Why 4 is not a primitive root modulo p for any prime p?
Solution 1:
As $2^2=4,$
$4^{\left(\frac{p-1}2\right)}=2^{p-1}\equiv1\pmod p$ for any prime $p>2$ using Fermat's Little Theorem
So, the ord$_p4\le \frac{p-1}2<p-1$
Solution 2:
For any relevant $p$, the order of 2 will be larger than the order of $4 = 2\cdot 2$.
Solution 3:
Hint $\ $ If $\rm\ 1 < d\mid p\!-\!1\ $ then $\rm\:mod\ p\!:\ a\not\equiv 0 \:\Rightarrow\: (a^d)^{(p-1)/d}\! \equiv a^{p-1}\! \equiv 1,\:$ therefore $\rm\:a^d$ has order at most $\rm\:(p-1)/d < p-1.\:$ Thus $\rm\:d$'th powers are not primitive roots. Your case is $\rm\:d = 2.$