View source for Pell's equation (simple solutions)

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

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

During the solution we need:

a) to construct a recurrent sequence <math>\{x_{i+1}, y_{i+1} \} = f({x_i, y_i});</math>

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

==Equation of the form <math>x^2 - 2y^2 = 1</math>==
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\} = \{1,0\}.</math>

<i><b>Proof</b></i>

<math>\boldsymbol{a.}</math> Let integers <math>(x_i, y_i)</math> are the solution of the equation <math>\hspace{10mm}   x_i^2 - 2 y_i^2 = 1,</math>

<cmath>\[ \begin{cases}
 x_{i+1} &= 3 x_i + 4 y_i ,\\
y_{i+1} &= 2 x_i + 3 y_i .
  \end{cases} \]</cmath>
Then <cmath>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.</cmath>
 
Therefore integers <math>(x_{i+1}, y_{i+1})</math> are the solution of the given equation. If <math>i > 0</math> then <cmath>x_{i+1} > y_{i+1}  \ge 2(x_i + y_i) > x_i > y_i > 0.</cmath>
<cmath>\{(x_i, y_i) \} = \{(1,0), (3,2), (17,12), (99,70), (577,408), (3363, 2738),...\}.</cmath>

<math>\boldsymbol{b.}</math> Suppose that the pare of the positive integers <math>(x_j, y_j)</math> is the solution different from founded in <math>\boldsymbol{a.}</math>

<math>x_j^2 - 2 y_j^2 = 1.</math> Let

<cmath>\[ \begin{cases}
 x_{i+1} &= 3 x_i - 4 y_i ,\\
y_{i+1} &= - 2 x_i + 3 y_i .
  \end{cases} \]</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.

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

There is no integer solution if <math> y_j = 1. y_j = 0</math> is impossible. So <math>y_i > 1.</math>
<cmath>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.</cmath>

<cmath> 0 \le y_{i+1} = y_i - 2 (x_i - y_i) < y_i. </cmath>

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

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

<i><b>Proof</b></i>

<cmath>x(x + 1) = 2y^2 \implies 4x^2 + 4x + 1 = 8y^2 + 1 \implies (2x+1)^2 - 2(2y)^2 = 1.</cmath>
It is the form of Pell's equation, therefore
<cmath>\[ \begin{cases}
2x_{i+1}+1 &= 3 (2x_i + 1)+ 4 (2y_i) ,\\
2y_{i+1} &= 2 (2x_i + 1) +3 (2y_i) .
  \end{cases} \]</cmath>
<cmath>\[ \begin{cases}
x_{i+1} &= 3x_i + 4y_i + 1,\\
y_{i+1} &= 2x_i + 3y_i + 1.
  \end{cases} \]</cmath>

<cmath>\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}</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>

'''vladimir.shelomovskii@gmail.com, vvsss'''

==Equation of the form <math>x^2 - 2y^2 = - 1</math>==
Prove that all positive integer solutions of the equation <math>x^2 - 2y^2 = -1</math> can be found using recursively transformation
<cmath>\[ \begin{cases}
 x_{i+1} &= 3 x_i + 4 y_i ,\\
y_{i+1} &= 2 x_i + 3 y_i .
  \end{cases} \]</cmath>
of the pare <math>\{x_0, y_0\} = \{1,1\}.</math>

<i><b>Proof</b></i>

The solution is similar to that discussed in the previous section.

<cmath>\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}</cmath>

'''vladimir.shelomovskii@gmail.com, vvsss'''

==Pythagorean triangles with almost equal legs==
Find all triangles with integer sides one leg of which is <math>1</math> more than the other.

