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.

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*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


*Supposed $m=n^k$is also a primitive root. Then by 1., $k$ must be odd, as $p-1$ is even.

*But then $mn=n^{k+1}$ is not primitive root by 1., since $k+1$ is even.

