Results 1 to 3 of 3

Math Help - Prime & Congruences

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    404

    Prime & Congruences



    I know x^2 ≡ -2 (mod p) means that p|(x^2 -2), but I can't figure out how to continue with this.
    Any help is appreciated!


    [also under discussion in math links forum]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by kingwinner View Post


    I know x^2 ≡ -2 (mod p) means that p|(x^2 -2), but I can't figure out how to continue with this.
    Any help is appreciated!


    [also under discussion in math links forum]
    by Dirichlet's approximation theorem there exist integers 1 \leq k \leq \lfloor \sqrt{p} \rfloor and m such that |\frac{kx}{p} - m| \leq \frac{1}{\lfloor \sqrt{p} \rfloor + 1} < \frac{1}{\sqrt{p}}. which gives us |kx - mp| < \sqrt{p}.

    now put a=|kx - mp| and b=k and show that p \mid a^2 + 2b^2 and 0 < a^2 + 2b^2 < 3p.
    Last edited by NonCommAlg; February 22nd 2010 at 12:03 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Quote Originally Posted by NonCommAlg View Post
    by Dirichlet's approximation theorem there exist integers 1 \leq k \leq \lfloor \sqrt{p} \rfloor and m such that |\frac{kx}{p} - m| \leq \frac{1}{\lfloor \sqrt{p} \rfloor + 1} < \frac{1}{\sqrt{p}}. which gives us |kx - mp| < \sqrt{p}.

    now put a=|kx - mp| and b=k and show that p \mid a^2 + 2b^2 and 0 < a^2 + 2b^2 < 3p.
    hmm...I haven't learnt that theorem yet...
    Is it possible to solve this problem with more basic concepts?

    thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 22nd 2011, 12:37 PM
  2. Replies: 1
    Last Post: June 19th 2011, 12:56 PM
  3. Replies: 6
    Last Post: August 27th 2010, 11:44 PM
  4. Some congruences
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: January 3rd 2010, 03:42 PM
  5. congruences
    Posted in the Number Theory Forum
    Replies: 13
    Last Post: March 30th 2008, 10:11 AM

Search Tags


/mathhelpforum @mathhelpforum