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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?

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

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

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

fkraiem
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    Note that there is no reason to exclude $p=2$ in your statement, the result remains valid in that case. – fkraiem Jan 27 '14 at 20:01
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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 $p-1$.

Woria
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