Product of two primitive roots $\bmod p$ cannot be a primitive root.

Solution 1:

A primitive root of an odd prime $p$ must be a quadratic non-residue of $p$, and the product of two non-residues is a residue.

Solution 2:

Take any generator $g$ of the multiplicative group $\Bbb F_p^\times$. If $r$ and $s$ are primitive roots then $r=g^u$ and $s=g^v$ where $u$ and $v$ are coprime with $p-1$, so $u$ and $v$ are odd. Then, $rs=g^{u+v}$ and $u+v$ is even, so $rs$ is not a primitive root.

Esentially it is the same reasoning than André's.

Solution 3:

1. If $d:=\gcd(k,p-1)>1$, then $n^k$ is not a primitive root modulo $p$.

   $(n^k)^{\frac{p-1}d} \equiv 1 \pmod p$ already

2. Supposed $m=n^k$is also a primitive root. Then by 1., $k$ must be odd, as $p-1$ is even.
3. But then $mn=n^{k+1}$ is not primitive root by 1., since $k+1$ is even.