square roots of 1, congruences Show that the congruence:
$$x^2  - 1  ≡  0(2^s)$$
has only $4$ solutions if $s\ge3$ .  I can find $4$ solutions, but want to show there are no more.
Yet, if $2$ is replaced by an odd prime there are only $2$ solutions, $1$ and $-1$.
(That seems to be based on the fact that if $x$ is a solution of the 
congruence:
$$x^2  - 1  ≡  0(p^s)$$, for $p$ an odd prime and $s>1$, it is also a solution of
$$x^2 - 1 ≡  0(p)$$
This doesn't seem to depend on $p$ being odd.
What am I missing?
 A: We want $2^s$ to divide $(x-1)(x+1)$. Note that $x-1$ and $x+1$ are even. Since they differ by $2$, one of them is divisible by $2$ but by no higher power of $2$, and the other is divisible by $2^{s-1}$.
Let us take the case $x-1$ divisible by $2$ but by no higher power of $2$, and you can take care of the other case. 
So $x+1$ has to be divisible by $2^{s-1}$. Thus $x=-1+k\cdot 2^{s-1}$ for some integer $k$. If $k$ is odd, that gives the solution $x\equiv -1+2^{s-1}\pmod{2^s}$. If $k$ is even that gives the solution $x\equiv -1\pmod{2^s}$. 
The reason that we need $s\ge 3$ is that if $s\le 2$ the "four" roots we have found will not be distinct modulo $2^s$. 
Remark: The difference between $2$ and an odd prime $p$ is that $2$ can divide both $x-1$ and $x+1$, while an odd prime cannot.
A: There's a non trivial theorem which you could use. Notice any such $x$ must have odd residue mod $2^{s}$. Now, if $s \geq 3$, $(\mathbf{Z}/2^{s}\mathbf{Z})^{*} \cong \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2^{s-3} \mathbf{Z}$. This group has 3 elements of order 2.