Find all natural solutions of the equation <math>x^2 + (x+1)^2 = y^2.</math>

<i><b>Solution</b></i>

<cmath>x^2 +(x + 1)^2 = 2x^2 + 2x + 1 = y^2 \implies 4x^2 + 4x + 1 = 2y^2 - 1 \implies (2x+1)^2 - 2y^2 = -1.</cmath>
All positive integer solutions of the equation <math>(2x+1)^2 - 2y^2 = -1</math> can be found using recursively transformation
<cmath>\[ \begin{cases}  2x_{i+1}+1 &= 3 (2x_i + 1) + 4 y_i ,\\
y_{i+1} &= 2(2x_i + 1) + 3 y_i .  \end{cases} \]</cmath>
<cmath>\[ \begin{cases} x_{i+1} &= 3x_i + 2y_i + 1 ,\\
y_{i+1} &= 4x_i + 3y_i + 2 . \end{cases} \]</cmath>
of the pare <math>\{x_0, y_0\} = \{0,1\}.</math>

<cmath>\begin{array}{c|c|c|c|c|c|c} 
& & & & & & \\ [-2ex]
\boldsymbol{i} & 0 & 1 & 2 & 3 & 4 & 5 \\ [0.5ex] \hline  
& & & & & & \\ [-1.5ex] 
\boldsymbol{x_i} & 0 & 3 & 20 & 119 & 696 & 4059 \\ [1ex]     
\boldsymbol{y_i} & 1 & 5 & 29 & 169 & 985 & 5741 \\ [1ex]    
\end{array}</cmath>

'''vladimir.shelomovskii@gmail.com, vvsss'''

==Equation of the form <math>x^2 - 3y^2 = - 1</math>==
Prove that the equation <math>x^2 - 3y^2 = -1</math> have not any solution.

<i><b>Proof</b></i>

<cmath>x = 3k \implies x^2 \pmod{3}= 0.</cmath>
<cmath>x = 3k \pm 1 \implies x^2 \pmod{3}= 1.</cmath>
<cmath>(- 3y^2 + 1) \pmod {3} = 1 \implies (x^2 - 3y^2 + 1) \pmod {3} = \{1, 2 \} \ne 0 \implies</cmath>
<cmath>x^2 - 3y^2 + 1 \ne 0.</cmath>
'''vladimir.shelomovskii@gmail.com, vvsss'''

==Equation of the form <math>x^2 - 2y^2 = 7</math>==
Prove that all positive integer solutions of the equation <math>x^2 - 2y^2 = 7</math> can be found using recursively transformation
<cmath>\[ \begin{cases} x_{i+1} &= 3 x_i + 4 y_i ,\\
y_{i+1} &= 2 x_i + 3 y_i . \end{cases} \]</cmath>
of the pares <math>\{x_1, y_1\} = \{3,1\}</math> and <math>\{\hat{x}_1, \hat{y}_1 \} = \{5,3\}</math>.

<i><b>Proof</b></i>

<math>\boldsymbol{a.}</math> Let integers <math>(x_i, y_i)</math> are the solution of the equation <math>\hspace{10mm}   x_i^2 - 2 y_i^2 = 7,</math>
<cmath>\[ \begin{cases} x_{i+1} &= 3 x_i + 4 y_i ,\\
y_{i+1} &= 2 x_i + 3 y_i . \end{cases} \]</cmath>

Then <cmath>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 = 7.</cmath>
 
Therefore integers <math>(x_{i+1}, y_{i+1})</math> are the solution of the given equation.
<cmath>x_{i+1} > y_{i+1}  \ge 2(x_i + y_i) > x_i > y_i > 0.</cmath>
<cmath>\begin{array}{c|c|c|c|c|c} 
& & & & & \\ [-2ex]
\boldsymbol{i} & 1 & 2 & 3 & 4 & 5 \\ [0.5ex] \hline  
& & & & & \\ [-1.5ex] 
\boldsymbol{x_i} & 3 & 13 & 75 & 437 & 2547 \\ [1ex]     
\boldsymbol{y_i} & 1 & 9 & 53 & 309 & 1801 \\ [1ex]    
\end{array}</cmath>
<math>\boldsymbol{b.}</math> The pare of the positive integers <math>(\hat {x}_i, \hat{y}_i)</math> is the solution different from founded in <math>\boldsymbol{a.}</math>

