I have many study tasks, but I do not have any example. Therefore, I do not know, how to solve these tasks.
For example, I need prove, that: $\{ b \in \mathbb{Z}_{p^n} \mid b^2 =1\} = \{-1, 1 \}$, where $p \in P \setminus \{{2}\}, n \in \mathbb{N}$
I'll be glad to any advice.
Since $(\pm 1)^2 = 1$, $\{b \in \Bbb Z_{p^n}\, |\, b^2 = 1\}\supset \{-1,1\}$ for all $n$. The reverse containments can be proven by induction on $n$. Since $\Bbb Z_p$ is an integral domain, $b^2 \equiv 1 \pmod p$ implies $b \equiv \pm 1 \pmod p$. So the result holds when $n = 1$. Now assume $n > 1$ and the assertion is true for $n$. Let $b$ be an integer such that $b^2 \equiv 1 \pmod{p^{n+1}}$. Then $b^2 \equiv 1 \pmod{p^n}$, so by the induction hypothesis, $b$ is of the form $\pm 1 + p^nk$, $k\in \Bbb Z$. Therefore
\begin{equation} b^2 = 1 \pm 2p^nk + p^{2n}k^2 \equiv 1\pm 2p^nk \pmod{p^{n+1}}. \end{equation}
Since $b^2 \equiv 1 \pmod{p^{n+1}}$ as well, it follows that $2p^nk \equiv 0\pmod{p^{n+1}}$. This implies $2k \equiv 0\pmod{p}$. Since $p$ is odd, the latter congruence yields $k \equiv 0 \pmod{p}$. So $b = \pm 1 + p^{n+1}m$, where $m = \frac{k}{p}\in \Bbb Z$. Consequently $b \equiv \pm 1 \pmod{p^{n+1}}$.