Question: Show that if $a$ is a quadratic residue mod $p$, then $a$ is not a primitive root mod $p$.

I know that a quadratic residue is a perfect square, and a primitive root has order equal to $\varphi(p)$, but I do not see any relationship between the 2 that would imply the desired conclusion.

Answer: 
If $a \equiv x^2 \pmod{p}$, then $a^{\phi(p)/2} \equiv x^{\phi(p)} \equiv 1 \pmod{p}$; the order of $a$ would not be $\phi(p)$.
If $a$ is a $k-$adic Residue, there exists $b$ such that $b^n\equiv a\pmod p$

We know, ord$\_mb=d,$ ord$\_m(b^k)=\dfrac d{(d,k)}$ ([Proof](http://archive.org/details/NumberTheory_862) @Page#95) 

What is the maximum value of $d$ here?