How to prove following statement :
Conjecture :
Let $p$ be a prime number of the form : ${\color{BlueViolet}{p=16 \cdot n^4+1}}$
If $n$ is an odd prime greater than $3$ then $3$ is a primitive root modulo $p$ .
I wrote small Maple program (see below) in order to find counterexample , but I haven't found anyone . For the first $204231$ primes of this form $3$ is a primitive root modulo $p$ .
There is theorem that states :
$a$ is a primitive root modulo $p$ iff $\operatorname{ord}_p(a) = \phi(p)$
One can show that only possible values of $\operatorname{ord}_p(3)$ are :
$16n , 16 n^2 , 16n^3 ~\text{or}~ 16n^4$
If we roll out first three possibilities then we have a proof , but I don't see any reason why $\operatorname{ord}_p(3)$ couldn't be : $ 16 n , 16n^2 ~\text{or}~ 16n^3$ .
with(numtheory): i:=0: for n from 4 to 44400000 do if isprime(n) then if isprime(16*n^4+1) then i:=i+1: if not(primroot(16*n^4+1)=3) then print(n); end if; end if; end if; end do; i;