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Difference between revisions of "Pell's equation (simple solutions)"

(→‎Equation of the form x^2 – 2y^2 = 1)
(→‎Equation of the form x^2 – 2y^2 = 1)
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then <math>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,</math>
 
then <math>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,</math>
 
   
 
   
 +
therefore integers <math>(x_{i+1}, y_{i+1})</math> are the solution of the given equation.
 +
 +
If <math>i > 0</math> then <math>x_{i+1} > y_{i+1}  \ge 2(x_i + y_i) > 0.</math>
 +
<cmath>\{(x_i, y_i) \} = \{(1,0), (3,2), (17,12), (99,70),...\}.</cmath>
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<math>\boldsymbol{b.}</math> Let integers <math>(x_i, y_i)</math> are the solution, <math>\hspace{10mm}  x_i^2 - 2 y_i^2 = 1,</math>
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<cmath>\begin{equation} \left\{ \begin{aligned}
 +
  x_{i+1} &= 3 x_i - 4 y_i ,\\
 +
  y_{i+1} &= - 2 x_i + 3 y_i .
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\end{aligned} \right.\end{equation}</cmath>
 +
then <math>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,</math>
 
therefore integers <math>(x_{i+1}, y_{i+1})</math> are the solution of the given equation.
 
therefore integers <math>(x_{i+1}, y_{i+1})</math> are the solution of the given equation.
−
<cmath>\{(x_i, y_i) \} = \{(1,0), (3,2), (17,12), (99,70),...\}.</cmath>
+
 
 +
If <math>x_i > 0, y_i > 0, x_{i+1} > 0, y_{i+1} > 0 </math> then <math>x_{i+1} > y_{i+1}, x_i > y_i .</math>

Revision as of 03:46, 17 April 2023

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

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