P-adic expansion construction

Question

Can anyone teach me about p-adic expansion? especially the case where we have to expand a square root.

I need to know how to construct them. for example: the 7-adic expansion of $\sqrt{305}$.

This is a general question and not an assignment or anything, so i cannot post the progress.

That is just a random example, please use any other example to explain if it's easier to understand.

edit: I'm stuck with the cases where i have to expand the square roots, expanding rational numbers is fine though

Answer 1

The $p$-adic number will be ....dcba
Start with units, $a^2=305=4 \bmod 7$, so $a=2$.
Next, the sevens $(7b+2)^2=305=11\bmod 49$, so $28b=7\bmod 49$, and $b=2$.
Next, the 49s: $(49c+7*2+2)^2=305\bmod 343$, and so on.
Although it wasn't guaranteed that $a$ would exist, because $305$ might not have been a quadratic residue $\bmod 7$, all the other digits must exist.

Answer 2

The quotients $\Bbb Z/p^n\Bbb Z$ should be thought of as "approximations" to $\Bbb Z_p$. Indeed, if $x\in\Bbb Z_p$, then there is a residue of $x$ mod $p\Bbb Z_p$ which sits inside $\Bbb Z/p^n\Bbb Z$ $-$ and this residue, when its canonical representative is written in base $p$, gives the first $n$ digits of $x$'s canonical $p$-adic expansion.

Therefore, to compute the expansion of $\sqrt{x}$ in $\Bbb Z_p$ to an accuracy of $n$ digits, one need only compute the square root of $x$'s residue mod $p^n$ and write in base $p$. If you do this with $n=1$, you should be able to lift to all other $n$ via lifting provided the hypotheses of Hensel's lemma apply.

You might as well assume $x\in\Bbb Z_p^\times$. If so, then let $0\le s<p$ be the integer which represents the square root of $x$ mod $p$. Then we can stipulate $\sqrt{x}=s\sqrt{x/s^2}$. This is useful because then $x/s^2\in1+p\Bbb Z_p$ (the so-called $1$-units) and you can apply the Newton-Binomial series if $p\ne2$:

$$(1+a)^{1/2}=\sum_{n\ge0}\binom{1/2}{n}a^n.$$

The formal binomials are defined by $\displaystyle\binom{x}{n}=\frac{x(x-1)\cdots(x-(n-1))}{n!}$.

I should mention that everything has two square roots, and there is no notion of "positive" or "negative" outside of the rational numbers in this context. The NB series will give you one of the roots. (In fact Koblitz has conditions for the square root of a positive rational perfect square to be the positive or negative rational root; see Gouvea's problem 179.) If you follow the route of Hensel lifting, which root you end up with depends on which square root mod $p$ was chosen as the "seed."

Answer 3

The most naive way:

305=4(mod 7) this gives 2 or 5 for last digit.

305=11(mod 49) and we have 16^2 or 33^2=11(mod 49). this gives 22 or 45 written in 7-adic of last two digit.

continuing this procedure we have $\sqrt{305}=\dots22$ or $...45$

Of course you can use Taylor expansion:

$\sqrt{305} = \sqrt{7*43+4}=2+\frac{1}{4}*(7*43)-\frac{1}{64}*(7*43)^2+\frac{1}{512}*(7*43)^2-\dots$

it's not hard to find it converge w.r.t 7-adic topology