Twin Primes (2)

When one accepts that the sum of twin primes (apart from the first pairing 3 and 5) is divisible by 12, this then implies that the product of these twin primes when divided by 12 will always leave a remainder of 1.

This is due tho the fact that (n + 1)(n – 1) = n² – 1. And if the sum of twin primes = 2n is divisible by 12, (i.e. n is divisble by 6), then n² is thereby divisible by 12; so when n² – 1 is divided by 12, it leaves a remainder of 1.

Expressed in an alternative fashion the result = k – 1/12 (where k is an integer).

This suggests a possible connection with the result of the Riemann zeta function when s = – 1,

i.e. ζ( – 1) = – 1/12.

In fact, in this context it is interesting to note that the denominators of the Riemann zeta function for all negative odd integers (– 1, – 3, – 5,......) - as with the sum of twin primes - appear to be always divisible by 12!        
 
Thus if n – 1 and n + 1 represent twin primes (other than 3 and 5) then n² is divisible by 12.

In fact it always appears to be divisible by 36!

When we calculate the result of  n²/36 with the value of  n here relating to  the average of successive twin primes, an interesting pattern results.

For 5 and 7,           n²/36 = 36/36      = 1 

For 11 and 13,       n²/36 = 144/36    = 4

For 17 and 19,       n²/36 = 324/36    = 9

Now if we watch the series on the right the terms seem to be increasing by 1, 3, 5,... Alternatively these results represent the successive squares of the natural numbers.

For 23 and 25,       n²/36 = 576/36 = 16 (though of course 25 is not prime). Interestingly 25 also showed up as a "bogus prime" in yesterday's exercise. Then with 29 and 31 normal service is resumed.

For 29 and 31,      n²/36 = 900/36  = 25

Again we would expect that the next result on the right should be 36 and this occurs

For 35 and 37,     n²/36 = 1296/36 = 36 (even though 35 is not a prime).

With next pair,  normal service is again resumed.

For 41 and 43,   n²/36 = 1764/36  = 49

Now the next 2 squares of natural numbers are 64, 81 And these occur:

For 47 and 49,  n²/36 = 2304/36 = 64 (though 49 is not prime),

For 53 and 55,  n²/36 = 2916/36 = 81 (though 55 is not prime).

For 59 and 61  n²/36 =  3600/36 = 100.

What is remarkable here, just as in yesterday's exercise that the twin primes concur with the squares of 2, 3, 5 and 7 (for the first four pairs (after 5 and 7).

And even in all the other cases where the result is a square of a natural number that is not prime, one of the pairings is always prime (certainly up to 100).

This would suggest that a great way of searching for twin prime pairings would be to keep testing for
n²/36 = k² (where k = 1, 2, 3, ....).

Incidentally, it is again very interesting in view of the earlier observation that  ζ( – 1) = – 1/12, that the sum of all our results for k² (where k = 1, 2, 3, .....) = ζ( – 2).