Suppose $p$ is $3\bmod 4$. Show $a$ and $-a$ cannot both be primitive roots $\bmod p$.

Question

Suppose $p$ is $3\bmod 4$. Show $a$ and $-a$ cannot both be primitive roots $\bmod p$.

I think the idea behind my proof is right but not sure it's written out that clearly, would appreciate another look.

Suppose $a$ is a primitive root $\bmod p$. We know if a number is a primitive root, it is a quadratic non-residue $\bmod p$. We will attempt to show $-a$ is also a primitive root $\bmod p$.

$(\frac{-a}{p}) = (\frac{-1}{p})(\frac{a}{p}) = (\frac{-1}{p})\cdot -1 = -1 \cdot -1 = 1$ (following since $-1$ is a quadratic non-residue if $p \equiv 3\bmod 4$). So $-a$ cannot be a primitive root $\bmod p$ and therefore $a$ and $-a$ cannot both be primitive roots $\bmod p$.

Answer 1

A primitive element is for sure a non-quadratic residue (otherwise its order would be at most $\frac{p-1}{2}$), but if $p\equiv 3\pmod{4}$ then $-1$ is a quadratic non-residue, and $a,-a$ cannot be both non-quadratic residues since the Legendre symbol is multiplicative (i.e. quadratic residues form a subgroup with index $2$ of $\mathbb{Z}/(p\mathbb{Z})^*$). Your approach is fine.

Answer 2

You can prove it without quadratic residues at all. Assume $a$ is a primitive root mod $p$. It means $a^{p-1} \equiv 1 (\text{mod}p)$ and for any $0<m<p-1$ we have $a^m \not\equiv 1 (\text{mod}p)$. We know p is prime so we can conclude $a^{\frac{p-1}{2}} \equiv -1 (\text{mod}p)$. Now let's look at $(-a)^{\frac{p-1}{2}}$ in mod $p$. We have $p \equiv 3 (\text{mod}4)$, so $\frac{p-1}{2}$ is odd. Hence:

$(-a)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}}a^{\frac{p-1}{2}} \equiv (-1)(-1) \equiv 1 (\text{mod}p)$

So $(-a)$ is not a primitive root.