Solution of $x^2 = 2$ in $\mathbb{Q}_p$ Let $p \neq 2$. Show that the equation
$$x^2=2 \quad (*)$$
has a solution $x \in \mathbb{Q}_p$ iff exists $y \in \mathbb{Z}$ such that $y^2 \equiv 2 \, \pmod p$.
$\Rightarrow$ Let $x= y + x_1 p + x_2 p^2+ \ldots \in \mathbb{Q}_p$ solution of $(*)$, then $2=x^2 \equiv y^2 \pmod p$.
$\Leftarrow$ ?
Thanks!
 A: Hint:
Show that if $y^2\equiv 2\mod p^n$, there is a solution of $z^2\equiv 2\mod p^{n+1}$ with $z\equiv y\mod p^n$ (this technique is called Hensel lift).
A: There are so many ways of approaching this that one can get dizzy contemplating them all. You can use the Binomial series for $(1+z)^{1/2}$, plugging in $2/y^2-1$, for example, to get $\sqrt2/y$, then multiply that result by $y$.
Or you can use Newton-Raphson directly. I'll illustrate how it goes for $p=7$, where you can take $y=3$. You have a function $f(x)=x^2-2$, $f'(x)=2x$, and you want a root of $f$ that's close to $3$. Set $z_1=3$, and the first error is $f(z_1)=7$, and the first correction is $7/6$, so the second approximation is $z_2=3-7/6=11/6$. The second error is $f(z_2)=49/36$, so the second correction is $(49/36)\big/(22/6)=49/132$, and the third approximation is $z_3=z_2-49/132=193/132$. The third error is $f(z_3)=2401/17424$, in which the numerator is already $7^4$. Continue as long as you want.
A: Now, I want to say something more than just the run-of-the-mill answer, if that's ok with you.
There is a general principle which says that a polynomial $F(T_1,\ldots,T_n)\in\mathbb{Z}[T_1,\ldots,T_n]$ has a zero in $\mathbb{Z}_p$ if and only if $F(T_1,\ldots,T_n)$ has a zero in $\mathbb{Z}/p^n\mathbb{Z}$ for all $n\geqslant 1$.
How does this apply here? Note that if $\alpha$ is a solution to $x^2=2$, then
$$v_p(\alpha)=\frac{1}{2}v_p(\alpha^2)=\frac{1}{2}v_p(2)\geqslant 0$$
so any solution to $x^2=2$ in $\mathbb{Q}_p$ must actually lie in $\mathbb{Z}_p$.
So, how does one go about proving this principle? For the same of simplicity, let's stick the to the univariate case, so we're dealing with a polynomial $F(T)$. The one direction is trivial, since, as I noted in your last question:
$$\mathbb{Z}_p=\left\{(x_n)\in\prod_n\mathbb{Z}/p^n\mathbb{Z}:\text{for all }n\leqslant m,\,\, x_n\equiv x_m\mod p^{n+1}\right\}$$
Thus, you see that 
$$F((x_n))=(F(x_n))$$
and so clearly any solution to $F(x)=0$ in $\mathbb{Z}_p$ begets a solution to $F(x)\equiv 0 \mod p^n$ for all $n\geqslant 1$. 
Conversely, if you start with a sequence of solutions $(x_n)$ in $\displaystyle \prod_n \mathbb{Z}/p^n\mathbb{Z}$, where $F(x_n)\equiv 0 \mod p^{n+1}$, one can take this sequence and "build" a solution sequence in $\mathbb{Z}_p$. Basically, the issue with our original sequence is that it may not be consistent. To fix this, note that there have to exist infinitely many $n$ for which all $x_n$ are all equivalent to the same thing modulo $p^{1}$. Try taking this subsequence, and noting that there must exist another subsequence all equivalent to the same thing modulo $p^2$. Continue in this way to create a solution sequence that actually lies in $\mathbb{Z}_p$. 
So, how does this all factor back into our problem? Namely, the above would solve everything if the problem stated

Show that $x^2=2$ has a solution in $\mathbb{Q}_p$ for $p\ne 2$ if and only if $x^2=2\mod p^n$ has a solution for all $n\geqslant 1$

So, how do we pass from a solution modulo $p$ to a solution modulo higher powers of $p$? This is exactly the content of Hensel's Lemma which states that if $f(x)\equiv 0\mod p^r$ has a solution $\alpha$ which satisfies $f'(r)\not\equiv 0\mod p^r$, then we can "lift" the solution $\alpha$ to $f(x)\equiv 0\mod p^r$ to solutions $f(x)\equiv 0\mod p^s$ for $s\geqslant r$. 
Putting this all together gives us the problem.
A: You want the $p$-adic infinite sum $x=x_0+x_1p+x_2p^2+\dots$ to satisfy $x^2=2$.  That will require each of the finite partial sums $S_n=x_0+x_1p+x_2p^2+\dots+x_np^n$ to satisfy ${S_n}^2\equiv 2\pmod{p^{n+1}}$, because the omitted terms of the infinite series will contribute only terms divisible by $p^{n+1}$ to $x^2$.  I suggest choosing the $x_n$'s one after the other, inductively, so that, when you're trying to choose $x_n$, you already know $x_0,\dots,x_{n-1}$.  The $y$ that you're given serves as $x_0$ to start the induction.  Choosing $x_n$ will amount, after some manipulations, to solving a congruence modulo $p$ which (fortunately) turns out to be linear in the unknown $x_n$, with a coefficient of $2y$, which is (fortunately) not $0$ modulo $p$.  
