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  1. #1
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    method of Ascent

    Use the method of Ascent to prove there are infinitely many solutions to the Diophantine equation: x^2 - 3y^2 =1.

    I don't fully get the concept of the method of Ascent. Could you please show me how ? Thank you very much.
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    Quote Originally Posted by beta12 View Post
    Use the method of Ascent to prove there are infinitely many solutions to the Diophantine equation: x^2 - 3y^2 =1.

    I don't fully get the concept of the method of Ascent. Could you please show me how ? Thank you very much.
    This is the classical diophantine equation from number theory. It is inacurrately called, "Pellian Equation".

    Some History.
    ---
    The first first person to make use of it was Fermat in 17th Century. He claimed he had a proof of the infinitude of solutions using the method of infinite descent (not ascent). Not supprsingly, he never published it. Nor do I think there exists a proof using infinite descent.

    A century later Euler decided to solve this problem in general. But even he failed. Eventually, Lagrange succedded in showing that:
    x^2-ny^2=1
    Where n is a non-square integer has infinitely many solutions.

    I have only seen two proofs. One which I believe was worked out by Euler/Lagrange by the use of infinite continued fractions. [Every solution is a convergent of the continued fractional expansion of sqrt(n)]. It takes like 6 pages (if you already know the fundamental properties of continued fractions).

    The other proof I have only gazed upon. It was done by Dirichelt. Without the use of continued fractions. All he did was show the infininitude of solution (where Euler/Lagrange has an algorithm).

    Both the proofs are really ugly/hard/long using non-stop use of inequalities.

    But the basic theorem you need to know that, is "n" is NOT a square then there are infinitely many solutions. Otherwise, the only solution is trivial.
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    Hi Perfecthacker,

    Thanks for the information.
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  4. #4
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    Let me at least solve it for you.
    ---
    We expand sqrt(3) as a continued fraction.

    [1; 1,2,1,2,1,2,....]
    The length of this expansion is n=2 which is even.
    Thus, fundamental solution is:
    x=p_{n-1} y=q_{n-1}
    Thus,
    x=p_1 and y=q_1

    We need to compute the convergent C_1 which is,
    1+1/1=2/1
    Thus,
    x=2 and y=1
    Which is true.

    Thus all solutions are,
    [2+sqrt(3)]^n
    When you exand it you have an integer and an integer with a radical. The free integer is the x value and the integer with the radical is the y value.

    For example, if n=2 then,
    [2+sqrt(3)]^2=4+4sqrt(3)+3=7+4sqrt(3)
    Thus,
    x=7 and y=3

    If, n=3 then,
    [2+sqrt(3)][7+4sqrt(3)]=26+15sqrt(3)
    Thus,
    x=26 and y=15

    And so on...
    ~~~
    Alternatively, all convergents C_{2k-1} for k=1,2,3,4,...
    Are solution to this equation.
    This shows why there are infinitely many.
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    Hi Perfecthacker,
    Thanks for the solution with continued fractions approach.

    I must solve this question with the method of Ascent. I've got some more information for this question. Can you figure out the proof by the method of Ascent?
    ==================
    Use the method of Ascent to prove there are infinitely many solutions to the Diophantine equation: x^2 - 3y^2 =1.

    You do this by showing how, given one solution (u,v), we can compute another solution(w,z) that is "larger " is some suitable sense. Then your proof will involve finding a pair of formulas , something like:
    w = x +y and z = x-y

    (These formulas won't work. But there is a pair of second degree formulas which do work; one of them has a cross term and one of them involves the number 3)
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    Quote Originally Posted by beta12 View Post
    Use the method of Ascent to prove there are infinitely many solutions to the Diophantine equation: x^2 - 3y^2 =1.

    You do this by showing how, given one solution (u,v), we can compute another solution(w,z) that is "larger " is some suitable sense. Then your proof will involve finding a pair of formulas , something like:
    w = x +y and z = x-y

    (These formulas won't work. But there is a pair of second degree formulas which do work; one of them has a cross term and one of them involves the number 3)
    Ah!
    I see what you are saying. I was referring to Fermant's method of descent when you show no solutions exist and I believed you misspelled the word. Okay now I understand.

    You basically want to have a constructive proof, meaning from one solution to obtain another.
    But the trick is to show that at least one solution exists, which is the problem.

    I need time to think about it and respond back.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    Ah!
    I see what you are saying. I was referring to Fermant's method of descent when you show no solutions exist and I believed you misspelled the word. Okay now I understand.

    You basically want to have a constructive proof, meaning from one solution to obtain another.
    But the trick is to show that at least one solution exists, which is the problem.

    I need time to think about it and respond back.
    A simple solution exists: x=2, y=1.

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    A simple solution exists: x=2, y=1.

    RonL
    Right

    I was thinking when I wrote the post,
    x^2-ny^2=1
    In general
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  9. #9
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    If,
    x=a
    y=b
    Is a solution,
    Then,
    (a^2+3b^2)^2-3(2ab)^2=(a^2-3b^2)^2=1

    Thus an new solution is,
    x=a^2+3b^2=A
    y=2ab=B

    Thus a new solution is,
    x=A^2+3B^2
    y=2AB

    Continue doing this and you shall obtain a newer solution, each being differenct from the others because they form a strictly monotonic increasing sequence.

    Thus, we "ascent"

    Note: This not necessarily produce all the solutions but it shows there are infinitely many. I think when Lagrange proved this diophantine equation he did not use continued fractions rather he shows the existence of a non-trivial solution and then used this argument to how one can construct more solutions.
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  10. #10
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    Hi Perfecthacker,

    Thank you very much. I got it now.
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