Consider the prime numbers of the form :
$p=4 \cdot k^2+1~$ , where $~k~$ is an odd prime number .
For the first $~1200000~$ primes of this form except when $~p=4 \cdot 193^2+1~$
$2~$ is a primitive root modulo $~p$ .
Note that :
Euler's totient function is given by :
$\phi(p)=4 \cdot k^2$
One can show that :
$\operatorname{ord}_p(2) \neq k~ ,\operatorname{ord}_p(2) \neq 2k~ ,\operatorname{ord}_p(2) \neq k^2~ ,\operatorname{ord}_p(2) \neq 2k^2~ $ , hence :
$\operatorname{ord}_p(2) = 4k~ \text {or}~\operatorname{ord}_p(2) = 4k^2~ $ since it cannot be $2 ~\text{or}~ 4$ .
My question :
Is there some special reason why $~2~$ isn't primitive root modulo $~p~$ in case of number
$~p=4 \cdot 193^2+1~$ ?