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Thread: diophantine problem

  1. #1
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    diophantine problem

    Suppose $\displaystyle n$ is a fixed integer (positive or negative) $\displaystyle \neq 0$ and suppose

    $\displaystyle x_0^2 - 3y_0^2 = n$

    with $\displaystyle x_0 \geq 0,\ y_0 \geq 0$. Let $\displaystyle x_1 = 2x_0 +3y_0,\ y_1=x_0+2y_0$. Show that

    $\displaystyle x_1^2-3y_1^2=n,\ \ \ \ \ x_1>x_0\geq 0,\ \ \ y_1>y_0\geq 0$.

    Conclude that the quation $\displaystyle x^2-3y^2=n$ either has no solutions or infinitely many solutions.

    All I can think of to do is plug in the given values of $\displaystyle x_1$ and $\displaystyle y_1$ to show that they come out to the same equation. No idea what to do after that or even if that is correct to do at all.
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  2. #2
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    Hello,

    Substitute $\displaystyle x_1$ and $\displaystyle y_1$ in $\displaystyle x_1^2-3y_1^2$ and prove it is equal to n if $\displaystyle x_0^2-3y_0^2=n$.

    This proves that if $\displaystyle \{x_0,y_0\}$ is solution, then $\displaystyle \{x_1,y_1\}$ is solution too.

    If there is no solution, we're done.
    If there is one solution, $\displaystyle \{x_0,y_0\}$, and therefore $\displaystyle \{x_1,y_1\}$ is solution too. Since $\displaystyle x_1^2-3y_1^2=n$, we can prove the same way as above that $\displaystyle \{x_2=2x_1+3y_0 ~,~ y_2=y_1+2y_1 \}$ is solution too. And so on...

    Isn't it the infinite ascent ?

    So either there is no solution, either there's an infinity of solutions.
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  3. #3
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    so then it would be
    $\displaystyle x_1^2 - 3y_1^2 = n$ with $\displaystyle x_1 = 2x_0 +3y_0,\ y_1=x_0+2y_0$
    $\displaystyle (2x_0 + 3y_0)^2 - 3(x_0 + 2y_0)^2 = n$
    $\displaystyle (4x_0^2 + 12x_0y_0 + 9y_0^2) - 3(x_0^2 + 4x_0y_0 + 4y_0^2) = n$
    $\displaystyle x_0^2 - 3 y_0^2 = n$
    which is equivalent to $\displaystyle x_1^2 - 3y_1^2 = n$

    and then from there you say that $\displaystyle x_{n+1}=2x_n+3y_n,\ y_{n+1}=x_n+2y_n$ is also a solution and so forth, so infinite solutions?
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    Quote Originally Posted by Proggy View Post
    so then it would be
    $\displaystyle x_1^2 - 3y_1^2 = n$ with $\displaystyle x_1 = 2x_0 +3y_0,\ y_1=x_0+2y_0$
    $\displaystyle (2x_0 + 3y_0)^2 - 3(x_0 + 2y_0)^2 = n$
    $\displaystyle (4x_0^2 + 12x_0y_0 + 9y_0^2) - 3(x_0^2 + 4x_0y_0 + 4y_0^2) = n$
    $\displaystyle x_0^2 - 3 y_0^2 = n$
    which is equivalent to $\displaystyle x_1^2 - 3y_1^2 = n$

    and then from there you say that $\displaystyle x_{n+1}=2x_n+3y_n,\ y_{n+1}=x_n+2y_n$ is also a solution and so forth, so infinite solutions?
    Yes, because every time you substitute these expressions, you get back to the initial equation !
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  5. #5
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    would changing the 3 in the equation to be a positive integer d that is not a perfect square change the outcome in a significant way?
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  6. #6
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    Quote Originally Posted by Proggy View Post
    would changing the 3 in the equation to be a positive integer d that is not a perfect square change the outcome in a significant way?
    Yes, because it's not sure you'd find an expression in terms of $\displaystyle x_n$ and $\displaystyle y_n$ for $\displaystyle x_{n+1}$ and $\displaystyle y_{n+1}$ such that it yields the initial equation !
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  7. #7
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    hmmm, just to make sure I was clear in what I was asking, the original problem would be restated as:

