The Product of Primitive Roots Modulo p Cannot Be Primitive
\[r, \; s\]
are primitive roots of a prime \[p\]
so that \[r^{p-1} \equiv s^{p-1} \equiv 1 \; (mod \; p)\]
and \[p-1\]
is the smallest power to do this for either \[r\]
or \[s\]
. All primitive roots can be expressed as powers of any one of them, so \[s=r^k\]
for some odd \[2 \lt k \lt p-1\]
- \[k\]
is not even else then \[(r^2)^{\frac{p-1}{2}} \equiv r^{p-1} \equiv \; (1 \mod \; p)\]
.\[(rs)=rr^k=r^{k+1}\]
then \[k+1\]
is even and the order of \[rs\]
is at most \[\frac{p-1}{2}\]
for the reason explained above. 
