Question: Given $p$, $q$ both primes such that $q = 2p + 1$, I need to prove that $-4$ is a primitive root mod $q$. 
So far haven't found a direction that could lead me to the solution. 
Any suggestion or short solution would be helpful, thanks in advance.

Answer: 
The order of $-4$ can only be $1,2,p,2p$. It is easy to whos that it is not $1,2$ thus we only need to show it is not $p$. 

Since $(-4)^{\frac{q-1}{2}}=\pm 1 \pmod q$ you need to prove that $(-4)^{\frac{q-1}{2}} \equiv -1 \pmod q$. This is the same as showing that $4^{\frac{q-1}{2}} \equiv 1 \pmod q$. 

**Hint** $4=2^2$.
To prove that $-4$ is a primitive root modulo $q$, we need to show that the order of $-4$ is $q-1$, which is equivalent to showing that $(-4)^{\frac{q-1}{2}} \equiv\_q -1$.

Assume that $p\neq2$. It follows that

$$(-4)^{\frac{q-1}{2}} \equiv\_q -1 \iff 4^{\frac{q-1}{2}} \equiv\_q 1 \iff 2^{q-1} \equiv\_q 1$$

But this is true by Fermat's little theorem, since $q=2p+1$ is odd and thus does not divide $2$.

However, if $p=2$, then $q=5$ and $-4 \equiv\_5 1$, so in this case the statement is not true.