Check if the equation $x^2=-1 \text{ in } \mathbb{Z}_2$ has a solution, and if it has, calculate the three first positions of the solution.
So, we are looking for a solution $\pmod 2$, one solution $\pmod{2^2}$ and one $\pmod {2^3}$, right?
That's what I have tried:
$$\mathbb{Z}_2=\{0,1\}$$
$$x \equiv 0 \pmod 2 \Rightarrow x^2 \equiv 0 \pmod 2$$
So, when $x \equiv 0 \pmod 2$, the equation $x^2 \equiv -1 \pmod 2$ is not satisfied.
$$x \equiv 1 \pmod 2 \Rightarrow x^2 \equiv 1 \pmod 2 \equiv -1 \pmod 2$$
So, the equation $x^2 \equiv -1 \pmod 2$ is satisfied, when $x \equiv 1 \pmod 2$.
Therefore, $1$ is the only solution of $x^2=-1$ in $\mathbb{Z}_2$.
$$x_0=1$$
We are looking for a $x_1 \in \mathbb{Z}$ such that:
$$x_1^2 \equiv -1 \pmod {2^2} \text{ such that } x_1 \equiv 1 \pmod 2$$
$$x_1 \equiv 1 \pmod 2 \Rightarrow 2 \mid x_1-1 \Rightarrow \exists a_1 \in \mathbb{Z} \text{ such that } x_1=1+2a_1$$
$$x_1^2+1=(1+2a_1)^2+1 \equiv 2 \pmod {2^2}$$
So: $x_1^2+1 \equiv 0 \pmod{2^2} \Leftrightarrow 2 \equiv 0 \pmod {2^2}$, that is a contradiction.
Is it right ? If so, does this mean that we cannot apply the Hensel Lifting Lemma in this case? Why not?
EDIT: Is there only one position of the solution? If so, how can we justify it?
Consider the equation modulo $4$.