<math>\hat {x}_i^2 - 2 \hat {y}_i^2 = 7.</math> Let

<cmath>\[ \begin{cases} \hat{x}_{i+1} &= 3  \hat{x}_i - 4  \hat{y}_i ,\\
\hat{y}_{i+1} &= - 2  \hat{x}_i + 3  \hat{y}_i . \end{cases} \]</cmath>

then <cmath>\hat{x}_{i+1}^2 - 2  \hat{y}_{i+1}^2 = (3  \hat{x}_i - 4  \hat{y}_i)^2 - 2 (-2  \hat{x}_i + 3  \hat{y}_i)^2 =   \hat{x}_i^2 - 2  \hat{y}_i^2 = 7,</cmath>
therefore integers <math>( \hat{x}_{i+1},  \hat{y}_{i+1})</math> are the solution of the given equation.

<math>\hat{x}_i^2 = 2 \hat{y}_i^2 +1 >  2 \hat{y}_i^2 \implies \hat{x}_i > \hat{y}_i > 0.</math> Similarly <math>\hat{x}_{i +1} > \hat{y}_{i+1}.</math>

If <math> \hat{y}_i > 5</math> then
<cmath>9 \hat{y}_i^2 \ge 8 \hat{y}_i^2 + 28  = 4 \hat{x}_i^2 \implies 3 \hat{y}_i \ge 2 \hat{x}_i  \implies \hat{y}_{i+1} \ge 0.</cmath>

<cmath> 0 \le \hat{y}_{i+1} = \hat{y}_i - 2 (\hat{x}_i – \hat{y}_i) <\hat{y}_i. </cmath>

There is no member <math>\hat{y}_j  = 0</math> in the sequence <math>\{\hat{y}_i \},</math> hence it is infinitely decreasing sequence of natural numbers. There is no such sequence. Contradiction.

We need to check <math>\hat{y}_j = \{2,4,5\}</math> (no solution), but <math>\hat{y}_j = 3</math> gives the integer solution, so there is the second sequence of the integer solutions
<cmath>\begin{array}{c|c|c|c|c|c} 
& & & & & \\ [-2ex]
\boldsymbol{i} & 1 & 2 & 3 & 4 & 5 \\ [0.5ex] \hline  
& & & & & \\ [-1.5ex] 
\boldsymbol{\hat{x}_i} & 5 & 27 & 157 & 915 & 5333 \\ [1ex]     
\boldsymbol{\hat{y}_i} & 3 & 19 & 111 & 647 & 3771 \\ [1ex]    
\end{array}</cmath>
'''vladimir.shelomovskii@gmail.com, vvsss'''

==Equation of the form <math>x^2 - xy - y^2 = 1</math>==
Prove that all positive integer solutions of the equation <math>x^2 - xy - y^2 = 1</math> are <math>\{x_i, y_i \} = \{F_{2i +1}, F_{2i}\}.</math> They can be found using recursively transformation <math>y_{i+1} = x_i + y_i , x_{i+1} = 2x_i + y_i = y_{i+1} + x_i </math> of the pares <math>\{x_0, y_0\} = \{1,0\}.</math> 

