# Pell's Equations

• May 13th 2009, 10:40 PM
Aquafina
Pell's Equations
Hi I haven't done Pell's Equations but i did read up the method to solve them. However, I'm not sure how i can show that these 2 equations each have infinite integers solutions for x and y.

http://mathworld.wolfram.com/images/...dEquation2.gif

and

http://mathworld.wolfram.com/images/...dEquation5.gif
• May 14th 2009, 12:04 AM
Opalg
Quote:

Originally Posted by Aquafina
Hi I haven't done Pell's Equations but i did read up the method to solve them. However, I'm not sure how i can show that these 2 equations each have infinite integers solutions for x and y.

http://mathworld.wolfram.com/images/...dEquation2.gif

and

http://mathworld.wolfram.com/images/...dEquation5.gif

The idea is to show that given any solution, you can always find another solution where the numbers are bigger. More precisely, if (x,y) is a solution then there are positive integers a,b,c,d such that (ax+by,cx+dy) is a solution. The theory of Pell's equation provides systematic ways of finding these constants. But if you don't know the theory, you can often guess them by looking at the first few solutions.

For the equation \$\displaystyle x^2-2y^2=1\$, the smallest solution is x=3, y=2. The next one is x=17, y=12. From that, you might guess that if (x,y) is a solution then so is (3x+4y,2x+3y). You can then confirm that conjecture by verifying that \$\displaystyle (3x+4y)^2 - 2(2x+3y)^2 = x^2-2y^2\$.

Similarly for the other equation: the first two solutions are (5,2) and (49,20). If (x,y) is a solution then so is (5x+12y,2x+5y), because \$\displaystyle (5x+12y)^2 - 6(2x+5y)^2 = x^2-6y^2\$.
• May 14th 2009, 12:16 AM
Aquafina
Thank you :)

I have a similair question with linear Diophantine equations if you have the time to look at it:

its on the thread Sum of Unknown..

http://www.mathhelpforum.com/math-he...tml#post316217