Recurrence formula in Pell's equation

[basil]

Hello,

I am trying to solve a Pell's equation

X² + 2Y² =1

I understand that the (3,2) is a fundamental solution to the equation. However, I am having difficulty to obtain the recurrence formula in order to generate further solutions. Can anybody help?

 

[phinds]

I don't know anything about this, but I do question how you can get (3,2) as a solution to x^2 + 2*y^2 = 1, since I get 3^2 + 2*2^2 = 17 which is most emphatically not 1. What am I missing?

 

[RamaWolf]

It works with: X^2 - 2*Y^2 = 1

 

[disregardthat]

The equation is (x-sqrt(2)y)(x+sqrt(2)y) = 1. Now, your solution is (3,2), so (3-sqrt(2)2)(3+sqrt(2)2) = 1, and by multiplying we have 1 = (x-sqrt(2)y)(3-sqrt(2)2)(x+sqrt(2)y)(3+sqrt(2)2) = (3x+4y-(2x+3y)sqrt(2))(3x+4y+(2x+3y)sqrt(2)) = (3x+4y)^2-2(2x+3y)^2. Hence if we denote (3,2) by (x_0,y_0), you can define (x_{n+1},y_{n+1}) = (3x_n+4y_n,2x_n+3y_n), and thus (x_n,y_n) will be a solution for each n.

Now, we have found some solutions, but is this all? It is a good exercise to prove this.

Hint: (Suppose (x,y) is a solution not in the sequence defined above, and use the recurrence formula backwards to find the "least" positive solution. (a,b) is the "least" solution if a+b is minimal. Try to find a contradiction.)

 

[basil]

Hey,

Yep its my mistake. Its supposed to be

x²-y²=1

Thanks for pointing out.

 

[basil]

The equation is (x-sqrt(2)y)(x+sqrt(2)y) = 1. Now, your solution is (3,2), so (3-sqrt(2)2)(3+sqrt(2)2) = 1, and by multiplying we have 1 = (x-sqrt(2)y)(3-sqrt(2)2)(x+sqrt(2)y)(3+sqrt(2)2) = (3x+4y-(2x+3y)sqrt(2))(3x+4y+(2x+3y)sqrt(2)) = (3x+4y)^2-2(2x+3y)^2. Hence if we denote (3,2) by (x_0,y_0), you can define (x_{n+1},y_{n+1}) = (3x_n+4y_n,2x_n+3y_n), and thus (x_n,y_n) will be a solution for each n.

Now, we have found some solutions, but is this all? It is a good exercise to prove this.

Hint: (Suppose (x,y) is a solution not in the sequence defined above, and use the recurrence formula backwards to find the "least" positive solution. (a,b) is the "least" solution if a+b is minimal. Try to find a contradiction.)

Awesome. Thanks heaps.

 

[RamaWolf]

Let x_{1}, y_{1} be basic solutions to

x^2-N*y^2 = 1, with N a non-square integer number

then we have the following recurrence relations for k>1:

x_{k+1}:=x_{1}*x_{k}+N*y_{1}*y_{k}
y_{k+1}:=x_{1}*y_{k}+y_{1}*x_{k}

For N=2, (2,3) being the basic solution, we have (17,12) as 2. and (99,70) as 3.solution

A trivial algorithm to find the basic solution could be:

y:=1, max:=999
Do While n < max
\cdotsCompute w=1+N*y^2
\cdotsIf w is an integral square x^2 then
\cdots\cdotsReturn "Basic solution: (x, y)" and Stop
\cdotsEnd_If
\cdotsIncr y
End_Do
Return max & " trials and no solution" and Stop

Because there are values of N, whrer this "Q&D"-algorithm fails,
who has experience with with non-trivial algorithms for basic solutions
to the Pell equation?