Let $p$ be an odd prime. Show that the product of the distinct primitive roots, $\mod{p}$, is $\equiv$ $1$ or $-1$ $\pmod{p}$.
I think this can be done by viewing the primitive roots as a elements of $(\mathbb{Z}/p\mathbb{Z})^*$ and then showing that their product is either $1$ or $-1$ in $(\mathbb{Z}/p\mathbb{Z})^*$.
I have this hint but dont know how to proceed.
For almost all primes, the product is actually congruent to $1$ modulo $p$.
If $p=2$, then the product of the primitive roots of $p$ is congruent both to $1$ and $-1$ modulo $p$. And if $p=3$, then the product of the primitive roots of $p$ is congruent to $-1$ modulo $p$.
We now show that if $p\gt 3$ is prime, then the product of the primitive roots of $p$ is congruent to $1$ modulo $p$.
Recall (or prove) that if $g$ is a primitive root of $p$, then the modular inverse $g^{-1}$ of $g$ is a primitive root of $p$. Note also that if $p\ne 3$, and $g$ is a primitive root of $p$, then $g^{-1}\not\equiv g\pmod{p}$. For $x^{-1}\equiv x\pmod{p}$ if and only if $x^2\equiv 1\pmod{p}$, that is, if and only if $x\equiv \pm 1\pmod{p}$.
Thus if $p\gt 3$, the primitive roots of $p$ can be divided into couples $\{g,g^{-1}\}$. The product of the elements in a couple is congruent to $1$, and therefore if $p\gt 3$ then the product of all the primitive roots is congruent to $1$.
