Recent content by Gaussianheart

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    Graduate Mertens function : new formulation

    The formula is correct! Good for me!
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    Graduate Factorize x^2 - z*y^2 with Gcd(x,y)=1, Gcd(x,z)=1, Gcd(y,z)=1, and z squarefree

    I did it now.
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    Graduate Factorize x^2 - z*y^2 with Gcd(x,y)=1, Gcd(x,z)=1, Gcd(y,z)=1, and z squarefree

    The answer is : no
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    Graduate Factorize x^2 - z*y^2 with Gcd(x,y)=1, Gcd(x,z)=1, Gcd(y,z)=1, and z squarefree

    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.
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    Graduate Factorize x^2 - z*y^2 with Gcd(x,y)=1, Gcd(x,z)=1, Gcd(y,z)=1, and z squarefree

    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...
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    Graduate Factorize x^2 - z*y^2 with Gcd(x,y)=1, Gcd(x,z)=1, Gcd(y,z)=1, and z squarefree

    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?
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    Graduate Factorize x^2 - z*y^2 with Gcd(x,y)=1, Gcd(x,z)=1, Gcd(y,z)=1, and z squarefree

    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.
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    Graduate Factorize x^2 - z*y^2 with Gcd(x,y)=1, Gcd(x,z)=1, Gcd(y,z)=1, and z squarefree

    Thank you.
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    Graduate Factorize x^2 - z*y^2 with Gcd(x,y)=1, Gcd(x,z)=1, Gcd(y,z)=1, and z squarefree

    You are right. Sorry I made mistake not precising x,y,z (positive integers)
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    Graduate Factorize x^2 - z*y^2 with Gcd(x,y)=1, Gcd(x,z)=1, Gcd(y,z)=1, and z squarefree

    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.
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    Graduate Mertens function : new formulation

    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
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    Graduate Mertens function : new formulation

    It is true! People checked it in other fora... So thanks for reading the post.
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    Graduate Mertens function : new formulation

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

    Is this identity true? Look at attachment Thank you.
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    Undergrad Is This New Algorithm Effective for Finding Common Factors?

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