Assume $a=-1$ or that $a$ is a perfect square. Prove that $a$ is not a primitive root modulo $p$ for any prime $p>3$.
Let $p>3$ be an arbitrary prime. First, we assume $a$ is a perfect square. Write $a=b^2$. Then $a^{\phi(p)/2}=b^{\phi(p)}$. How do I know that $\gcd(b,p)=1$ here? (Because $b=1,2,3,\dots$, so $b$ and $p$ must have some common factor). Anyway, if this is true, we would then get $b^{\phi(p)}\equiv 1\pmod p$, which makes $a$ not to be a primitive root modulo $p$. This case is done. For the other case, I have no clue where to begin.
Let $g$ be a primitive root mod $p$. Write $b=g^n$. Then $a=g^{2n}$, which has order $$\dfrac{p-1}{\gcd(2n,p-1)} \le \dfrac{p-1}{2} < p-1$$ Thus, $a$ cannot be a primitive root mod $p$.