The Pell Equation 2 2 2 = x Dy k

The equation 2 2 = x Dy N  , with given integers D and N and unknowns x and y , is called Pell’s equation. If D is negative, it can have only a finite number of solutions. If D is a perfect square, say 2 = D a , the equation reduces to    = x ay x ay N   and again there is only a finite number of solutions. The most interesting case of the equation arises when 1 D  be a positive non-square. Although J. Pell contributed very little to the analysis of the equation, it bears his name because of a mistake by Euler. Pell’s equation 2 2 = 1 x Dy  was solved by Lagrange in terms of simple continued fractions. Lagrange was the first to prove that 2 2 = 1 x Dy  has infinitly many solutions in integers if 1 D  is a fixed positive non-square integer. If the lenght of the periode of D is 1, all positive solutions are given by 2 1 = vk x P  and 2 1 = vk y Q  if k is odd, and by 1 = vk x P  and 1 = vk y Q  if k is


Introduction
The equation 2 2 = x Dy N  , with given integers D and N and unknowns x and y , is called Pell's equation.If D is negative, it can have only a finite number of solutions.If D is a perfect square, say 2 = D a , the equation reduces to   = x ay x ay N   and again there is only a finite number of solutions.The most interesting case of the equation arises when 1 D  be a positive non-square.
Although J. Pell contributed very little to the analysis of the equation, it bears his name because of a mistake by Euler.
Pell's equation 2 2 = 1 x Dy  was solved by Lagrange in terms of simple continued fractions.Lagrange was the first to prove that 2 2 = 1 x Dy  has infinitly many solutions in integers if 1 D  is a fixed positive non-square integer.If the lenght of the periode of D is 1, all positive solutions are given by  , are the positive solutions of 2 2 = 1 x Dy   provided that 1 is odd.
There is no solution of 2 2 = 1 x Dy   other than , : =1,2, v v x y v  given by   where 1 1 , x y is the least positive solution called the fundamental solution, which there are different method for finding it.The reader can find many references in the subject in the book [7].
For completeness we recall that there are many papers in which are considered different types of Pell's equation.when 1 D  be a positive non-square and 2 k  , we obtain some formulas for its integer solutions., and let

The Pell
for 1 n  .Then the integer solutions of the Pell equation x y , where Proof.We prove the theorem using the method of mathematical induction.For = 1 n , we have from (1), x y which is the fundamental solution of . Now, we assume that the Pell equation (3) and we show that it holds for   , n n x y .Indeed, by (1), it is easy to prove that hence we conclude that , then and Proof.By (1), we have In the other hand, we have x y satisfy the following recurrence relations Proof.The proof will be by induction on n .Using (5), we have Using ( 5) and (8), we get Then by ( 5) and (9), we find 4 x and 4 y .
Now, replacing (8) and ( 9) in (7), one obtains which are the same formulas as in (10).Therefore (7) holds for = 4 n .Now, we assume that (7) holds for 4 n  and we show that it holds for 1 n  .Indeed, by (5) and by hypothesis we have

The Negative Pell Equation
, where for 0 n  .Proof.We prove the theorem using the method of mathematical induction.For = 0 n , we have from (11), x y which is the fundamental solution of  .Now, we assume that the Pell equation and we show that it holds for 1 n  .Indeed, by (1), it is easily to seen that   Hence, by (*), we have is also a solution of the Pell equation 2 is arbitrary, we get all integer solutions of the Pell equation 2 and Proof.Using (1), we have . By (11), we have In the other hand, we have and hence x y satisfy the following recurrence relations Proof.The proof will be by induction on n .Using (14), we have Using ( 14), ( 17) and ( 18), we get Then by ( 19) and (20), we find 7 x and 7 y .
which are the same formulas as in (21).Therefore (16) holds for = 3 n Now, we assume that (16) holds for 3 n  and we show that it holds for 1 n  .Indeed, by ( 14) and by hypothesis we have completing the proof.

Acknowledgements
We would like to thank Saäd Chandoul and Massöuda Loörayed for helpful discussions and many remarks.