number of square roots of unity modulo a prime power

Question

Let $p\geq 3$ be a prime number and let $k\geq 1$ be some integer.

Is it always true that if $x^2\equiv 1\pmod{p^k}$ then $x\equiv\pm1\pmod{p^k}$ ?

For $k=1$ it is true since $x^2-1\in\mathbb{F}_p[x]$ is a degree $2$ polynomial with two distinct roots $-1,1$ and so any other root in $\mathbb{F}_p$ must be one of them.

I know that there exists a primitive root modulo $p^k$.

Answer 1

We know that $$p^k \mid a^2-1=(a-1)(a+1)$$we prove that it is impossible that $$p \mid a-1\\p \mid a+1$$but this is obvious since$$p \mid (a+1)-(a-1)=2$$which is a contradiction for $p>2$ therefore either $p\not\mid a-1$ or $p\not\mid a+1$ which means that either $p \mid a-1$ or $p \mid a+1$

Answer 2

Another way to say that there exists a primitive root modulo $p^k$ is to say that $U(p^k)$ is cyclic. Since it has even order, it has exactly one subgroup of order $2$: the set of all elements such that $x^2=1$.

Answer 3

Another approach using induction (inspired by Hensel’s lemma). Assume $p \neq 2$. For $k=1$ the statement holds. Now if $k > 1$ and $a^2\equiv 1 \pmod{p^k}$ then also $a^2\equiv 1 \pmod {p^{k-1}}$. So by induction $a \equiv \pm 1 \pmod{p^{k-1}}$, that is, $a = \pm 1 + m p^{k-1}$ for some integer $m$. Then $$1 \equiv a^2 = 1 \pm 2 m p^{k-1} + m^2 p^{2k-2}\equiv 1 \pm 2 m p^{k-1} \pmod{p^k}.$$ This means (since $p \neq 2$) that $p \mid m$. Then $a = \pm 1 + m p^{k-1}\equiv \pm 1 \pmod{p^k}$.

Answer 4

We know that $1$ and $-1$ are roots of $1\bmod p^k$.

Let $g$ be a primitive root $\bmod p^k$ and $m:=\phi(p^k)$ be its order. Then considering $(g^i)^2$ with $i\in [1..m]$ we note that only $i=m/2$ and $i=m$ give $(g^i)^2\equiv 1 \bmod p^k$. Thus there are only two roots of $1 \bmod p^k$.