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Wilson's sexy twin cousin? and new function?

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  • John W. Nicholson
    I am reading The Little Book of Big Primes by Paulo Ribenboim. I m on page 145 trying to figure out the steps for proving: (== means congruent) 4((n-1)!+1)+n
    Message 1 of 2 , Jan 6, 2004
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      I am reading The Little Book of Big Primes by Paulo Ribenboim. I'm on
      page 145 trying to figure out the steps for proving:
      (== means congruent)

      4((n-1)!+1)+n == 0 (mod n(n+2))

      the first step is stating Wilson's theorem as the following:

      (n-1)!+1 == 0 (mod n) which is has been proved earlier in the book.

      the next thing is one of the places where I am getting lost:

      4(n-1)!+2 == 0 (mod (n+2))

      It seems like this statement was pulled out of the blue. Where did it
      come from?

      I see that the '+2' inside the mod function and the (n-1)! comes from
      dealing with the prime (p+2) with Wilson's, but why where did the term
      4 (n-1)! and the +2 come in? It would seem like 'something'(n-1)! +3
      would be added but for what ever reason the 4(n-1)! and +2 is used
      instead of any other factor.

      Things get better for a little bit, then he starts talking about the
      converse.

      There he states Wilson's for n and (n+2) then states:
      "But n(n+1)=(n+2)(n-1)+2, so 2(n-1)! +1 = k(n+2), where k is an
      integer." On the first eq. I see that the n(n+1) is needed for the
      (n-1)! to be changed into the term(n+1)! and the right hand side on
      both eq's are needed for dividing by (n+2). The left hand of the
      second eq. is where I am lost the second time. Where did this come
      from? It looks like the first one times 2, which begs the question why
      not 2 instead of 4 for this term?

      Now for what I would like to do.

      My thought is that there is some way to gereralize this so it works
      for cousins (n+4), sexies (n+6), and even better sexy twin cousins.
      hee hee.

      I see that I can use n(n+1)(n+2)(n+3) = (n-1)(n+1)(n+2)(n+4)+4 for
      cousins. But what would is the form of second eq. from above?

      {my guess for the sexy is n(n+1)(n+2)(n+3)(n+4)(n+5) =
      (n-1)(n+1)(n+2)(n+3)(n+4)(n+6)+6 or (n+5)!/ n! =
      (n+6)!/((n-2)!*n*(n+5))+6 }??

      Is there a function simply reduces n! to factors which with work can
      use Fermat's Little Theorum as to use (mod P#)?

      Something like
      P = 11, and 6! = 2^4*3^2*5
      so 2^k(1) (mod 2310) + 3^k(2) (mod 2310) + ... 11^k(8) (mod 2310) == x
      (mod 2310) which just finds x? Note that P and n! are independent
      varibles.
    • richard_heylen
      ... It s a sloppy change of variables. Set m+2=n then (m+2-1)! +1 == 0 (mod m+2) (m+1)(m)(m-1)!+1 == 0 (mod m+2) (-1)(-2)(m-1)!+1 == 0 (mod m+2) 2(m-1)!+1
      Message 2 of 2 , Jan 7, 2004
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        --- In primenumbers@yahoogroups.com, "John W. Nicholson"
        <johnw.nicholson@s...> wrote:
        > I am reading The Little Book of Big Primes by Paulo Ribenboim. I'm
        > on page 145 trying to figure out the steps for proving:
        > (== means congruent)
        >
        > 4((n-1)!+1)+n == 0 (mod n(n+2))
        >
        > the first step is stating Wilson's theorem as the following:
        >
        > (n-1)!+1 == 0 (mod n) which is has been proved earlier in the book.
        >
        > the next thing is one of the places where I am getting lost:
        >
        > 4(n-1)!+2 == 0 (mod (n+2))
        >
        > It seems like this statement was pulled out of the blue. Where did
        > it come from?

        It's a sloppy change of variables. Set m+2=n then
        (m+2-1)! +1 == 0 (mod m+2)
        (m+1)(m)(m-1)!+1 == 0 (mod m+2)
        (-1)(-2)(m-1)!+1 == 0 (mod m+2)
        2(m-1)!+1 == 0 (mod m+2)
        QED

        Rick
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