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Let $a>0$.

I need to check following:

  1. Is $\sqrt{a}$ an element of $\mathbb{Q}_p$?
  2. How I can find the $p$-adic expansion for $\pm\sqrt{a}$?
  3. How I can distinguish $\sqrt{a}$ and $-\sqrt{a}$?

Example: $5$-adic expansion for $\sqrt{6},\sqrt{7}$.

I am a beginner and do not know much about specific calculations.

Anzu Ariake
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    See dias 12 and 13 here – Jean Marie Jun 27 '22 at 06:47
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    $\sqrt{6} = \sum_{k\ge 0} {1/2\choose k} 5^k$. There is no notion of positive/negative square root, $\sqrt{6},-\sqrt{6}$ are two different $5$-adic numbers and my choice of $\sqrt{6}$ is arbitrary. – reuns Jun 27 '22 at 07:45
  • The symbol $\sqrt a$ is, arguably, not even well-defined in this context, cf. https://math.stackexchange.com/a/4007515/96384. Did you make up this question or is it from some source -- if yes, which one? Then, a well-defined way to make sense of the first question is whether there exists an element in $\mathbb Q_p$ whose square is $a$, which sometimes is true and sometimes not. – Torsten Schoeneberg Jun 28 '22 at 04:32
  • It says that $\sqrt{7}\not\in\mathbb{Q}_5$ due to $7\equiv2\mod{5}$. I understand that $1^2\equiv4^2\equiv1,2^2\equiv3^2\equiv4$ so there is no 5-adic integer $x$ s.t. $x^2\equiv 2$. However, how about $\sqrt{14}$? This argument can be applied to other roots? – Anzu Ariake Jun 29 '22 at 09:55

1 Answers1

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First you should check if there even is a square root mod $p$.

Suppose $(n,p) = 1$ and $n$ is a square mod $p$, so the square root in ${\mathbb Q}_p$ should exist. One quick way to explicitly find good approximations of it is Newton's method. The iteration is $x_{k+1} = \frac{x_k + n/x_k}{2}$ where $x_0$ is an integer with $n \equiv x_0^2 \mod p$. Then $x_{k+1}^2 - n = (x_k - n)^2/4$ so $\|x_{k+1}^2 - n\|_p = \|x_k^2 - n\|_p^2$. Since $\|x_0^2 - n\|_p \le p^{-1}$ we get $\|x_k^2 - n\|_p \le p^{-2^k}$.

Guillermo García Sáez
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  • I don't understand your $x_{k+1}^2 - n = (x_k - n)^2/4$... – Jean Marie Jun 27 '22 at 07:08
  • I understand that this algorithm asymptotes on the order of a quadratic in ordinary decimal distances. But can the same claim be made in p-adic distances? I know that just because it is close at ordinary distances does not mean that it is close at p-adic distances. – Anzu Ariake Jun 29 '22 at 10:06