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Factorization : x^2 - z*y^2

  1. Apr 10, 2012 #1
    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.
     
    Last edited: Apr 10, 2012
  2. jcsd
  3. Apr 10, 2012 #2


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

    DonAntonio
     
  4. Apr 10, 2012 #3
    You are right. Sorry I made mistake not precising x,y,z (positive integers)
     
  5. Apr 10, 2012 #4

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

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

    DonAntonio
     
  6. Apr 10, 2012 #5

    Thank you.
     
  7. Apr 10, 2012 #6
    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.
     
  8. Apr 10, 2012 #7

    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
     
  9. Apr 11, 2012 #8
    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?
     
  10. Apr 11, 2012 #9


    No. I assumed you meant [itex]x^2-zy^2[/itex] 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 [itex] x = 3, y = z = 2\,,\,\, x^2-zy^2=9-2\cdot 4=1[/itex] , which is irreducible, but we get with

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

    DonAntonio
     
  11. Apr 12, 2012 #10
    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?
     
  12. Apr 13, 2012 #11
    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.
     
  13. Apr 14, 2012 #12

    morphism

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    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?
     
  14. Apr 16, 2012 #13
    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.
     
  15. Apr 16, 2012 #14


    You didn't answer morphism's question.

    DonAntonio
     
    Last edited: Apr 16, 2012
  16. Apr 17, 2012 #15
    The answer is : no
     
  17. Apr 17, 2012 #16
    I did it now.
     
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