If $p$ is a Sophie Germain prime, that is $q = 2p+1$ is also prime and $a\not\equiv \pm1$(mod $p$) then $a^2 \not\equiv 1$(mod $q$).

Question

I am trying to understand the proof that

If $p$ is a Sophie Germain prime, that is $q = 2p+1$ is also prime and $a\not\equiv 0,\pm1$(mod $p$)
$$\text{If }\biggl(\frac{a}{q}\biggr)=-1 \text{ then }a \text{ is a primitive root mod(q)}$$

where $\bigl(\frac{a}{q}\bigr)$ is Legendre's symbol

Since $a^{q-1} \equiv 1$ (mod $q$) and $q-1 = 2p$, $\operatorname{ord}_q(a) = \{1,2,p,2p\}$

Since $(\frac{a}{q})=-1$, we got $\operatorname{ord}_q(a) \neq1,p$

So, there remains to show $a^2 \not\equiv 1\pmod q$.