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

I recently proved that the product of all primitive roots of an odd prime $p$ is $\pm 1$ as an exercise. As a result, I became interested in how few distinct primitive roots need to be multiplied to guarantee a non-primitive product. Testing some small numbers, I have the following claim: "If $n$ and $m$ are two distinct primitive roots of an odd prime $p$, then $nm \bmod p$ is not a primitive root."

I've tried to make some progress by rewriting one of the primitive roots as a power of the other, but haven't been able to see any argument which helps me prove the result. Any help would be appreciated.

• you need to understand how $(\mathbb{Z}/p\mathbb{Z},\times)$ is just a permutation of $(\mathbb{Z}/(p-1)\mathbb{Z},+)$ (the two groups are isomorphic) – reuns Feb 5 '16 at 17:32

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.
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.
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.