Prove $\{x \mid x^2 \equiv 1 \pmod p\}=\{1, -1\}$ for all primes $p$

Question

One way to prove that $\{x \mid x^2 \equiv 1 \pmod p\}=\{1, -1\}$ is to use the fact that $\{1, -1\}$ is the only subgroup of the cyclic group of primitive residue classes modulo $p$ that has the order 2. Is there a more basic way to prove this?

Answer 1

$$x^2\equiv1\mod p\iff p\text{ divides }(x-1)(x+1)$$

But a prime divides a product if and only if it divides one of the two factors.

Answer 2

The integers modulo $p$ form a field. A polynomial of degree $n$ over a field has at most $n$ roots in the field.

Answer 3

You can use group theory method.

$(\Bbb Z_{p}^{*}, \otimes_{p})$ is a cyclic group of an even number, so just $\varphi(2)=1$ element has order $2$, so the equation $x^2=1$ has only two solutions identity and that unique element of order $2$.

---

You can use number theory method.

$x$ satisfies the given equation if and only if $p|x^2-1=(x-1)(x+1)$ so if we assume $$x\in\{0,1,2,\dots,p-1\}$$ then the only possible answers are $1$ and $p-1$.