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Thread: Diophantine Equations

  1. #1
    Member diddledabble's Avatar
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    Lightbulb Diophantine Equations

    Suppose the x=x0 > 0, y = y0 > 0 is a solution to x^2 - 3y^2 =6
    Show that x1 = 2x0 + 3y0, y1= x0 + 2y0 is also a solution with x1 > x0 > 0, y1 > y0 >0. Can you find one solution to x^2 - 3y^2 =6 that results in there being infinitely many solutions? What would the next three solutions be in the ascent?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Suppose that $\displaystyle x=x_0 > 0, y = y_0 > 0$ is a solution to $\displaystyle x^2 - 3y^2 =6$

    Then, $\displaystyle x_0^2 - 3y_0^2 =6$--------------(1)

    $\displaystyle x_1^2-3y_1^2=(2x_0+3y_0)^2-3(x_0+2y_0)^2$

    $\displaystyle x_1^2-3y_1^2=(4x_0^2+9y_0^2+12x_0y_0)-3(x_0^2+4y_0^2+4x_0y_0)$

    $\displaystyle x_1^2-3y_1^2=x_0^2-3y_0^2=6$ {Using (1)}

    Thus $\displaystyle x_1 = 2x_0 + 3y_0, y_1= x_0 + 2y_0$ is also a solution with $\displaystyle x_1 > x_0 > 0,\ y_1 > y_0 >0$

    --------------------------------------------------------------------------------------------------------------------

    $\displaystyle (3, 1)$ is a solution to $\displaystyle x^2 - 3y^2 =6$

    --------------------------------------------------------------------------------------------------------------------

    $\displaystyle x_0=3$ and $\displaystyle y_0=1$

    $\displaystyle x_1 = 2x_0 + 3y_0$ and $\displaystyle y_1= x_0 + 2y_0$

    $\displaystyle x_1=9$ and $\displaystyle y_1=5$

    $\displaystyle x_2= 2x_1 + 3y_1$ and $\displaystyle y_2= x_1 + 2y_1$

    $\displaystyle x_2=33$ and $\displaystyle y_2=19$

    $\displaystyle x_3= 2x_2 + 3y_2$ and $\displaystyle y_3= x_2+ 2y_2$

    $\displaystyle x_3=123$ and $\displaystyle y_3=71$

    --------------------------------------------------------------------------------------------------------------------

    The next three solutions in the ascent are $\displaystyle (9, 5), (33, 19), (123, 71)$.
    Last edited by alexmahone; Jul 15th 2009 at 05:50 AM.
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