Results 1 to 5 of 5

Math Help - ascent problem

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
    42

    ascent problem

    Suppose n is a fixed integer not equal to 0 and d is a positive integer not a perfect square, and suppose

    x_0^2 - dy_0^2 = n,

    where x_0 \geq 0,\ y_0 \geq 0. Let x_1 = ax_0 + bdy_0,\ y_1 = bx_0 + ay_0, where a>0, b>0 provides a solution to the Fermat-Pell equation, a^2 - db^2 = 1. Show that

    x_1^2 - dy_1^2 = n,\ \ \ \ \ x_1 > x_0 \geq 0,\ \ \ \ \ y_1 > y_0 \geq 0.

    Conclude that the equation x^2 - dy^2 = n either has no solutions or infinitely many solutions.

    ********************
    I do know that the problem is similar to http://www.mathhelpforum.com/math-he...od-ascent.html, but I did not understand something in that thread. How did TPH go from x^2 - 3y^2 = 1 to
    Quote Originally Posted by ThePerfectHacker View Post
    If,
    x=a
    y=b
    Is a solution,
    Then,
    (a^2+3b^2)^2-3(2ab)^2=(a^2-3b^2)^2=1
    Is there a method to that or is it just something you recognize as the right formula to use?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Doth this help?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2008
    Posts
    42
    Quote Originally Posted by ThePerfectHacker View Post
    Doth this help?
    honestly, no...the theorem he posted is word for word from the text I am using in my class and I did not really understand it any more than he did. But whereas your answer helped him to understand it, I am still apparently missing something in my comprehension of it...or of how to apply it to my posted problem. So I would still appreciate help on this
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2008
    Posts
    42
    okay, so in finding another example and then rereading what was here and in the book, I see that those are pretty much just formulas. I was looking for more than that, like you had gotten those from the problem itself somehow but now I see it is just a fomula to be used.

    Then back to the problem in my original post. I keep wanting to show the formula simply by substituing in the given formula values and am unsure of whether or not it is that simple or I need to show it in another way. Ideas? Hints? Blatant spoilers?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member diddledabble's Avatar
    Joined
    Jul 2009
    Posts
    80

    Is there anyone out there who can truly answer this question?

    Proggy,
    I am in the same boat. This class was super easy until this section and then the author just stopped writing. The questions are nothing like the examples. Did you survive the course?

    If anyone out there has a solution to this proof please post.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Path of Steepest Ascent
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 14th 2010, 08:07 PM
  2. [SOLVED] Method of Ascent for Diophantine Equations
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: June 7th 2010, 08:18 AM
  3. direction of steepest ascent
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 1st 2010, 09:50 AM
  4. clarification on descent vs ascent
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: September 4th 2008, 09:33 AM
  5. method of Ascent
    Posted in the Number Theory Forum
    Replies: 9
    Last Post: October 11th 2006, 10:26 AM

Search Tags


/mathhelpforum @mathhelpforum