<i><b>Proof</b></i>
<cmath>x^2 - xy - y^2 = 1 \implies 4x^2 - 4xy + y^2 - 5y^2 = 1 \implies (2x+y)^2 - 5y^2 = 1.</cmath>
Let the pare of the positive integers <math>(x_i, y_i)</math> be the solution of given equation <math> x_i^2 - x_i y_i - y_i^2 = 1</math> and  
<cmath>\[ \begin{cases} x_{i+1} &= 2x_i + y_i ,\\
y_{i+1} &= x_i + y_i .  \end{cases} \]</cmath>
Then
<cmath>x_{i + 1}^2 - x_{i + 1} y_{i + 1} - y_{i + 1}^2 = x_i^2 - x_i y_i - y_i^2 = 1.</cmath>
<cmath>\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} & 1 & 2 & 5 & 13 & 34 & 89 & 233 & 610 \\ [1ex]     
\boldsymbol{y_i} & 0 & 1 & 3 & 8 & 21 & 55 & 144 & 377 \\ [1ex]    
\end{array}</cmath>

'''vladimir.shelomovskii@gmail.com, vvsss'''

==Equation of the form <math>x^2 - xy - y^2 = -1</math>==
Prove that all positive integer solutions of the equation <math>x^2 - xy - y^2 = -1</math> are <math>\{x_i, y_i \} = \{F_{2i}, F_{2i-1}\}.</math> They can be found using recursively transformation
<cmath>\[ \begin{cases} x_{i+1} &= 2x_i + y_i = y_{i+1} + x_i,\\
y_{i+1} &= x_i + y_i .  \end{cases} \]</cmath>
of the pares <math>\{x_0, y_0\} = \{1,1\} = \{F_1, F_2 \}.</math> 
<cmath>\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} & 1 & 3 & 8 & 21 & 55 & 144 & 377 & 987 \\ [1ex]     
\boldsymbol{y_i} & 1 & 2 & 5 & 13 & 34 & 89 & 233 & 610 \\ [1ex]    
\end{array}</cmath>

'''vladimir.shelomovskii@gmail.com, vvsss'''

==Equation of the form <math>x^2 - 5y^2 = 4</math>==
Prove that all positive integer solutions of the equation <math>x^2 - 5y^2 = 4</math> are <cmath>\{x_i, y_i \} = \{F_{2i+1} + F_{2i - 1}, F_{2i}\}.</cmath>
It is clear that solutions can be found using recursively transformation
<cmath>\[ \begin{cases} x_{i+1} = \frac {3x_i }{2}+ \frac {5y_i }{2},\\
y_{i+1} = \frac {x_i }{2}+ \frac {3y_i }{2}  \end{cases} \]</cmath> 
of the pare <math>\{x_1, y_1\} = \{3, 1\} = \{F_3 +F_1, F_2\}.</math>
 
One can use the small transform for understanding 
<cmath>x_{i + 1} - y_{i + 1} = x_i + y_i = (x_i - y_i) + 2y_i  \Rightarrow 2F_{2i+1} = 2F_{2i-1} + 2F_{2i}, x_1-y_1 = 2F_1.</cmath>
<cmath>y_{i + 1} = \frac {x_i - y_i}{2}+ 2 y_i   \Rightarrow    F_{2i +2} = F_{2i-1} + 2F_{2i} =  F_{2i+1} + F_{2i}, y_1 = F_2.</cmath>
<cmath>\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} & 3 & 7 & 18 & 47 & 123 & 322 & 377 & 843 \\ [1ex]     
\boldsymbol{y_i} & 1 & 3 & 8 & 21 & 55 & 89 & 144 & 377 \\ [1ex]    
\end{array}</cmath>

'''vladimir.shelomovskii@gmail.com, vvsss'''

==Equation for binomial coefficients==
Find all positive integer solutions of equation  <math>\binom{x}{y-1} = \binom{x-1}{y}.</math>

