Diophantine Equations

• Jul 11th 2009, 09:28 AM
diddledabble
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?
• Jul 11th 2009, 10:16 AM
alexmahone
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 $(9, 5), (33, 19), (123, 71)$.