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

(→‎Equation of the form x(x + 1) = 2y^2)
(→‎Equation of the form x(x + 1) = 2y^2)
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<cmath>x_{i+1} = 3 x_i + 4 y_i + 1 = 2x_i + 2 \sqrt {2 x_i (x_i + 1)} + (x_i + 1) = (\sqrt {2x_i} + \sqrt{x_i + 1})^2</cmath>
 
<cmath>x_{i+1} = 3 x_i + 4 y_i + 1 = 2x_i + 2 \sqrt {2 x_i (x_i + 1)} + (x_i + 1) = (\sqrt {2x_i} + \sqrt{x_i + 1})^2</cmath>
 
<cmath>\implies \sqrt{x_{i+1}}= \sqrt{2x_i} + \sqrt{x_i + 1}.</cmath>
 
<cmath>\implies \sqrt{x_{i+1}}= \sqrt{2x_i} + \sqrt{x_i + 1}.</cmath>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==Equation of the form <math>x^2 – 2y^2 = - 1</math>==
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Prove that all positive integer solutions of the equation <math>x^2 – 2y^2 = -1</math> can be found using recursively transformation <math>x_{i+1} = 3 x_i + 4 y_i , y_{i+1} = 2 x_i + 3 y_i </math> of the pare <math>\{x_0, y_0\} = \{1,1\}.</math>
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<i><b>Proof</b></i>
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Similarly as for equation <math>x^2 – 2y^2 = 1.</math>
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<cmath>\begin{array}{c|c|c|c|c|c|c|c}
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& & & & & & & \\ [-2ex]
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\boldsymbol{i} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ [0.5ex] \hline 
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& & & & & & & \\ [-1.5ex]
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\boldsymbol{x_i} & 1 & 7 & 41 & 239 & 1393 & 8119 & 47321 \\ [1ex]   
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\boldsymbol{y_i} & 1 & 5 & 29 & 169 & 985 & 5741 & 33461 \\ [1ex]   
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\end{array}</cmath>
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 06:09, 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 $x^2 – 2y^2 = 1$

Prove that all positive integer solutions of the equation $x^2 – 2y^2 = 1$ can be found using recursively transformation $x_{i+1} = 3 x_i + 4 y_i , y_{i+1} = 2 x_i + 3 y_i$ of the pare $\{x_0, y_0\} = \{1,0\}.$

Proof

$\boldsymbol{a.}$ Let integers $(x_i, y_i)$ are the solution of the equation $\hspace{10mm} x_i^2 - 2 y_i^2 = 1,$ \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 \[x_{i+1} > y_{i+1} \ge 2(x_i + y_i) > x_i > y_i > 0.\] \[\{(x_i, y_i) \} = \{(1,0), (3,2), (17,12), (99,70),...\}.\]

$\boldsymbol{b.}$ Suppose that the pare of the positive integers $(x_I, y_I)$ is the solution different from founded in $\boldsymbol{a.}\hspace{10mm} x_I^2 - 2 y_I^2 = 1.$ Let \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.

$x_i^2 = 2 y_i^2 +1 > 2 y_i^2 \implies x_i > y_i >0.$ Similarly $x_{i +1}> y_{i+1}.$

There is no integer solution if $y_j = 1. y_j = 0$ is impossible. So $y_i > 1.$ \[9 y_i^2 \ge 8 y_i^2 + 4 = 4 x_i^2 \implies 3y_i \ge 2x_i \implies y_{i+1} \ge 0.\]

\[0 \le y_{i+1} = y_i - 2 (x_i – y_i) < y_i.\]

There is no member $y_j = 0$ in the sequence $\{y_i \},$ hence it is infinitely decreasing sequence of natural numbers. There is no such sequence. Contradiction.

vladimir.shelomovskii@gmail.com, vvsss

Equation of the form $x(x + 1) = 2y^2$

Prove that all positive integer solutions of the equation $x(x + 1) = 2y^2$ can be found using recursively transformation $x_{i+1} = 3 x_i + 4 y_i + 1, y_{i+1} = 2 x_i + 3 y_i + 1$ of the pare $\{x_0, y_0\} = \{0,0\}.$ In another form $\sqrt{x_{i+1}} = \sqrt{2x_i} + \sqrt{x_i + 1}.$

Proof

\[x(x + 1) = 2y^2 \implies 4x^2 + 4x + 1 = 8y^2 + 1 \implies (2x+1)^2 – 2(2y)^2 = 1.\] It is the form of Pell's equation, therefore \begin{equation} \left\{ \begin{aligned} 2x_{i+1} + 1 &= 3(2x_i + 1) + 4(2 y_i) ,\\ 2y_{i+1} &= 2(2 x_i + 1) + 3(2 y_i) . \end{aligned} \right. \implies \left\{ \begin{aligned} x_{i+1} &= 3 x_i + 4 y_i + 1 ,\\ y_{i+1} &= 2 x_i + 3 y_i + 1 . \end{aligned} \right. \end{equation}


\[\begin{array}{c|c|c|c|c|c|c|c|c} & & & & & & & & \\ [-2ex] \boldsymbol{i} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ [0.5ex] \hline & & & & & & & & \\ [-1.5ex] \boldsymbol{x_i} & 0 & 1 & 8 & 49 & 288 & 1681 & 9800 & 57121 \\ [1ex] \boldsymbol{y_i} & 0 & 1 & 6 & 35 & 204 & 1189 & 1681 & 40391 \\ [1ex] \end{array}\] \[x_{i+1} = 3 x_i + 4 y_i + 1 = 2x_i + 2 \sqrt {2 x_i (x_i + 1)} + (x_i + 1) = (\sqrt {2x_i} + \sqrt{x_i + 1})^2\] \[\implies \sqrt{x_{i+1}}= \sqrt{2x_i} + \sqrt{x_i + 1}.\]

vladimir.shelomovskii@gmail.com, vvsss

Equation of the form $x^2 – 2y^2 = - 1$

Prove that all positive integer solutions of the equation $x^2 – 2y^2 = -1$ can be found using recursively transformation $x_{i+1} = 3 x_i + 4 y_i , y_{i+1} = 2 x_i + 3 y_i$ of the pare $\{x_0, y_0\} = \{1,1\}.$

Proof Similarly as for equation $x^2 – 2y^2 = 1.$

\[\begin{array}{c|c|c|c|c|c|c|c} & & & & & & & \\ [-2ex] \boldsymbol{i} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ [0.5ex] \hline & & & & & & & \\ [-1.5ex] \boldsymbol{x_i} & 1 & 7 & 41 & 239 & 1393 & 8119 & 47321 \\ [1ex] \boldsymbol{y_i} & 1 & 5 & 29 & 169 & 985 & 5741 & 33461 \\ [1ex] \end{array}\]

vladimir.shelomovskii@gmail.com, vvsss

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