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Primality Criteria for Wagstaff numbers

  1. Mar 12, 2012 #1
    [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 =
    \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} [/tex]
    [tex]\text{Next define sequence}~S_i~\text{as :} [/tex]
    [tex]S_i =
    \begin{cases}
    S_0, & i=0 \\
    8S^4_{i-1}-8S^2_{i-1}+1, & i>0
    \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]
     
    Last edited: Mar 12, 2012
  2. jcsd
  3. Mar 12, 2012 #2
    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.
     
  4. Mar 12, 2012 #3
    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 .
     
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