$p$ is an odd primitive, Show why there are no primitive roots of $\bmod 3p$ if $p$ is an odd primitive, Prove there are no primitive roots of $\bmod 3p$
Where I'm at: $a^{2(p-1)}=1 \pmod{3p}$ where a is a primitive root of $3p$ (by contradiction) 
$(a/3p)=(a/3)(a/p)$ are the Legendre symbols, and stuck here..tried a couple of things, but got nowhere, could use a helping hand :]

Note that when $p=3$ the theorem does not hold! 2 is a primitive root.
So supposing $p$ is not 3...
Since $p$ is odd let $p = 2^r k+1$ with $k$ odd.
The group of units is $\mod {3p}$ is $$(\mathbb Z/(3p))^\times \simeq \mathbb Z/(2) \times \mathbb Z/(2^r) \times (\mathbb Z/(k))^\times$$ by Sun Zi's theorem.
There can be no primitive root for this because of the two $\mathbb Z/(2^i)$ parts.
