Formula for prime numbers of the form (m)*(n)+1

[Hugo1177]

Proof of the Twin primes Conjecture and Goldbach's conjecture We can find infinite prime numbers with the separation we want and we can express every even number as the sum of two prime numbers. http://www.academia.edu/43581083/Pro...chs_conjecture

[CRGreathouse]

The crux of the proof is on p. 3:

I have tried it with high numbers and it seems that there is no problem, which would be to say that there are infinite prime numbers separated the quantity that we want.

I'll note that the method requires taking large symbolic derivatives and then factoring integers far larger than the twin primes generated, so this is not an algorithmic improvement.

[jnml]

Originally Posted by Hugo1177

Proof of the Twin primes Conjecture and Goldbach's
conjecture We can find infinite prime numbers with the separation we want and we can
express every even number as the sum of two prime numbers
http://www.academia.edu/43581083/Pro...chs_conjecture

It's fine if you prefer to keep the PDF to just yourself.

It's fine if you prefer to sell it for money.

It's bad to "publish" it for free - but at a site that requires registration and or login. Some, if
not most people, may chose to simply go away.

[sweety439]

Originally Posted by Hugo1177

Proof of the Twin primes Conjecture and Goldbach's conjecture We can find infinite prime numbers with the separation we want and we can express every even number as the sum of two prime numbers. http://www.academia.edu/43581083/Pro...chs_conjecture

It is known that there are infinite prime numbers with separation n for an even number <= 246, but we don't known what this n is!!! (like that, we know that at least one of zeta(5), zeta(7), zeta(9), and zeta(11) is irrational, but we don't known which number is irrational)

Also, there is no infinite (at most one pair) prime numbers with separation n when n is odd!!!

[Dr Sardonicus]

This magnum opus does not begin well.

The function that generates those two prime numbers is:

dn/dx = x^(4/p) - 3x^(2/p) + 1

Where n is the derivative of order n and p is the distance in units of the separation that we want to find.

So, "n" is self-referencing, and the symbol p is used to denote the separation, or distance between two primes, rather than a prime number.

It continues into the demonstrably false:

The relationship between this function and the Lucas numbers is that in the undifferentiated function

x^(4/p) - 3x^(2/p) + 1 it's zeros are the Lucas numbers.


For example x^(4/7) - 3x^(2/7) + 1 = 0

One of its zeros is 29 which is the 7th number of Lucas

Er, no. Substituting 29 for x in the given expression gives -0.00198416, approximately.

The zeros of y^2 - 3*y + 1 are



so, taking x = y^(p/2), we have

[Hugo1177]

https://www.researchgate.net/publica...f_the_form_mn1

[Hugo1177]

MODERATOR NOTE: Moved from Lounge.

https://www.researchgate.net/profile..._Garcia_Pelaez

[Hugo1177]

Formula that returns prime numbers of the form (m)*(n)+1 how many prime numbers between 1 and 1000000 I used a polynomyal that it´s roots are the golden ratio squared and the golden ratio conjugate. When I aply fractional exponents to the x the polinomyal it´s roots returns lucas and fibonacci numbers exactly. And when I derivate this function I obtain prime numbers included in one number of the structure of the derivative, but there are more that prime numbers come in relation with the order of the derivative multiplied for the fractional exponent. Is like there a relation between fibonacci and lucas numbers and prime number where the original polynomial derivated adapted his form to returns special prime numbers.

https://www.researchgate.net/publica...f_the_form_mn1

[Batalov]

Mod warning:

Stop spam-posting. Next duplicate posts (and crossposts) will be deleted altogether,

[CRGreathouse]

Originally Posted by Hugo1177

prime numbers of the form (m)*(n)+1

As opposed to what other kind of prime numbers?