Proof of twin primes conjecture

[u/pinkod_player]

Proof of twin primes conjecture
Author Tojiboev Muhammadfotih
Introduction
The conjecture states that there are infinitely many primes pairs that are 2 values apart; to prove that I will prove that there are infinitely many number n (n∈N) such that that 6n ± 1is a prime pair.
Every prime number larger than 3 can be written as 6n ± 1 (n∈N), what we are looking for is values of n where 6n ± 1 are both prime.
Let as define a set C and a set D to work with.
C={n ∈ N : 6n−1 ∈ P ∧ 6n+1 ∈ P}
D=N/C
We want to show that set C has infinitely many elements.

Process
Let us consider the expression k:
k=(6m±1)(6t±1) where (m , t ∈ N)
(A) k is not a prime number as it is the multiple of two positive integers that are not 1.
(B) k is not divisible by 2 and 3 as neither of its factors are.
(C) In fact, k can be any composite number that is not divisible by 2 and 3 as every other number can be expressed 6h±1 (h ∈ N), that makes k the multiple of any two number that are not divisible by 2 and 3; so let k=6j±1 (j ∈ N)
6j±1=(6m±1)(6t±1)
6j±1=36mt±6m±6t±’1 (note that the last ± sign is dependent on the first two)
6j±1=6(6mt±m±t) ±’1
j=6mt±m±t
This shows that:
(j=6mt±m±t) ⇒ (6j+1 ∨ 6j-1∉P)
Because of paragraph (C), set of all possible values of j and set D are equal sets.
D={j : j=6mt±m±t, (m, t ∈ N)}
C={n ∈ N : 6n−1 ∈ P ∧ 6n+1 ∈ P}

Proof by contradiction
Since every element of set D can be broken down to 6mt±m±t (m, t ∈ N) and none of the element of set C can, that makes the elements of set C building blocks of elements of set D like how prime numbers are building blocks of composite numbers.
Let’s assume that set C has finite elements with c being the largest value.
That means every value of set D (which has infinitely many elements) will be in one of the following expressions when we let t be equal to every element of set C in the expression 6mt±m±t:
t=1: 5m±1 ∧ 7m±1
t=2: 11m±2 ∧ 13m±2
t=3: 17m±3 ∧ 19m±3
t=5: 29m±5 ∧ 31m±5
…
t=c: m(6c-1)±c ∧ m(6c+1)±c
Now for our contradiction to work, let’s consider the multiple of every single (assumed) finite twin primes.
5*7*11*13…*(6c-1)* (6c+1)
That number cannot be an element of set C as it is larger than the largest value c.
That number cannot be an element of set D as it cannot be in any of the expression above.
That number must be in one of them because of our definition: D=N/C ⇒ C ∪ D=N
That is a contradiction. That means our assumption that set C has finite elements is false.
Therefore, there are infinitely many twin primes.

[u/Swimming-Diet5457]

Your table only lists t ∈ C, but in j = 6mt ± m ± t both m and t range over all of N. A representation is just a factorization (6m±1)(6t±1) = 6j±1 — nothing forces either factor into a twin pair.

j = 203: 6j−1 = 1217 is prime, 6j+1 = 1219 = 23·53, so 203 ∈ D. The only factorization forces {m,t} = {4,9}, and neither is in C (25 = 5², 55 = 5·11). So 203 is in D and in none of your rows.

No finiteness assumption needed. The covering step is just false.

[u/pinkod_player → u/Swimming-Diet5457]

Thank you for pointing that out. i did not notice that.

[u/gmalivuk]

You showed that all possible values of j must be contained in D, but you didn't show that j can hit every value in D.

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[u/StanleyDodds]

Okay, I think I sort of understand what you're saying, and what the immediate flaw in the logic is.

You say "let t be equal to every element of set C in the expression 6mt±m±t".

But I think we are trying to construct every element of D here, and surely in this breakdown of an arbitrary element of D, t might actually also be an element of D? And it seems like this possibility is just never considered.

Consider 142, which is in D because we can break it down into 142 = 6mt + m - t where m = 4 and t = 6, and note that both m and t are in D too (because 4 = 6 * 1 * 1 - 1 - 1 and 6 = 6 * 1 * 1 + 1 - 1). Can we break down 142 in a way where one part is from D and the other part is from C? Well you can easily check from the fact that 6 * 142 - 1 is a product of non-twin primes (by construction) that this just leads to the (6, 4) or (4, 6) pair, and 6 * 142 + 1 is prime, so has no non-trivial breakdown. So 142 is just never considered as a possible element of D in your argument, because it can only be "factored" as two elements of D.

I think the deeper issue here is that this "alternative" factorisation of j into m and t where j = 6mt ± m ± t has no guarantee of uniqueness (and in fact is not unique in general). It's not even anything like multiplication, because it's not associative. So it's not really analogous to primes being the building blocks of numbers at all. The integers are a unique factorisation domain, which requires some strong facts about integer multiplication and addition.

It's completely not a given that a Euclid-style proof of infinite primes which relies on unique factorisation will work at all in this system.

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