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For $d=2$ or $3$ the continuous fractions for $\sqrt 2$ and $\sqrt 3$ help (at least partially), but for $d\ge 5$, this doesn't seem to work.

Is there any known solution or technique to solve this equation?

ajotatxe
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1 Answers1

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here is a good one, several orbits, your $d=13.$ There are six orbits, with $w_n^2 - 13 v_n^2 = -12,$ and recurrences $$ w_{n+12} = 1298 w_{n+6} - w_n \; , \; $$ $$ v_{n+12} = 1298 v_{n+6} - v_n \; . \; $$

The orbit we knew about begins with $w$ in $$ 1, 2989, 3879721, 5035874869, $$ and $v$ in $$ 1, 829, 1076041, 1396700389, $$

The next (out of six) orbit begins $$ 14, 18446, 23942894, $$ $$ 4, 5116, 6640564, $$

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:  
    649   2340
    180   649
  Automorphism backwards:  
    649   -2340
    -180   649

  649^2 - 13 180^2 = 1

 w^2 - 13 v^2 = -12 =   -1 * 2^2 3

Thu May  9 12:08:38 PDT 2019

w:  1  v:  1  SEED   KEEP +- 
w:  14  v:  4  SEED   KEEP +- 
w:  25  v:  7  SEED   KEEP +- 
w:  155  v:  43  SEED   BACK ONE STEP  -25 ,  7
w:  274  v:  76  SEED   BACK ONE STEP  -14 ,  4
w:  1691  v:  469  SEED   BACK ONE STEP  -1 ,  1

w:  2989  v:  829
w:  18446  v:  5116
w:  32605  v:  9043
w:  201215  v:  55807
w:  355666  v:  98644
w:  2194919  v:  608761

w:  3879721  v:  1076041
w:  23942894  v:  6640564

Thu May  9 12:08:53 PDT 2019

 w^2 - 13 v^2 = -12 =   -1 * 2^2 3

well, there are several details.

jagy@phobeusjunior:~$ ./Pell_Lubin_tableaux  13

Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$

$$ \sqrt { 13} = 3 + \frac{ \sqrt {13} - 3 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {13} - 3 } = \frac{ \sqrt {13} + 3 }{4 } = 1 + \frac{ \sqrt {13} - 1 }{4 } $$ $$ \frac{ 4 }{ \sqrt {13} - 1 } = \frac{ \sqrt {13} + 1 }{3 } = 1 + \frac{ \sqrt {13} - 2 }{3 } $$ $$ \frac{ 3 }{ \sqrt {13} - 2 } = \frac{ \sqrt {13} + 2 }{3 } = 1 + \frac{ \sqrt {13} - 1 }{3 } $$ $$ \frac{ 3 }{ \sqrt {13} - 1 } = \frac{ \sqrt {13} + 1 }{4 } = 1 + \frac{ \sqrt {13} - 3 }{4 } $$ $$ \frac{ 4 }{ \sqrt {13} - 3 } = \frac{ \sqrt {13} + 3 }{1 } = 6 + \frac{ \sqrt {13} - 3 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccccccccc} & & 3 & & 1 & & 1 & & 1 & & 1 & & 6 & & 1 & & 1 & & 1 & & 1 & & 6 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 3 }{ 1 } & & \frac{ 4 }{ 1 } & & \frac{ 7 }{ 2 } & & \frac{ 11 }{ 3 } & & \frac{ 18 }{ 5 } & & \frac{ 119 }{ 33 } & & \frac{ 137 }{ 38 } & & \frac{ 256 }{ 71 } & & \frac{ 393 }{ 109 } & & \frac{ 649 }{ 180 } \\ \\ & 1 & & -4 & & 3 & & -3 & & 4 & & -1 & & 4 & & -3 & & 3 & & -4 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 13 \cdot 0^2 = 1 & \mbox{digit} & 3 \\ \frac{ 3 }{ 1 } & 3^2 - 13 \cdot 1^2 = -4 & \mbox{digit} & 1 \\ \frac{ 4 }{ 1 } & 4^2 - 13 \cdot 1^2 = 3 & \mbox{digit} & 1 \\ \frac{ 7 }{ 2 } & 7^2 - 13 \cdot 2^2 = -3 & \mbox{digit} & 1 \\ \frac{ 11 }{ 3 } & 11^2 - 13 \cdot 3^2 = 4 & \mbox{digit} & 1 \\ \frac{ 18 }{ 5 } & 18^2 - 13 \cdot 5^2 = -1 & \mbox{digit} & 6 \\ \frac{ 119 }{ 33 } & 119^2 - 13 \cdot 33^2 = 4 & \mbox{digit} & 1 \\ \frac{ 137 }{ 38 } & 137^2 - 13 \cdot 38^2 = -3 & \mbox{digit} & 1 \\ \frac{ 256 }{ 71 } & 256^2 - 13 \cdot 71^2 = 3 & \mbox{digit} & 1 \\ \frac{ 393 }{ 109 } & 393^2 - 13 \cdot 109^2 = -4 & \mbox{digit} & 1 \\ \frac{ 649 }{ 180 } & 649^2 - 13 \cdot 180^2 = 1 & \mbox{digit} & 6 \\ \end{array} $$

Will Jagy
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