square root in 2-adics $\newcommand\Q{\mathbb Q} \newcommand\Z{\mathbb Z}$I am a bit confused on the square root of 2-adics. I am pretty sure I am mixing some steps in an algorithm. To be precise, I am trying to solve an exercise in Koblitz' book p-adic numbers, p-adic analysis and zeta functions (this is not a homework, I just want to do it). In exercise of I.7 the reader is asked to solve the square root of -7 in $\Q_2$ up to 5-digits and the hint is to use a generalization of Hensel's Lemma:
Let $f(x)$ be a polynomial with coefficients in $\Z_p$. If $a_0$ in $\Z_p$ satisfies

*

*$f(a_0) = 0 \bmod p^{2m+1}$

*$f'(a_0) = 0 \bmod p^m$

*$f'(a_0) \ne 0 \bmod p^{m+1}$
then there is a unique $a\in\Z_p$ such that $f(a) = 0$ and $a=a_0 \bmod p^{m+1}$
Obviously the procedure to solve comes from an algorithm to be seen in the proof of this generalized Hensel's lemma. For my particular problem $f(x)=x^2+7$ and $p=2$, I choose $a_0$ either $1$ or $0$ (so in fact an integer) and then I iterate to get, for a chosen $b_1 =1$ or $b_1=0$, $a_1 = a_0 + b_1 p^{m+1}$ for which $f(a_1)=0 \bmod p^{2m+2}$ and so on i.e. I need to get $b_i$ which gives $a_i$ such that $f(a_i) = 0 \bmod p^{2m+1+i}$.
I might be missing something, but for me the answer for square root of -7 would be of the form
$a_0 + b_12+ b_22^2 + \dots $
But I have a problem getting the correct answer. So one of the square root (using Pari to check) is $(a_0,b_1,b_2,\dots)=(1,1,0,1,0,\dots)$.
Is there a cleaner way to think of this algorithm. I get easily confused and make mistakes with the iteration steps.
Maybe I have mixed something in the procedure. Could someone explain to me why for $a_0=1$, we get $(1,1,0,1,0,\dots)$?
From this I feel that manually taking roots and even doing simple arithmetic is not practical in the p-adics.
EDIT:
The algorithm I have in mind is actually also written here in this online calculator:
http://www.numbertheory.org/php/2adic.html
I still fail to get the correct $b_4$ using the algorithm. However, the online calculator gives me the correct $b_4$. If I follow the procedure, my $b_4$ is $(a_3^2+7)/64 \bmod 2 = 7\bmod 2$ so it is $1$, while the correct answer is $0$.

Since $11^2\equiv-7\bmod2^6$, why not render $...01011_2$ and call it a day?
