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(n,n+4) prime pairs' rule

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  • torasso.flavio@enel.it
    By the same method I used in obtaining a twin primes rule I worked on (n,n+4) pairs and I found the following: for n not equal to 9, (n,n+4) is a prime pair
    Message 1 of 1 , Apr 5, 2001
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      By the same method I used in obtaining a twin primes' rule
      I worked on (n,n+4) pairs and I found the following:
      for n not equal to 9, (n,n+4) is a prime pair iff
      36[(n-1)/2]!^2 = +/-(7n-36) mod n(n+4)
      the sign being "-" when n=4k-1, "+" when n=4k+1.

      Could it be a new result or was already proved in the past?

      Here is the reasoning that leads to this result:
      (A) n is prime iff [(n-1)/2]!^2 = +/-1 mod n
      (B) for n not equal to 9, (n,n+4) is a prime pair iff
      9[(n-1)/2]!^2 = +/-9 mod n = +/-16 mod (n+4)
      (note that the sign on the left side of / holds when n=4k-1
      and the sign on the right side of / holds when n=4k+1)
      (C) when n=4k-1
      consider the identity [(n-3)/4]n + 9 = [(n-7)/4](n+4) + 16
      then [(n-3)/4]n + 9 = +9 mod n = +16 mod (n+4)
      hence, by (B), 9[(n-1)/2]!^2 = [(n-3)/4]n + 9 mod both n and
      (n+4)
      this means that 9[(n-1)/2]!^2 = [(n-3)/4]n + 9 mod n(n+4)
      which leads to 36[(n-1)/2]!^2 = -(7n-36) mod n(n+4)
      (D) when n=4k+1
      consider the identity [(n+11)/4]n - 9 = [(n+7)/4](n+4) - 16
      and proceed as in (C) obtaining
      36[(n-1)/2]!^2 = +(7n-36) mod n(n+4).

      Thanks for any comments
      Best regards
      Flavio Torasso
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