Primality Criteria for Wagstaff numbers

[pedja]

$$\text{Let} ~ W_p ~ \text{be a Wagstaff number of the form :} W_p = \frac{2^p+1}{3}~, \text{where}~p>3$$
$$\text {Let's define }~S_0~ \text{as :}$$
$$S_0 = \begin{cases} 3/2, & \text{if } p \equiv 1 \pmod 4 \\ 11/2, & \text{if } p \equiv 1 \pmod 6 \\ 27/2, & \text{if} ~p \equiv 11 \pmod {12} ~\text{and}~p \equiv 1,9 \pmod {10} \\ 33/2, & \text{if}~ p \equiv 11 \pmod {12} ~\text{and}~p \equiv 3,7 \pmod {10} \\ \end{cases}$$
$$\text{Next define sequence}~S_i~\text{as :}$$
$$S_i = \begin{cases} S_0, & i=0 \\ 8S^4_{i-1}-8S^2_{i-1}+1, & i>0 \end{cases}$$

$$\text{How to prove following statement :}$$
$$\text{Conjecture :}$$
$$W_p=\frac{2^p+1}{3}~\text{is a prime iff}~S_{\frac{p-1}{2}} \equiv S_0 \pmod {W_p}$$

 

[dodo]

Hi, pedja,
I don't have an answer, but it may be easier to get help if you can clear up a bit the definition of S_0: it's undefined for p=3, and the cases are not mutually exclusive.

 

[pedja]

Hi, pedja,
I don't have an answer, but it may be easier to get help if you can clear up a bit the definition of S_0: it's undefined for p=3, and the cases are not mutually exclusive.

Hi ,
I forgot to point out that p has to be greater than three . I know that first two cases are not mutually exclusive . Both values of S_0 can be used in order to prove primality of corresponding Wagstaff number .