How to compute 2-adic square roots? 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.

Because the derivative of $x^2-17$, i.e. $2x$ is $0 \bmod{2}$ Hensel's Lemma doesn't work very cleanly.  In this situation when going from $p$ to $p^2$ either there is no lift, or every lift will work$\bmod p^2$.  Let's look at what happens here -
$x^2\equiv 17 \bmod 2 \text{ has the solution }x\equiv 1 \bmod 2$
$(2y+1)^2 \equiv 17 \bmod 4 \text { is always true, telling us } x\equiv 1,3 \bmod 4 \text{ both work}$
When we lift to$\bmod 8$ we find $1$ and $5$ (lifts of $1 \bmod 4\,$) both work$\bmod 8$ as well as $3$ and $7$ (the lifts of $3 \bmod 4$).  Note that we seem to have 4 solutions!  Let's look at$\bmod 16$ and beyond.
$$
\begin{array}\\
1,5\pmod 8 & 1^2 \equiv (1+16) \equiv 17 \pmod{16} & 5^2\equiv 9 \not \equiv 17 \pmod{16} \\
3,7\pmod{ 8} & 3^2 \equiv 9 \not\equiv 17 \pmod{16} & 7^2\equiv 49 \equiv 17 \pmod{16} \\
\end{array}
$$
So of our 4 solutions only $1$ and $7\bmod 8$ will lift to$\bmod 16$.  We lift those and try$\bmod 32$. 
$$
\begin{array}\\
1,9\pmod{16} & 1^2 \not\equiv 17 \pmod{32} & 9^2\equiv 81 \equiv 17 \pmod{32} \\
7,15\pmod{16} & 7^2 \equiv 49 \equiv 17 \pmod{32} & 15^2\equiv 225 \not\equiv 17 \pmod{32} \\
\end{array}
$$
So of our 4 solutions only $9$ and $7\bmod 16$ will lift to$\bmod 32$.  We lift those and try$\bmod 64$. 
\begin{array}\\
9,25\pmod{32} & 9^2 \equiv 81 \equiv 17 \pmod{64} & 25^2\equiv 625 \not\equiv 17 \pmod{64} \\
7,23\pmod{32} & 7^2 \equiv 49 \not\equiv 17 \pmod{64} & 23^2\equiv 529 \equiv 17 \pmod{64}\end{array}
Fairly tedious stuff for humans, but nothing a computer algebra system won't whip out in no time.  We have found 2 roots, $1 + 2^3 + O(2^5)$ and $1 + 2+ 2^2 + 2^4 + O(2^5)$.
When Doing the calculations by hand it would probably make more sense to find only one root and multiply by $-1=\frac{1}{1-2}=1+2+2^2+...$ for the other root.