    Suppose $\displaystyle n$ is a fixed integer $\displaystyle \neq 0$ and d is a positive integer not a perfect square, and suppose

    $\displaystyle x_0^2 - dy_0^2 = n$

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

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

    Conclude that the quation $\displaystyle x^2-dy^2=n$ either has no solutions or infinitely many solutions.
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  8. #8
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    what I get:


    $\displaystyle x_1^2 - dy_1^2 = n$ with $\displaystyle x_1 = ax_0 + bdy_0,\ y_1=bx_0+ay_0$
    $\displaystyle (ax_0 + bdy_0)^2 - d(bx_0 + ay_0)^2 = n$
    $\displaystyle (a^2x_0^2 + abdx_0y_0 + b^2d^2y_0^2) - d(b^2x_0^2 + abx_0y_0 + a^2y_0^2) = n$
    $\displaystyle a^2x_0^2 + abdx_0y_0 + b^2d^2y_0^2 - b^2dx_0^2 - abdx_0y_0 - a^2dy_0^2 = n$
    $\displaystyle a^2(x_0^2 - dy_0^2) - b^2d(x_0^2 - dy_0^2) = n$
    $\displaystyle (a^2 - db^2)(x_0^2 - dy_0^2) = n$

    i am just not sure what to do from here
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  9. #9
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    Quote Originally Posted by Proggy View Post
    hmmm, just to make sure I was clear in what I was asking, the original problem would be restated as:

    Suppose $\displaystyle n$ is a fixed integer $\displaystyle \neq 0$ and d is a positive integer not a perfect square, and suppose

    $\displaystyle x_0^2 - dy_0^2 = n$

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

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

    Conclude that the quation $\displaystyle x^2-dy^2=n$ either has no solutions or infinitely many solutions.
    Oh ok

    Then it's exactly the same reasoning. I didn't know about the Fermat-Pell stuff
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  10. #10
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    Quote Originally Posted by Proggy View Post
    what I get:


    $\displaystyle x_1^2 - dy_1^2 = n$ with $\displaystyle x_1 = ax_0 + bdy_0,\ y_1=bx_0+ay_0$
    $\displaystyle (ax_0 + bdy_0)^2 - d(bx_0 + ay_0)^2 = n$
    $\displaystyle (a^2x_0^2 + abdx_0y_0 + b^2d^2y_0^2) - d(b^2x_0^2 + abx_0y_0 + a^2y_0^2) = n$
    $\displaystyle a^2x_0^2 + abdx_0y_0 + b^2d^2y_0^2 - b^2dx_0^2 - abdx_0y_0 - a^2dy_0^2 = n$
    $\displaystyle a^2(x_0^2 - dy_0^2) - b^2d(x_0^2 - dy_0^2) = n$
    $\displaystyle (a^2 - db^2)(x_0^2 - dy_0^2) = n$

    i am just not sure what to do from here
    heh, your comment on fermat-pell made me notice that they are saying that $\displaystyle (a^2 - db^2)=1$ so
    $\displaystyle (a^2 - db^2)(x_0^2 - dy_0^2) = n$
    becomes
    $\displaystyle 1(x_0^2 - dy_0^2) = n$
    which is the same and therefore has infinite solutions! i think?
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  11. #11
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    Quote Originally Posted by Proggy View Post
    heh, your comment on fermat-pell made me notice that they are saying that $\displaystyle (a^2 - db^2)=1$ so
    $\displaystyle (a^2 - db^2)(x_0^2 - dy_0^2) = n$
    becomes
    $\displaystyle 1(x_0^2 - dy_0^2) = n$
    which is the same and therefore has infinite solutions! i think?
    Yes !
    I thought you used it (I didn't see your resolution)

    But it's not good starting with $\displaystyle x_1^2 - dy_1^2 = n$
    Start with $\displaystyle x_1^2 - dy_1^2$ and then prove it equals n. (in fact, remove all the =n at the end of each line ^^)
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  12. #12
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    got it, thanks, and sorry for the trick question by not including the pell-fermat part. I barely noticed that extra wording in the problem and it turned out to be the deal breaker! Thanks so much for your help. One more problem to go and my homework for my number thoery class will be over.
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  13. #13
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    Another question along these lines is what would the next three ascent be in the solution if the original equation was changed to:
    x^2 - 3y^2 = 6
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