Results 1 to 4 of 4

Math Help - Jacobian problem

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    2

    Jacobian problem

    The problem is

    "Suppose n=pq where pq are distinct odd primes. prove that there always exist positive integers a,b, such that

    \left(\frac{a}{n}\right)=-1 and \left(\frac{b}{n}\right)=-1, and such that ab is not a square modulo n."

    Does for ab to not be a square modulo n mean \left(\frac{ab}{n}\right)=-1?

    I think I'm doing it wrong because I keep getting \left(\frac{ab}{n}\right)=1.

    Could anyone give me a hand with this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Well we always have \left(\frac a n\right) \left(\frac b n\right)=\left(\frac {ab}{n}\right). So in this case, by assumption we have \left(\frac {ab}{n} \right)=1. You are asked to show that, although \left(\frac {ab}{n} \right)=1, ab is not a square modulo n.

    The point of the exercise is to realize that \left(\frac {m}{n} \right)=-1 implies that ab is not a square, but that \left(\frac {m}{n}\right)=1 does not imply that m is a square.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2009
    Posts
    2
    So since I keep getting \left(\frac {ab}{n} \right)=1, ab can be either a square not a square mod n. Do I use quadradic reciprocity to somehow find integers a and b so ab's not a square mod?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Here is why. Let \mathbb{Z}_n,\mathbb{S}_n denote respectively the integers mod n and the multiplicative group of nonzero squares mod n. We have

    S_n \cong \mathbb{Z}_n/\{\pm 1\}

    and also

    \mathbb{S}_{n} \cong \mathbb{S}_p \times \mathbb{S}_q.

    Therefore |\mathbb{S}_{n}|=|\mathbb{S}_{p}||\mathbb{S}_{q}|=  (p-1)(q-1)/4. Since we have \left(\frac{a}{n}\right)=1 for (p-1)(q-1)/2 = \phi(pq)/2 elements a of \mathbb{Z}_n, there must be some of those which are not elements of \mathbb{S}_{n}, i.e. not squares.


    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: December 7th 2011, 05:00 PM
  2. Jacobian Integration Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 23rd 2010, 12:02 PM
  3. Jacobian
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 3rd 2009, 07:02 PM
  4. The problem of Jacobian determinant
    Posted in the Calculus Forum
    Replies: 0
    Last Post: May 1st 2009, 05:36 AM
  5. Stoke's Theorem - Jacobian Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 7th 2009, 02:33 AM

Search Tags


/mathhelpforum @mathhelpforum