Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
The p-adic norm of $x$ is denoted by $|x|_p$ and defined to be $p^{-e}$ if $p^e$ is the power of $p$ appearing in prime decomposition of $x$.
Then suppose x and a are integers and $x^2 \equiv a$ mod $p$ with $p\neq2$ and $p$ does not divide $a$. Show that for any real $\epsilon$, there exists integer $y$ such that $|y^2 - a|_p < \epsilon$
I know that the set of $x$ such that $|x-a|_p < p^{-e}$ is given by $x \equiv a$ mod $p^{e+1}$ and i feel like thats useful but im not sure how to apply that.
This is a special case of Hensel's lemma, but here's a direct proof.
For any $\varepsilon>0$, there exists some $n\in\mathbb{N}$ such that $p^{-n}<\varepsilon$. Thus it suffices to show that, for every $n\in\mathbb{N}$, there exists an integer $b$ such that $p^n$ divides $b^2-a$; then $|b^2-a|_p\leqslant p^{-n}<\varepsilon$, as needed. We will show this by induction on $n$; note that the base case of $n=1$ is given by hypothesis, with $b=x$.
For the inductive step at stage $n$, we suppose we have such $b$, say with $b^2-a=p^n\mu$; we wish to find $c$ such that $p^{n+1}$ divides $c^2-a$.
Note that, since $p\nmid a$, also $p\nmid b$. Moreover, we have $p\neq 2$. Thus $2b$ is invertible modulo $p$; hence we may find $\lambda$ such that $\mu+2b\lambda$ is divisible by $p$. Now let $c=b+\lambda p^n$. Then
\begin{align} c^2-a &= b^2-a+2b\lambda p^n+\lambda^2p^{2n} \\ &= p^n(\mu+2b\lambda+\lambda^2p^n). \end{align} But $p$ divides $\mu+2b\lambda$ and $p$ divides $\lambda^2 p^n$, and hence $p^{n+1}$ divides $c^2-a$, as needed.
