Square roots of unity modulo a power of a prime

Question

I need clarification on the proof of the following claim:

if $x^{2}\equiv1\pmod {p^\alpha}$ then $ x\equiv\pm1$ where $p$ is an odd prime and $\alpha\geq 1$.

Proof: $x^2\equiv1\pmod{p^\alpha} \rightarrow p^\alpha\mid x^2-1=(x-1)(x+1)$

but not both $(x-1)$ and $(x+1)$ can be divisible by $p^\alpha$

So either:

$$p^\alpha \mid x-1\rightarrow x\equiv1\pmod{p^\alpha}$$

or

$$p^\alpha\mid x+1\rightarrow x\equiv-1 \pmod{p^\alpha}$$

I am sure it is obvious, but I am not understanding why $(x-1)$ and $(x+1)$ cannot both be divisible $p^\alpha$. Is it because they differ by $2$ and power of odd primes cannot do that?

Answer 1

If $x-1$ and $x+1$ were both divisible by $p^\alpha$, then their difference $(x+1)-(x-1)=2$ would also be divisible by $p^\alpha$. This implies $p^\alpha$ is either $1$ or $2$, which is impossible if $p>2$ and $\alpha\geq 1$.

Or, very intuitively, if $x-1$ were a multiple of $p^\alpha$, then the next multiple of $p^\alpha$ would be $x-1+p^\alpha$ which is larger than $x+1$ since $p^\alpha>2$. So $x+1$ cannot be a multiple of $p^\alpha$.

Answer 2

Here’s a rather advanced approach with the advantage that it’s relatively conceptual: We can say that you’re asking for elements of order two in the group $(\Bbb Z/p^n\Bbb Z)^\times$. If $p>2$, this group is cyclic of order $(p-1)p^{n-1}$, and consequently there are only two elements of the type desired. Specifically, the group in question is direct sum of two subgroups, the roots of $1$ in $\Bbb Z_p$ ($p$-adic integers) of which there are $p-1$, and the cyclic $p$-group generated by $1+p$, this group being of order $p^{n-1}$. Direct sum of two cyclic groups of relatively prime order is again cyclic.

Very special, but rather more fun to analyse is the case $p=2$. Here, the structure of $(\Bbb Z/2^n\Bbb Z)^\times$ is $\{\pm1\}\times C$, where again $C$ is a (multiplicative) cyclic group of order $2^{n-2}$, generated by $5$. Now in the case $n\ge3$, this latter group, being cyclic of even order, does have two elements of order two, namely $1$ and $1+p^{n-1}$, so that the square roots of unity in $\Bbb Z/2^n\Bbb Z$ are $\{\pm1,\pm1+2^{n-1}\}$ provided $n\ge3$.