How to prove or disprove following statement :
Let $K_n(b)$ be generalized Kynea prime of the form :
$K_n(b)=(b^n+1)^2-2$ , where : $b=2^k\cdot 3^l , ~k,l \in \mathbb{Z^{*}}$ then
If $~n \equiv 0 \pmod 2~$then $5$ is primitive root modulo $K_n(b)$
I have checked statement for following prime numbers :
$b=2~ ; ~n \in \{2,8,12,18,32,...\}$
$b=6~ ; ~n \in \{2,4,12,30,...\}$
$b=12~ ; ~n \in \{2,8,...\}$
$b=18~ ; ~n \in \{10,...\}$
Maple code that examines this conjecture :
with(numtheory): i:=0: b:=2; for n from 2 to 100 by 2 do if isprime((b^n+1)^2-2) then if not(pprimroot(4,(b^n+1)^2-2)= 5) then i:=i+1; print((b^n+1)^2-2,"is counterexample"); break; end if; end if; end do; if i = 0 then print("no counterexample"); end if;
Hint :
If $~ord_{K_n(b)}(5) = \phi(K_n(b))= b^{2n}+2\cdot b^{n}-2~$ then
$5$ is a primitive root modulo $K_n(b)$