Exercise 2.2.8 (Solutions for $x^2 \equiv 1 \pmod{p^\alpha}$)

beginproof Let $p$ be an odd prime. Consider the proposition

• If $\alpha =1$, $x^2\equiv 1(modp)$ implies $p\mid (x-1)(x+1)$. Since $p$ is prime, $p\mid x-1$ or $p\mid x+1$, so $x\equiv \pm 1modp$ . This proves $𝒫(1)$.

• Assume for induction the proposition $𝒫(\alpha )$, for $\alpha \ge 1$. Now suppose that $x^2\equiv 1(mod{p}^{\alpha +1})$. A fortiori, $x^2\equiv 1(mod{p}^{\alpha })$, and the induction hypothesis shows that $x\equiv x_0(mod{p}^{\alpha })$, where $x_0\in \{-1,1\}$. Thus $x=x_0+k{p}^{\alpha }$ for some integer $k$. Since $\alpha \ge 1$, $2\alpha \ge \alpha +1$, thus $p^{2\alpha }\equiv 0(mod{p}^{\alpha +1})$, therefore $x^2\equiv 1(mod{p}^{\alpha +1})$ implies

  $$\begin{array}{llll}\hfill & {(x_0+k{p}^{\alpha })}^2\equiv 1(mod{p}^{\alpha +1}),& \hfill & \\ \hfill & x_0^2+2k{x}_0{p}^{\alpha }+k^2{p}^{2\alpha }\equiv 1(mod{p}^{\alpha +1}),& \hfill & \\ \hfill & 2k{x}_0{p}^{\alpha }\equiv 0(mod{p}^{\alpha +1})(\text{since }{x}_0^2=1\text{ and }{p}^{2\alpha }\equiv 0),& \hfill & \\ \hfill & 2k{x}_0\equiv 0(modp),& \hfill & \\ \hfill & p\mid 2k(\text{since }{x}_0=\pm 1),& \hfill & \\ \hfill & p\mid k(\text{since }p\text{ is odd}).& \hfill & \end{array}$$

  Thus $k=\lambda p$ for some integer $\lambda$, so $x=x_0+\lambda p^{\alpha +1}$, where $x_0=\pm 1$, which gives

  $$x\equiv 1(mod{p}^{\alpha +1})\text{ or }x\equiv -1(mod{p}^{\alpha +1}).$$

• The induction is done. Moreover $1$ and $-1$ are solutions of the congruence. This proves that, for any $\alpha \ge 1$, and for any odd prime $p$,

  $$x^2\equiv 1(mod{p}^{\alpha })⟺x\equiv 1(mod{p}^{\alpha })\text{ or }x\equiv -1(mod{p}^{\alpha }).$$