Prove ab is a quadratic residue mod p 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.
 A: 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 $p$. 
A: We want to show that $\left(\frac{a}{p}\right) \left(\frac{b}{p}\right) = \left(\frac{ab}{p}\right)$. Can split it up into cases:
1: $p \mid a$ or $p \mid b$. Then $p \mid ab$ so we get $0=0$ by the definition of the Legendre symbol.
2: $p \not \mid a$ and $p \not \mid b$. Let $g$ be a primitive root mod $p$. Then $g^m \equiv a \pmod p$ and $g^n \equiv b \pmod p$ for some $m$ and $n$. We have three cases:


*

*Both $m$ and $n$ are even. Then $\left(\frac{a}{p}\right) = \left(\frac{b}{p}\right) = 1$ since even powers are always quadratic residues. Since $ab = g^{m+n}$, we see that $ab$ is also an even power of $g$, hence $\left(\frac{ab}{p}\right) = 1$ as well.

*Both $m$ and $n$ are odd. Then  $\left(\frac{a}{p}\right) \left(\frac{b}{p}\right) = (-1) \cdot (-1) = 1$. Similar to above, we can show $ab$ is an even power and hence $\left(\frac{ab}{p}\right) = 1$.

*One of $m$, $n$ is odd, the other is even. Then the LHS of the product is $1 \cdot (-1) = -1$. We can show $ab$ is an odd power since the sum of an even and an odd is odd. Therefore the RHS is $-1$ as well.


In all cases we have shown $\left(\frac{a}{p}\right) \left(\frac{b}{p}\right) = \left(\frac{ab}{p}\right)$, completing the proof.