<i><b>Solution</b></i>
<cmath>\binom{x}{y-1} = \frac{x!}{(y-1)!(x-y+1)!}, \binom{x-1}{y} = \frac{(x-1)!}{y!(x-y-1)!} \implies </cmath>
<cmath>xy=(x-y+1) \cdot (x-y) \implies x^2 -3xy +y^2 +x - y = 0.</cmath>
It is known that every real quadratic form under a linear change of variables may be transformed in a "diagonal form".
<cmath>(x + \frac {1}{5})^2 + (y - \frac {1}{5})^2 - 3 (x + \frac {1}{5})(y - \frac {1}{5})^2 = \frac {1}{5}</cmath>
<cmath>(5y - 1)^2 - 5 (2x - 3y + 1)^2 = - 4</cmath>
<cmath>u = 5y - 1, v = 2x - 3y + 1 \implies u^2 - 5 v^2 = - 4 \implies</cmath>
<cmath>u_i = 5y_i - 1 = F_{2i} + F_{2i - 2},  v_i = 2x_i - 3y_i + 1 = F_{2i-1}.</cmath>
<cmath>y_i = \frac {F_{2i} + F_{2i - 2} + 1}{5} = \frac {L_{2i - 1} + 1}{5},</cmath> where L is Lucas number,
<math>L_0 = 2, L_1 = 1, L_i = F_{i-1} + F_{i+1} = L_i + L_{i - 1}.</math>
It is clear that 
<cmath>L_1 \pmod 5 = 1, L_2 \pmod 5 = 3 \pmod 5 = 3,</cmath>
<cmath>L_3 \pmod 5 = (L_1 \pmod 5 + L_2 \pmod 5 ) \pmod 5 = 4 \pmod 5 = 4,</cmath>
<cmath>L_4 \pmod 5 = (L_2 \pmod 5 + L_3 \pmod 5 ) \pmod 5 = 2,</cmath>
<cmath>L_5 \pmod 5 = (L_3 \pmod 5 + L_4 \pmod 5 ) \pmod 5 = 1 = L_1 \pmod 5,</cmath>
<cmath>L_6 \pmod 5 = (L_4 \pmod 5 + L_5 \pmod 5 ) \pmod 5 = 3 = L_2 \pmod 5,</cmath> 
The sequence of Lucas numbers modulo 5 is periodic, the period is 4, and the numbers with index 4i-1 are 4 modulo 5.

Therefore numbers <cmath>y_i = \frac {F_{4i} + F_{4i – 2} + 1}{5}</cmath> are integer.

We using <cmath>F_{2i-1}^2- F_{2i} \cdot F_{2i-2} = 1, F_{4i} = F_{2i} (F_{2i+1} + F_{2i-1}),
F_{4i – 2} = F_{2i-1}(2F_{2i} - F_{2i-1})</cmath> and get  <cmath>y_i = F_{2i} F_{2i-1}.</cmath>

We using <cmath>F_{4i-1}= F_{2i}^2 + F_{2i-1}^2, F_{2i-1}^2 - F_{2i} \cdot F{2i-2} = 1,</cmath>
and get 
<cmath>x_i = \frac{ 3y_i - 1 + F_{4i-1}}{2} =  \frac{ F_{2i}(3F_{2i-1} + F_{2i} +F_{2i-2})}{2} =  F_{2i} \cdot F_{2i+1}.</cmath>

<cmath>\begin{array}{c|c|c|c|c|c|c|c} 
& & & & & & & \\ [-2ex]
\boldsymbol{i} & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ [0.5ex] \hline  
& & & & & & & \\ [-1.5ex] 
\boldsymbol{u_i} & 1 & 4 & 11 & 29 & 76 & 199 & 521  \\ [1ex]     
\boldsymbol{v_i} & 1 & 2 & 5 & 13 & 34 & 89 & 233 \\ [1ex]    
\boldsymbol{x_i} & 2 & 15 & 104 & 714 & 4895 & 33552 & 229970 \\ [1ex]     
\boldsymbol{y_i} & 1 & 6 & 40 & 273 & 1870 & 12816 & 87841 \\ [1ex]    
\end{array}</cmath>
<cmath>\binom{15}{5} = \binom{14}{6} =  3003</cmath>
<cmath>\binom{104}{39} = \binom{103}{40} =  61218182743304701891431482520</cmath>

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==Equation of the form==