Primality Criteria for Wagstaff numbers

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

[tex]\text{How to prove following statement :}[/tex]
[tex]\text{Conjecture :}[/tex]
[tex]W_p=\frac{2^p+1}{3}~\text{is a prime iff}~S_{\frac{p-1}{2}} \equiv S_0 \pmod {W_p}[/tex]
 
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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.
 
Dodo said:
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 .