Loading ...
Sorry, an error occurred while loading the content.

15510Re: [PrimeNumbers] Closest n^2+1 twin primes

Expand Messages
  • Jens Kruse Andersen
    Oct 28, 2004
    • 0 Attachment
      Robin Garcia wrote:
      > Define Closest n^2+1 twin primes as numbers
      > n =m+10*t /+2 t=0,1,2,3,... (m=4 mod 10) such that n^2+1 is prime
      > This naming could be changed: a name is a name is.......

      It sounds like you require "Closest" to have consecutive t's.
      Then there are no solutions for t>5 as shown below.
      My suggestion was "the closest they can be together".
      I think you have to define something like that for large sets.
      My page currently goes to k=18. A larger set would require 10 twin pairs.

      2 divides n^2+1 for all odd n.
      5 divides n^2+1 for even n = 2,8,12,... (mod 10)
      Then a n^2+1 twin must have n = 4 and 6 (mod 10) to avoid the factor 5.

      Of n values = 4 or 6 (mod 10), 17 divides n^2+1 for
      n = 4,64,106,166,174,... (mod 170)

      This means there can be at most 5 consecutive t values giving n^2+1 twins
      on the form n = m+10*t+(0 and 2).
      Those twins can start at n = 14 or 114 (mod 170).
      I have not computed any but they should be plenty.

      Of n values = 4 or 6 (mod 10), 13 divides n^2+1 for
      n = 34,44,86,96,164,... (mod 130)
      For n = 104 to 156 (mod 130), there are 6 consecutive twin candidates
      avoiding 13 as factor, so 17 is actually more limiting than 13 in our case.

      --
      Jens Kruse Andersen
    • Show all 4 messages in this topic