Residue classes of Z/pZ where p is odd?

Question

can someone help me with this problem. I have no idea where to start.

What I have so far:

$[a]^{(p-1)} = [1] \Rightarrow [a]^{(p-1)/2} = [1]^{1/2} = [1]$ ?

In the problem it should read. a is not divisible by p. ( I am assuming)

Answer 1

Let $x=[a]^{\frac{p-1}{2}}$ then $x^2=[1]$ Now $x^2-[1]=(x-[1])(x-[-1])$ thus since $\mathbb{Z}_p$ is a integral domain, actually a field, either $x-[1]=0$ or $x-[-1]=0$.

Answer 2

Let $g$ be a primitive root of $\Bbb{Z}_p$ and let $e = g^{\frac{p-1}{2}}$, then whe have $e^2 = 1$. Since $e \neq 1$ because $\frac{p-1}{2} \neq p-1$ we must have $e = -1$. For an arbitrary alement $a = g^r$ we have $a^{\frac{p-1}{2}} = (-1)^r$ establishing the correspondence with quadratic residues.