TITLE: How to compute 2-adic square roots?
QUESTION [8 upvotes]: I know that, for a $2$-adic unit to be a perfect square, it must be of the form $\cdots001.$, for example the number $17$ ($10001.$) is a $2$-adic square. How would I go about finding the $2$ adic expansion of its square roots? There ought to be two, either of which is $-1$ times the other, but I don't know how to find either one.
I've tried setting up long multiplication and guessing digits that work, but there seem to be too many degrees of freedom. Any insights are appreciated.

REPLY [5 votes]: What seems like another way to approach this (but is not -- see below) is to use the $p$-adic exp and log functions. For an element $y = 1+8x$ (with $x\in \Bbb{Z}_2$), by the usual arguments everything makes sense and converges and 
$$\alpha := \exp\left(\frac{1}{2}\log(y)\right)$$
is a square root of $y$. (It is the one that is $\equiv 1$ mod $4$. Its negative is the other one.)
Concretely, for $y=17$ we have $8x=16 =2^4$, and working out just the first few terms gives
$$\begin{array} \displaystyle\frac{1}{2} \log(17) &=& 2^3-2^6+\frac{1}{3}2^{11}-2^{13} + \dots\\
\exp(\frac{1}{2}\log(17))&=&1\\
&&+\:(2^3-2^6+\frac{1}{3}2^{11}-2^{13} +\dots) \\
&&+ \frac{1}{2}(2^6-2\cdot 2^9 +2^{12}+\dots )\\
&&+\frac{1}{6}(2^9 +\dots)\\
&&+\frac{1}{24}(2^{12} +\dots)\\
&&+\dots\\
&=& 1+2^3+2^5-2^6+\frac{1}{3}2^8-\frac{1}{3}2^{10} \dots
\end{array}$$
which is 
$$\dots 011011101001$$
and that matches the value in ccorn's comment to his answer. And of course, this method works better the closer $y$ is to $1$ ($\Leftrightarrow$ the closer $x$ is to $0$) in $\Bbb{Z}_2$.
In general, one could concretely work out
$$\exp\left(\frac{1}{2}\log(1+8x)\right) = 1 +4x -8x^2 +\dots$$
as a series in $x$ which converges $2$-adically for $|x|_2\le 1$. However, now one sees that what we get here is actually the binomial formula in mercio's answer in disguise.