# p-adic norm, show $|y^2 - a|_p < \epsilon$

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.

• Have you heard of Hensel's Lemma yet? If not, this is a nice special case that can and maybe should be done by hand. The crucial thing is an induction step, showing that if you have a solution modulo $p^n$, you can "improve it" to a solution modulo $p^{n+1}$. For that, you will need both assumptions $p\neq 2$ and $p$ does not divide $a$. Mar 31, 2022 at 2:26

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.

• Im having a hard time seeing the connection here haha. What is the y we are looking for? And how is $|y^2 - a|_p < \epsilon$.
– CHTM
Mar 31, 2022 at 2:34
• hi @CHTM; the $y$ we are looking for is what I denote by $b$. I have rewritten the answer a little bit to clarify; let me know if it's still unclear :) Mar 31, 2022 at 2:36