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RE Primality tests and Wolstenholme's Theorem (typo)

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  • Jose Ramón Brox
    ... From: Jose Ramón Brox = SUM_i=1,q_{c_i · m·p^(i-1) } / (q-1)! = m· SUM_i=1,q_{c_i · p^(i-1) } / (q-1)! = = m· SUM_i=0,q-1_{c_i
    Message 1 of 1 , Dec 20 7:36 PM
      ----- Original Message -----
      From: "Jose Ramón Brox" <ambroxius@...>

      = SUM_i=1,q_{c_i · m·p^(i-1) } / (q-1)! = m· SUM_i=1,q_{c_i · p^(i-1) } / (q-1)! =
      = m· SUM_i=0,q-1_{c_i · p^i } / (q-1)! = B(q)

      Since m divides p, B(q) = = 0 (mod p) <--> SUM_i=0,q-1_{c_i · p^i } / (q-1)! = = 0 (mod p)

      ------------------------------------------------------------------------

      Should read:

      "Since m=p/q , B(q) = = 0 (mod p) <--> SUM_i=0,q-1_{c_i · p^i } / (q-1)! = = 0 (mod q)"

      and it is (mod q) from this part on, but I think the rest is OK (the calculations are
      actually done modulo q, I had another notation in my former document, where I called p1 to
      q, but I thought it was a bit misleading).

      Jose Brox
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