## (n,n+4) prime pairs' rule

Expand Messages
• 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
• 0 Attachment
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).