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

[Gaussianheart]

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.

 

[DonAntonio]

Let x,y,z > 0
Gcd(x,y)=1
Gcd(x,z)=1
Gcd(y,z)=1

z squarefree

Factorize

x^2 - z*y^2

Thank you.

Factorize...over what ring or field? What kind of beings are x,y,z, anyway?

DonAntonio

 

[Gaussianheart]

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

 

[DonAntonio]

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

The expresion can't be factorized over the integers, or over the rationals, of course. Over the reals though

we have that x^2-zy^2=(x-y\sqrt{z})(x+y\sqrt{z}) .

DonAntonio

 

[Gaussianheart]

The expresion can't be factorized over the integers, or over the rationals, of course. Over the reals though

we have that x^2-zy^2=(x-y\sqrt{z})(x+y\sqrt{z}) .

DonAntonio

Thank you.

 

[Gaussianheart]

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.

 

[DonAntonio]

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.

As I told you, there is not such factorization over the integers, but you now say you have a method to do it, so either you are wrong or I am, and the easiest and fastest way to find out is for you to write down your method.

DonAntonio

 

[Gaussianheart]

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?

 

[DonAntonio]

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?

No. I assumed you meant x^2-zy^2 is an integer polynomial in two (or even three) variables, and then I said it can't be reduced.

Of course that if you mean number it can, or not, be reduced, according as if it is a prime or composite number.

For example, with x = 3, y = z = 2\,,\,\, x^2-zy^2=9-2\cdot 4=1 , which is irreducible, but we get with

x=6, y=z=1\,,\,\,x^2-zy^2=36-1=35=5\cdot 7 , which is reducible...

DonAntonio

 

[ramsey2879]

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.

1*(x^2 - z*y^2). Other than that there is no algorithm to factor the expression. Of course many composite are of this form, such as 2*n where x,y and z are each odd. What method do you have in mind?

 

[Gaussianheart]

1*(x^2 - z*y^2). Other than that there is no algorithm to factor the expression. Of course many composite are of this form, such as 2*n where x,y and z are each odd. What method do you have in mind?

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 right and provable.
Implemeting the method is not easy.
I will post it as soon as I finsh the tests.

 

[morphism]

Let's fix an integer z. Does your method give an answer to the question: what primes are of the form x^2-zy^2?

 

[Gaussianheart]

Let's fix an integer z. Does your method give an answer to the question: what primes are of the form x^2-zy^2?

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.

 

[DonAntonio]

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.

You didn't answer morphism's question.

DonAntonio

 

[Gaussianheart]

Let's fix an integer z. Does your method give an answer to the question: what primes are of the form x^2-zy^2?

The answer is : no

 

[Gaussianheart]

You didn't answer morphism's question.

DonAntonio

I did it now.