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
Best Answer
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 $32$.