## Saturday, November 23, 2013

### Even simpler idea for a proof of twin prime conjecture

The first lemma will be: "The second of a pair of twin primes does not have a remainder of 2 when divided by any prime up to its square root, and a number is the second of a pair of twin primes if it doesn't have a remainder of 2 or 0 when divided by any prime up to it's square root."
Proof: Suppose P did divide a prime p less than it's square root with a remainder of 2. Then the number 2 less than P would divide it exactly. Thus the number 2 less than P is not prime, which contradicts the fact that P was the second of a pair of twin primes. Also if P does not divide any number less than it's square root with a remainder of 0 then it is prime, and if it also does not divide any of those with a remainder of 2 then the number 2 less than P does not divide them either, which means it is also prime. Hence, P is the second of a pair of twin primes.

Theorem: Let N = (p+1)^2-1, a way to roughly count the number of primes less than N is:
N*(1/2)*(2/3)*(4/5)...*(p-1)/p
Proof: Roughly 1/2 of all the numbers up to N won't divide 2, 2/3 won't divide 3, and so on up to p-1. Whether a number divides p or divides another prime q happen independently of one another so the percentages are multiplied. (p+1)^2 is also the first number where factors of p+1 need to be considered as any number that divides p+1 evenly before that also divides a smaller prime.

Theorem 2: Let N = (p+1)^2-1, a way to count one of each of every pair of twin primes less than N is roughly:
N*(1/2)*(1/3)*(3/5)...*(p-1)/p:     The same as the formula from theorem 1 but the numerators of all but the 1/2 term are decreased by 1.

Proof: Because of the lemma, we know that numbers that have a non-zero remainder when divided by every p up to the square root, and also not 2 when divided by each p are one of a pair of twin primes. For every p but 2 this is one less possible remainder so every term other than 1/2's numerator is reduced by 1. For example there are 3 possibilities for the remainder when divided by 5, 1,3, and 4. So 3/5 appears in the fraction. This formula thus roughly counts the number of twin primes less than N.

Theorem 3: There are an infnite number of twin primes estimated by the formula in Theorem 2:
Proof: Suppose the number of twin primes up to N used theorem 2's formula and counted P pairs of twin primes by:
((p)^2-1)*(1/2)*(1/3)*(3/5)...*(p-2)/p
for now we can write that as:
((p)^2-1)*X
We can show that if we replace p with the next bigger prime q, the number of twin primes estimated always  increases because the ratio of estimated primes is:

This ratio is always greater than 1 as p gets larger because the ratio of q^2-1 to p^2-1 is larger than 1 by more than the ratio of (q-2) to q is smaller than 1.  This means there must be a positive number of estimated pairs of twin primes between p^2-1 and q^2-1 And since there are an infinite number of primes to choose p and q from there must be an infinite number of pairs of twin primes...

#### 1 comment:

1. maybe it will help i read it like this what yo uare giving
so like you thought square root 2 or 0.....
N equals (p+1)^2-1 for a prime
bit later N increasing, infinite number twin primes
your estimate for a prime p N*(1/2)*(3/5)*...........
look your angle (p-2)/p or am i wrong here? maybe it helped maybe not however best regards