Why does the product of all the elements in $\mathbb{Z} _{m}^{*} $ equal to $\pm {1} \mod{m}$?

Question

I am asked to prove that the product of all the elements in $ \mathbb{Z}_{m}^{*}$ for $ m>2$ equals to $ \pm {1} \mod{m} $. So far I tried the following direction. I defined: $$ S_{1} = \{x \in Z^*_{m} : x^2\equiv 1\mod{m}\} $$ and $$ S_{2} = \{x \in Z^*_{m} : x^2\not \equiv 1\mod{m}\} $$ If $P_{i}$ is the product of all the elements in $S_{i}$ for $i=1,2$, I proved that it holds $$ P = P_1\cdot P_2 $$ I already proved that $P_2=1 \mod m$. I need to prove $P_{1}= \pm 1 \mod{m} $. I already know that $P_{1}^2 = 1\mod{m} $ but unfortunately this doesn't imply the required. Can I get a hint on how to prove that $P_{1}= \pm 1 \mod{m} $? Thank you

Answer 1

On $S_1$, introduce the equivalence relation

$$x \sim y \iff (x = y \lor x = -y)\,.$$

Then every equivalence class contains exactly two elements of $\mathbb{Z}_m^{\ast}$ (if $m > 2$ is even, then $m/2$ is not coprime to $m$, so we have $x \neq -x$ for all $x \in \mathbb{Z}_m^{\ast}$), and

$$x\cdot (m-x) = mx - x^2 \equiv -x^2 \equiv -1 \pmod{m}\,.$$

Hence

$$P_1 \equiv (-1)^{(\operatorname{card} S_1)/2} \pmod{m}\,.$$