I'm trying to prove the following statement:
We consider the group $\mathbb{Z}_{p_k}^*$. I want to show for any element $a$ of this group that the number of square roots of $a$ is a power of $2$.
In particular, I am having some trouble invoking Hensel's lemma.
Proof attempt:
For $p\geq 3$, we can invoke Hensel's lemma which states that we can uniquely lift roots. The exact statement of Hensel's lemma found in my text is:
Let $f(x)$ be a polynomial with integer coefficients. Let $k$ be a positive integer and $r$ an integer such that $f(r)=0\text{ mod }p^k$. Suppose $m$ is a positive integer, and $p\geq 3$ a prime. If $f'(r)\neq 0\text{ mod }p$, there is a unique integer $s$ so $f(s)=0\text{ mod }p^{k+m}$ and $s=r\text{ mod }p^k$.
Here is my doubt: Let $r_1,r_2$ be a solution to the polynomial $x^2-a\text{ mod }p$. (I know there are $0$ or $2$ such solutions.) We then apply Hensel's lemma to "lift" these solutions to solutions of $x^2-a$ modulo $p^2$. I can see why the $s$ we get don't give us extra solutions to the initial polynomial modulo just $p$, but I can't see why the original $r$'s have to still be roots modulo $p^2$ now. If they were, we could just iterate this procedure to end up with $2^k$ solutions. In particular, why is it that if $r^2=a\text{ mod }p$, then $r^2=a\text{ mod }p^2$?
This may be very trivial, to be honest. I just can't see it. Or are the other 2 roots coming from elsewhere? ETA: Nevermind, I can see this now since $s=r\text{ mod }p$. But my other question still holds...
Moreover, how can we use this in the case that $p=2$?
