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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 :


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
print((b^n+1)^2-2,"is counterexample");
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 $K_n(b)$

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I assume by your use of $b=2$ as an example that you meant the nonnegative integers rather than $\mathbb{Z}^*.$ – Charles Feb 9 '12 at 19:27
up vote 2 down vote accepted

The conjecture is false. With $b=2^{11}3^2=18432$ and $n=2$, $K=(b^n+1)^2-2=115422333316890623$ which is prime. But the order of 5 mod $K$ is $(K-1)/11$ so it is not a primitive root.

Code used (PARI/GP):


Time used: 2 ms.

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