Recent content by Gaussianheart

The formula is correct! Good for me!

I did it now.

The answer is : no

For any fixed z squarefree >1 you will always have primes of the form x^2-zy^2. My goal is to find a way to factorize odd semi-prime (n) starting writing it as equal to some (x,y,z) in relation x^2-zy^2. There are an infinite number of writing it like above by using the Brahmagupta identity.

Many composite could be expressed in many ways as x^2 - z*y^2. That is why there always a way to factor those numbers. My problem now is how to choose (x,z,y) such as the factorization will be easy to do. I did not finished yet. I'm testing testing testing. But the core of my method is...

If I understood you seems to say that we can not find : u*v=x^2-zy^2 with u and v integers and assuming that some known number A is equal to x^2-zy^2 It that right?

So there is no algorithm or some method to factorize over the integers the equation above? I just want to be sure because I have found a way to do it. Not finished yet to be published.

Thank you.

You are right. Sorry I made mistake not precising x,y,z (positive integers)

Let x,y,z > 0 (x,y,z naturals numbers) Gcd(x,y)=1 Gcd(x,z)=1 Gcd(y,z)=1 z squarefree Factorize x^2 - z*y^2 Thank you.

n=8=2*4 an 4 is not odd Read the condition : o must be odd >=3 then you can compute M(2*o) My formula holds. Someone in another forum just proved it. I have a proof but it is little bit long. Thank you for your comment

It is true! People checked it in other fora... So thanks for reading the post.

Example : o=25 M(50) = mu(27)+mu(29)+mu(31)+...+mu(49) = 0+(-1)+(-1)+...+0= -3 mu(n) is Mobius function

Is this identity true? Look at attachment Thank you.

? a and b have to be different. If b=a then no need to search the gcd.