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Primenumbers

Member

Registered: 2013-01-22

Posts: 149

Infinite No. of twin primes! (attempt 3)

Assumptions:

1. There are regular gaps of 4 between primes
2. These gaps of 4 have varying remainders for 4, (either 1 or 3)

A= 3x5x7x11x13x17............(x next prime in series)
x = the next prime above highest prime in A used
p and p+4 are prime

(A - p)/2 = twin prime
(A-(p+4))/2 = twin prime

Example:

A= 3x5x7= 105
x=11
A has remainder=1 for 4
2 Primes = 67 and 71
67 and 71 have remainder=3 for 4, which is different from A
(105-67)/2 = 19
(105-71)/2 = 17
Twin primes!

Last edited by Primenumbers (2015-11-08 06:32:20)

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