- #1
e2theipi2026
- 9
- 1
I can prove the twin prime counting function has this form:
[tex]\pi_2(n)=f(n)+\pi(n)+\pi(n+2)-n-1,[/tex]
where [tex]\pi_2(n)[/tex] is the twin prime counting function, [tex]f(n)[/tex] is the number of twin composites less than or equal to [tex]n[/tex] and [tex]\pi(n)[/tex] is the prime counting function.
At [tex]n=p_n,[/tex] this becomes
[tex]\pi_2(p_n) = f(p_n) + \pi(p_n) + \pi(p_n + 2) - p_n - 1.[/tex]
With this form, can I make the following argument?: Assume the twin prime counting function becomes a constant [tex]c[/tex], then I can change the twin prime counting function to [tex]c[/tex] in the equation. The prime counting function [tex]\pi(n)[/tex] at the prime sequence [tex]p_n[/tex] is just [tex]n[/tex], so I can change that to [tex]n[/tex]. Because I'm assuming no more twin primes, [tex]p_n+2[/tex] is not a prime so [tex]\pi(p_n+2)[/tex] will also become [tex]n[/tex], the equation directly above this paragraph can therefore be simplified to:
[tex]c = f(p_n) + 2n - p_n - 1.[/tex]
Adding [tex]1[/tex] to both sides of this and rearranging it gives,
[tex]p_n - f(p_n) = 2n - b[/tex], where [tex]b=c+1.[/tex]
The right side of [tex]p_n - f(p_n) = 2n - b[/tex]
has only one possible parity, either odd or even because it is an even number [tex]2n[/tex] minus a constant [tex]b.[/tex]
But, the left side can be both odd and even many times over because [tex]f(p_n)[/tex] can be odd or even and is subtracted from [tex]p_n[/tex] which is odd for [tex]p>2.[/tex]
So, the left side will change parity for different values of [tex]n,[/tex] while the right side of the equation will remain one parity. Therefore, the two sides cannot be equal for all [tex]n.[/tex]
This seems to show the twin prime counting function cannot become constant and therefore, there are infinite twin primes. Now assuming I can prove the form of the twin prime counting function given at the beginning of this question, does that argument hold water?
[tex]\pi_2(n)=f(n)+\pi(n)+\pi(n+2)-n-1,[/tex]
where [tex]\pi_2(n)[/tex] is the twin prime counting function, [tex]f(n)[/tex] is the number of twin composites less than or equal to [tex]n[/tex] and [tex]\pi(n)[/tex] is the prime counting function.
At [tex]n=p_n,[/tex] this becomes
[tex]\pi_2(p_n) = f(p_n) + \pi(p_n) + \pi(p_n + 2) - p_n - 1.[/tex]
With this form, can I make the following argument?: Assume the twin prime counting function becomes a constant [tex]c[/tex], then I can change the twin prime counting function to [tex]c[/tex] in the equation. The prime counting function [tex]\pi(n)[/tex] at the prime sequence [tex]p_n[/tex] is just [tex]n[/tex], so I can change that to [tex]n[/tex]. Because I'm assuming no more twin primes, [tex]p_n+2[/tex] is not a prime so [tex]\pi(p_n+2)[/tex] will also become [tex]n[/tex], the equation directly above this paragraph can therefore be simplified to:
[tex]c = f(p_n) + 2n - p_n - 1.[/tex]
Adding [tex]1[/tex] to both sides of this and rearranging it gives,
[tex]p_n - f(p_n) = 2n - b[/tex], where [tex]b=c+1.[/tex]
The right side of [tex]p_n - f(p_n) = 2n - b[/tex]
has only one possible parity, either odd or even because it is an even number [tex]2n[/tex] minus a constant [tex]b.[/tex]
But, the left side can be both odd and even many times over because [tex]f(p_n)[/tex] can be odd or even and is subtracted from [tex]p_n[/tex] which is odd for [tex]p>2.[/tex]
So, the left side will change parity for different values of [tex]n,[/tex] while the right side of the equation will remain one parity. Therefore, the two sides cannot be equal for all [tex]n.[/tex]
This seems to show the twin prime counting function cannot become constant and therefore, there are infinite twin primes. Now assuming I can prove the form of the twin prime counting function given at the beginning of this question, does that argument hold water?