Pell's equation (simple solutions)

Revision as of 02:46, 17 April 2023 by Vvsss (talk | contribs) (Equation of the form x^2 – 2y^2 = 1)

Pell's equation is any Diophantine equation of the form $x^2 – Dy^2 = 1,$ where $D$ is a given positive nonsquare integer, and integer solutions are sought for $x$ and $y.$

Denote the sequence of solutions $\{x_i, y_i \}.$ It is clear that $\{x_0, y_0 \} = \{1,0 \}.$

During the solution we need:

a) to construct a recurrent sequence $\{x_{i+1}, y_{i+1} \} = f({x_i, y_i})$ or two sequences $\{x_{i+1} \} = f({x_i}), \{y_{ i+1} \} = g({y_i});$

b) to prove that the equation has no other integer solutions.

Equation of the form $x^2 – 2y^2 = 1$

$\boldsymbol{a.}$ Let integers $(x_i, y_i)$ are the solution, $\hspace{10mm}   x_i^2 - 2 y_i^2 = 1,$ \begin{equation} \left\{ \begin{aligned}    x_{i+1} &= 3 x_i + 4 y_i ,\\   y_{i+1} &= 2 x_i + 3 y_i . \end{aligned} \right.\end{equation} then $x_{i+1}^2 - 2 y_{i+1}^2 = (3 x_i + 4 y_i)^2 - 2 (2 x_i + 3 y_i)^2 =  x_i^2 - 2 y_i^2 = 1,$

therefore integers $(x_{i+1}, y_{i+1})$ are the solution of the given equation.

If $i > 0$ then $x_{i+1} > y_{i+1}  \ge 2(x_i + y_i) > 0.$ \[\{(x_i, y_i) \} = \{(1,0), (3,2), (17,12), (99,70),...\}.\] $\boldsymbol{b.}$ Let integers $(x_i, y_i)$ are the solution, $\hspace{10mm}   x_i^2 - 2 y_i^2 = 1,$ \begin{equation} \left\{ \begin{aligned}    x_{i+1} &= 3 x_i - 4 y_i ,\\   y_{i+1} &= - 2 x_i + 3 y_i . \end{aligned} \right.\end{equation} then $x_{i+1}^2 - 2 y_{i+1}^2 = (3 x_i - 4 y_i)^2 - 2 (-2 x_i + 3 y_i)^2 =  x_i^2 - 2 y_i^2 = 1,$ therefore integers $(x_{i+1}, y_{i+1})$ are the solution of the given equation.

If $x_i > 0, y_i > 0, x_{i+1} > 0, y_{i+1} > 0$ then $x_{i+1} > y_{i+1}, x_i > y_i .$