Results 1 to 3 of 3

Math Help - Infinitely many solutions to: x^2-dy^2=+/-1

  1. #1
    Member
    Joined
    Jul 2008
    Posts
    78

    Infinitely many solutions to: x^2-dy^2=+/-1

    I need help understanding the following theorem.

    Theorem. Suppose d is a positive integer. If there is one solution to the Fermt-Pell equation (1): x^{2}-dy^{2}=1;\ x, y>0, then there are infinitely many solutions. If there is one solution to the Fermat-Pell equation (2): x^{2}-dy^{2}=-1;\ x, y>0, then there are infinitely many solutions to both (1) and (2).

    Proof.
    Suppose there are integers a and b,
    a^{2}-db{2}=c; \ a,b>0,
    where c will shortly be chosen to be \pm1. Suppose further that for some n\ge 1 there are integers x_{n} and y_{n} such that
    x_{n}^{2}-dy_{n}^{2}=c^{n};\ x_{n},y_{n}>0.
    When n = 1 this is possible with x_{1}=a and y_{1} = b. Set
    [This is where I'm confused, how do they get these equations? I'm lost from here on.]
    x_{n+1}=ax_{n}+dby_{n},

    y_{n+1}=ay_{n}+bx_{n}.
    These values are legal since
    x_{n+1}^{2}-dy_{n+1}^{2}={\left( a^{2}x_{n}^{2}+2dabx_{n}y_{n}+d^{2}b^{2}y_{n}^{2}\  right) - d\left ( a^{2}y_{n}^{2}+2abx_{n}y_{n}+b^{2}x_{n}^{2}\right)  }

    ={a^{2}x_{n}^{2}+d^{2}b^{2}y_{n}^{2}-da^{2}y_{n}^{2}-db^{2}x_{n}^{2}}

    ={\left( a^2-db^2\right)\left(x_{n}^{2}-dy_{n}^{2}\right)}

    ={c\cdot c^{n}}={c^{n+1}}
    and
    x_{n+1}={ax_{n}+dby_{n}>1\cdot x_{n}+d\cdot 0\cdot y_{n}}={x_{n}\left(>0\right)},

    y_{n+1}={ay_{n}+bx_{n}>1\cdot y_{n}+0\cdot x_{n}}={y_{n}\left(>0\right)}.
    [...] The rest of the proof follows, but I think I posted the relavent parts.

    Thanks.
    Last edited by Pn0yS0ld13r; August 18th 2008 at 04:45 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    If x_0,y_0>0 solve x^2 - dy^2 = 1 and x_1,y_1>0 is a solution then (x_0x_1 + dy_0y_1)^2  - d(x_0y_1+x_1y_0)^2=1. Thus, x_0x_1+dy_0y_1 and x_0y_1+x_1y_0 is a solution. We can use this to generate infinitely many solutions.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2008
    Posts
    78
    Ah, I get it now, thank you very much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 4th 2011, 09:21 PM
  2. Replies: 1
    Last Post: September 23rd 2011, 04:39 AM
  3. unique and infinitely solutions problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 18th 2009, 07:56 PM
  4. Replies: 2
    Last Post: September 7th 2009, 03:01 PM
  5. Matrices with infinitely many solutions
    Posted in the Advanced Algebra Forum
    Replies: 11
    Last Post: September 17th 2007, 06:52 PM

Search Tags


/mathhelpforum @mathhelpforum