Quadratic residue and the p-adic numbers

Show that the equation $x^2 = 2$ has a solution in $\mathbb{Z}_7$.

We prove a more generalized form of the problem:

Let $p$ be an odd prime and $(a,p) = 1$ and $a$ be a quadratic residue modulo $p$, then $x^2 = a$ has a solution in $\mathbb{Z}_p$.

Lemma1 Let $F(x)$ be a polynomial with integer coefficient, then if $F(x)$ is solvable in $\mathbb{Z}/p^n \mathbb{Z}$ for all $n$, then $F(x)$ is solvable in $\mathbb{Z}_p$.

Proof)
Let $x_n$ be the solution to $F(x)$ in $\mathbb{Z}/p^n \mathbb{Z}$. If $(x_n) \in \mathbb{Z}_p$ then we are done; however, it is not true in general. We will have to somehow obtain a $p$-adic integer from $(x_n)$.

First observe that if $n \geq m$, then $x_n \equiv 0 ~\bmod{~p^m}$ because $x_n \equiv 0 ~\bmod{~p^n}$. In particular, $(x_n)_{n \geq m}$ has the property that $x_n \equiv 0 ~\bmod{~p^m}$ for all $n$. Considering $(x_n)$ as a sequence in $\mathbb{Z}$, we can always find a subsequence such that all its terms are congruent to $0$ mod $p^n$ for any $n$ by deleting the first $m-1$ terms.

Because $\mathbb{Z}/p \mathbb{Z}$ is finite, again considering $(x_n)$ as a sequence, there exists $y_1 \in \mathbb{Z}/p\mathbb{Z}$ such that $x_n \equiv y_1 ~\bmod{~p}$ for infinitely many $n$. We, then, have a subsequence $(x_n^{(1)})$ of $(x_n)$ satisfies the following two properties,

1) $x_n^{(1)} \equiv y_1 ~\bmod{~p}$
2) $F(x_n^{(1)}) \equiv 0 ~\bmod{~p}$

In particular, $F(y_1) \equiv F(x_n^{(1)}) \equiv 0 ~\bmod{~p}$.

Similarly, we take $(x_n^{(2)})$ a subsequence of $(x_n^{(1)})$ satisfying

1) $x_n^{(2)} \equiv y_2 ~\bmod{~p^2}$
2) $F(x_n^{(2)}) \equiv 0 ~\bmod{~p^2}$

where $y_2 \in \mathbb{Z}/p^2 \mathbb{Z}$ which is equivalent to infinitely many terms in $(x_n^{(1)})$. The second can be easily obtained by removing the first term of $(x_n^{(1)})$ if necessary. Because $x_n^{(2)}$ are terms in $(x_n^{(1)})$, $x_n^{(2)} \equiv y_1 ~\bmod{~p}$, thus $y_2 \equiv y_1 ~\bmod{~p}$. We again have $F(y_2) \equiv 0$ mod $p^2$. Continuing this process we obtain $y_k \in \mathbb{Z}/p^k \mathbb{Z}$ such that

1) $x_n^{(k)} \equiv y_k ~\bmod{~p^k}$
2) $F(x_n^{(k)}) \equiv 0 ~\bmod{~p^k}$

with $F(y_k) \equiv 0 ~\bmod{~p^k}$ and $y_k \equiv y_{k-1} ~\bmod{~p^{k-1}}$. This means that $y= (y_k) \in \mathbb{Z}_p$. Furthermore $F(y_k) \equiv 0 ~\bmod{~p^k}$ is equivalent to saying that $F(y) = 0$ in $\mathbb{Z}_p$.

Lemma2 Let $p$ be an odd prime and $(a,p) = 1$. If $a$ is a quadratic residue modulo $p$, then $a$ is a quadratic residue modulo $p^n$ for any $n \geq 1$.

The converse is obviously true.

Proof) We assume that $x^2 \equiv a ~\bmod{~p}$ for some $x$. This proves the base case for induction. Now suppose that $x^2 \equiv a ~\bmod{~p^n}$, then $x^2 = a + p^nk$ for some $k$. We consider an element of the form $c= x + p^n l$. Then
$$\begin{array}{lcl} c^2 & = & x^2 + 2xp^nl + p^{2n}l^2 \\ & = & a + p^n(k+2xl) + 2^{2n}l^2 \\ & \equiv & a + p^n(k+2xl) ~\bmod{~p^{n+1}} \end{array}$$
The last equivalence follows from the fact that $n > 1 \Rightarrow n^2 \leq 2n > n+1$. If we can choose an appropriate $l$ such that $p | k+2xl$, then we are done. Suppose $(x,p) \neq 1$ which means that $p | x$, then $p | x^2 = a + p^nk \Rightarrow p |a$. This leads to a contradiction. Thus $(x,p) = 1 \Rightarrow (2x, p^{n+1}) =1$ as $p$ is an odd prime. Thus we can let $l$ be congruent to $-k(2x)^{-1}$. Then $c^2 \equiv a ~\bmod{~p^n}$ hence the result.

If we have a coprime quadratic residue $a$ modulo an odd prime $p$, then $a$ is a quadratic residue modulo $p^n$ for all $n$ by Lemma 2. This is tantamount to saying that $F(x) = x^2 - a$ has solution in $\mathbb{Z}/p^n \mathbb{Z}$. By Lemma 1, $F(x)$ has a solution in $\mathbb{Z}_p$.