Welcome back RD-BH
Thank you 8)
=== still working on correct wording ===
Let us examine Integers and Indivisibility.
x/n is an integer when n is a divisor of x
1i^(4x/n) equals 1 when n is a divisor of x [1i = (-1)^(1/2)]
numberOfDivisors(x) = sum_(n=1)^(x){floor(abs(1i^(4x/n)+1)/2)}
numberOfDivisors(x) = 2 for prime integers [1, P]
numberOfDivisors(P^x) = x+1 for prime integer P [P^0..P^x]
Integer/Prime leaves a remainder of 0..(Prime-1)
If we form a matrix where the number of columns is indivisible by a prime number of rows, then fill the matrix with sequential integers, we note Integer/Prime (number of rows equal prime) leave uninterrupted vertical series of indivisible integers.
Consider:
a vertical series of 1s representing the number line
now zero integers divisible by two and form 2 columns
column 0: 2*I + 0, column 1: 2*I + 1 (ie vertical column with each row 01)
now zero integers in column 1 divisible by 3 and form 6 columns
column 0: 2*3*I + 0, column 1: 2*3*I + 1,
, column 5: 2*3*I + 5 (ie each row 010001)
we note Integer/Prime leaves uninterrupted vertical (ie linear) series of indivisible integers
this pattern of indivisibility repeats through infinite periods (ie Period*I + remainder; 6*I+1,6*I+5)
What can we glean from this behavior?
1) Because there are infinite Primes, dividing an infinite series by a prime leaves infinite linear series of indivisible integers.
2) There are infinite indivisible twin pairs
2 leaves an infinite series of indivisible twin pairs with remainders (1,1)
3 leaves 1 infinite series of indivisible twin pairs with remainders (2,1) [zeroing remainders from 2]
5 leaves 3 infinite series of indivisible twin pairs with remainders (1,3),(2,4),(4,1) [zeroing remainders from 2 and 3]
7 leaves 5 infinite series of indivisible twin pairs with remainders (1,3),(2,4),(3,5),(4,6),(6,1) [zeroing remainders from 2,3, and 5]
if we continue you will see the number of infinite series is P-2 for P>2
This is because we make P copies (1 per P rows) of the previous indivisible twin pairs (101) and P can only divide 1 integer in each column per P rows.
We can now show the total number of indivisible twin pairs.
ITP(x) = (product_(n=1)^(x){(P_n) 2}) / (product_(n=1)^(x){P_n}) * infinity
note the actual number is > ITP(x) as we are not counting combos where P-2 or P+2 are prime
This is a/b * c where a is infinite, b is infinite, and c is infinite.
If we assume a finite number of twin primes, a/b * c should be finite.
note: c > product_(n=1)^(P_x){I_n} > b for I_n = nth integer, therefore c/b > 1
Therefore, there must be infinite twin primes.
Consider:
(2*I + 1) is not divisible by 2
(2*I + 1) 2^J is not divisible by 2 for integer J > 0
(2*I + 1) + 2^J is not divisible by 2 for integer J > 0
since all primes > 2 are odd integers they all resolve to (2*I + 1)
since odd integer * odd integer is also odd then all multiplicative combos of primes > 2 also resolve to (2*I + 1)
... since all odd primes are greater than 2, and there exist integers divisible by only odd primes, then none of those odd prime divisors can divide an integer +- 2^J from that integer
Therefore f(x) = (product_(n=1)^(x){P_n}/2) * (2*I + 1) +- 2^J + K*product_(n=1)^(x){P_n} for integers I, J > 0, K has infinite number of infinite series of indivisible twin pairs solutions
If we assume a finite number of twin primes, f(x) should have finite solutions, it has infinite solutions.
Therefore there must be infinite twin primes.
=== hopefully no typos ===
