, to pinpoint at which point there is no solution modulo $2^n$.

Here's the one line proof: if $x^2 = 20$ has a solution in $\mathbb Z_2$, it has a solution modulo $2^n$ for all $n$. But it has no solution modulo $32$.
A: It seems like your attempt probably works.
A slightly more general version of Result 2 that could help you is that squares in the $2$-adic integers are of the form $2^{2k}u$ where $u\in \mathbb Z_2^*$ is congruent to $1$ mod $8$. Since $20$ is not of this form, it is not a square.
One can see this follows immediately from Result 2 and your observation that the valuation of a square is a square: if $x$ is a square then it has valuation $2^{2k}$ for some integer $k$, and so $x2^{-2k}=u$ is a square (product of squares is a square) with unit valuation and so by Result 2, $u=1$ mod $8$. Conversely anything of that form is clearly a square by applying Result 2 again (and observing the product of squares is a square).
A: Here is an elementary solution, but which has the advantage to show "experimentally" why $2
$ is a "distinguished" prime. If $20$ is a square in $\mathbf Z_2$, we can factor out $4$ to get $5=2^{2n}u^2$, where $u$ is a $2$-adic unit. But $5=1+4$ is a "principal unit", i.e. it belongs to the multiplicative subgroup $1+2\mathbf Z_2$, so that $n=0$ and $u$ is also a principal unit, say $u=1+ 2^aw$ and $u^2=1+ 2^{a+1} w+ 2^{2a} w^2$ . Writing $v_2$ for the $2$-adic valuation, we have $v_2 (5-1)=2$ and $v_2 (u^2-1)=a+1$ if $a\ge 2$, a contradiction. If $a=1, u^2 -1 = 4w(1+w)$. Taking classes modulo $2\mathbf Z_2$, denoted by brackets[.], we have $[1+w]=[1]+[w]=[1]+[1]=[0]$ because the residual field is just $\mathbf Z/2\mathbf Z$. This means that $v_2(u^2-1) \ge 3$, a contradiction again. The last phenomenon could not have happene for $p$ odd.
A: I want to show a more elementary approach by directly grinding out algebra steps just for the fun of showing how Hensel's lemma and the ultrametric inequality can be used. Let's begin by supposing there is an $x \in \mathbb{Z}_2$ such that $x^2=20$.
$$x^2=20$$
$$|x^2|_2 = |20|_2$$
$$|x|_2^2 = 1/4$$
$$|x|_2 = 1/2$$
So we know $x = 2u$ for $|u|_2=1$.
$$(2u)^2 = 20$$
$$u^2 = 5$$
Now let's use Hensel's lemma in its more general form with $f(u) = u^2-5=0$. Simply, for there to be a $u \in \mathbb{Z}_2$ we need 
$$|f(u)|_2 < |f'(u)|_2^2$$
$$|u^2-5|_2 < |2u|_2^2$$
$$|u^2-5|_2 < 1/4$$
At this point we can use with $|k|_2\le 1$,
$$u = 1+2k$$
$$u^2 = 1 + 4k + 4 k^2 $$
$$u^2 = 1 + 4k(k+1) $$
At this point we notice no matter if k is "even or odd" (in $2\mathbb{Z}_2$ or $1+2\mathbb{Z}_2$), we must have $|k(k+1)|_2<1$.
$$u^2 = 1 + 4k(k+1) $$
$$u^2 - 1 = 4k(k+1) $$
$$|u^2 - 1|_2 = |4k(k+1)|_2 $$
$$|u^2 - 1|_2 < 1/4 $$
But wait, earlier we showed $|u^2-5|_2 < 1/4$ is required for there to be a solution in $\mathbb{Z}_2$. Because $|5-1|_2 = 1/4$ it's impossible for $u^2$ to be that close to both! Let's directly show the contradiction now, hinging upon the ultrametric inequality:
$$|u^2-1|_2 = |u^2-5 + 4|_2 \le \max(|u^2-5|_2 , |4|_2)$$
The ultrametric inequality's "greatest wins" property gives us the maximum whenever there's a competition. Since here $|u^2-5|_2 < 1/4$ and $|4|_2 = 1/4$ we have,
$$|u^2-1|_2 = 1/4$$
This clearly contradicts what we showed a second ago with $|u^2-1|_2 <1/4$, so there are no solutions.
