Let $p$ be an odd prime
Suppose $a$ and $b$ are primitive roots mod $p$
Prove that $ab$ is a quadratic residue mod $p$
I know that $ord_{p}{a}=\phi( p)=p-1$ (i.e.) $a^{p-1}\equiv 1$ mod p and same for b.
I try to use the Legendre symbol but I don't know how to prove $(\frac{a}{p})$ have the same sign as $(\frac{b}{p})$.
Thanks.
Best Answer
The primitive root condition is kind of silly, since the product of any two quadratic non-residues is a quadratic residue.
But let's play along. Let $a$ be a primitive root of $p$. Since $b$ is a primitive root, $b\equiv a^k \pmod{p}$ for some odd $k$. (And in fact for some $k$ relatively prime to $p-1$.) Then $ab\equiv a^{k+1}\pmod{p}$. So $ab\equiv (a^{(k+1)/2})^2\pmod{p}$, and therefore $ab$ is a quadratic residue of $p$.