# 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 \$\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\$

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

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\$\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\$

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The next three solutions in the ascent are \$\displaystyle (9, 5), (33, 19), (123, 71)\$.