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Math Help - 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 x=x_0 > 0, y = y_0 > 0 is a solution to x^2 - 3y^2 =6

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

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

    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)

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

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

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

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

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

    x_0=3 and y_0=1

    x_1 = 2x_0 + 3y_0 and y_1= x_0 + 2y_0

    x_1=9 and y_1=5

    x_2= 2x_1 + 3y_1 and y_2= x_1 + 2y_1

    x_2=33 and y_2=19

    x_3= 2x_2 + 3y_2 and y_3= x_2+ 2y_2

    x_3=123 and y_3=71

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

    The next three solutions in the ascent are (9, 5), (33, 19), (123, 71).
    Last edited by alexmahone; July 15th 2009 at 06:50 AM.
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