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If I have a solution $(x_0, y_0)$ to $x^2 - Dy^2 = k$ where D > 1 is not a square, how do I find another solution $(x_1,y_1)$ to $x^2 - Dy^2 = k$ without knowing the solution $(u,v)$ to $u^2 - Dv^2 = 1$ ?

I can find the first solution with Conway's topograph quite easily. However, the second solution shows up at the end of the period in the river of Conway's topograph (the river shows up in the topograph of indefinite binary quadratic forms).

I have read multiple answers to similar questions, but the prerequisite for solving the problem is to first find a fundamental solution to $x^2 - Dy^2 = 1$, which I do not have.

Edit 1: Forgot to mention that $D$ is too large to try computing the continued fraction. I tried this code implementing the algorithm for this, and it took a long time. Also, $k$ is not necessarily prime, I am looking for a general solution.

Here's what I have tried on my own:

I used this code to test out the chakravala method but it was slow.

ender
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  • Things are pretty straightforward when $k$ is $\pm$ a prime. If $|k|$ is small enough, all solutions to $x^2 - Dy^2 = k$ show up in the continued fraction for $\sqrt D,$ which is also how you solve $x^2 - Dy^2 = 1$ and sometimes $x^2 - Dy^2 = -1.$ – Will Jagy Jul 17 '23 at 23:57
  • Well, see what you can do for $x^2 - 5 y^2 = 209 = 11 \cdot 19 .$ The numbers are small enough....... – Will Jagy Jul 18 '23 at 00:04
  • It looks like $(\pm 17,4)$ is a fundamental solution to $^2−5^2=209$. Is there a reason I should look into this? – ender Jul 18 '23 at 00:10
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    Alright; you've edited. You should be aware of methods for Pell that use only integer arithmetic; it is necessary just to find the floor of $\sqrt D;$ this can be done with Newton's method over tyhe integers. – Will Jagy Jul 18 '23 at 00:10
  • I am looking into Newton's method! I will update my question with the solution once I have understood this approach. – ender Jul 18 '23 at 00:15
  • alright, you did the example. Note that $9^2 - 5 \cdot 4^2 = 1$ Composing solutions for $1$ and $209$ gives a new solution, $(73, 32) $ However, there are other solutions in between. What are those? – Will Jagy Jul 18 '23 at 00:18
  • @vvg That gives the same solution or the same solution with a sign changed. – NDB Jul 18 '23 at 03:19
  • To get all solutions of Pell equation need only fundamental solutions and fundamental unit of quadratic form of equation. – Dmitry Ezhov Jul 18 '23 at 17:39
  • @DmitryEzhov Can you do this in polynomial time? – ender Jul 18 '23 at 22:17
  • @WillJagy Is there a resource you recommend for Newton's method? I may have not tried or attempted to use that method correctly. – ender Jul 18 '23 at 22:18
  • @WillJagy Also, it appears that $( \pm 23, 8)$ is another solution to $x^2−5y^2=209$. – ender Jul 18 '23 at 22:45
  • @ender if you use Theorem 3.3 in the first file I linked to in my answer below (not the Lenstra article) and use $u = 9+4\sqrt{5}$ there, it tells you each $x+y\sqrt{5}$ such that $x^2 - 5y^2 = 209$ is a power of $u$ times a solution where $|x| \leq 32$ and $|y| \leq 12$. So just take $y = 0, 1, \ldots, 12$ to see when $209+5y^2$ is a perfect square and you'll see that it happens only when $y$ is $4$ or $8$: those are the solutions you already found. So the solutions are $(x+y\sqrt{5})u^k$ where $(x,y)$ is $(\pm 17, \pm 4)$ or $(\pm 23, \pm 8)$ (independent choices of signs). – KCd Jul 19 '23 at 20:53

3 Answers3

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Why do you want to find more integral solutions without ever using information about integral solutions to the equation $x^2 - Dy^2 = 1$?

The main structure theorem about integral solutions to $x^2 - Dy^2 = k$ is threefold:

(i) when $(x_0,y_0)$ is an integral solution to $x^2 - Dy^2 = k$, so are the coefficients of $(x_0 + y_0\sqrt{D})(a+b\sqrt{D})$ where $(a,b)$ is an arbitrary solution to $x^2 - Dy^2 = 1$,

(ii) there are finitely many integral solutions $(x_0,y_0),\ldots,(x_m,y_m)$ to $x^2 - Dy^2 = k$ such that every other integral solution to this equation occurs as coefficients of $(x_i+y_i\sqrt{D})(a+b\sqrt{D})$ for some $i = 1,\ldots,m$ and $a^2 - Db^2 = 1$,

(iii) in terms of one $(a,b)$ where $a^2 - Db^2 = 1$ and $a, b \geq 1$, you can find the finitely many solutions $(x_i,y_i)$ for (ii) in an explicit box in the plane.

See Theorem 3.3 and Corollary 3.5 here. Section 4 there has several worked examples.

You say in your answer that "$D$ is too large to try computing the continued fraction" of $\sqrt{D}$. What kind of $D$ are you using? If you find enough integral solutions to $x^2 - Dy^2 = k$, then taking a suitable ratio $(x+y\sqrt{D})/(x'+y'\sqrt{D})$ for some of them is going to give you a solution to $x^2 - Dy^2 = 1$, so bypassing the direct use of solutions to the Pell equation in solving he generalized Pell equation is going to amount to solving the Pell equation by another method. So maybe you should look at a survey paper on ways to solve Pell equations, like Lenstra's AMS Notices article https://www.ams.org/notices/200202/fea-lenstra.pdf.

KCd
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1

All solutions of equation $a^2-91b^2=-10$ derives from modulo polynomial $a+bx\equiv(-x\pm9)(-1574-165x)^j\pmod{x^2-91}$ where $j=$0,1,2,....

pari/gp code:

{
  D= 91; C= -10;
  Q= bnfinit('x^2-D,1);
  fu= Q.fu[1]; print("Fundamental Unit: "fu);
  N= bnfisintnorm(Q, C); print("Fundamental Solutions (Norm): "N);
  for(k=1, #N, n= N[k];
   print("\n\nNorm: "n"\n");
   for(j=0, 8,
    s= lift(n*fu^j);
    X= abs(polcoeff(s, 0)); Y= abs(polcoeff(s, 1));  
    if(X^2-D*Y^2==C, if(Y, if(X==floor(X), if(Y==floor(Y),
     y= Y; x= X; 
     print("("x", "y")    j = "j);
    ))))
   )
  )
};

Output:

Fundamental Unit: Mod(-165*x - 1574, x^2 - 91)
Fundamental Solutions (Norm): [-x + 9, -x - 9]

Norm: -x + 9


(9, 1)    j = 0
(849, 89)    j = 1
(2672661, 280171)    j = 2
(8413535979, 881978219)    j = 3
(26485808589231, 2776467153241)    j = 4
(83377317025363209, 8740317716424449)    j = 5
(262471767510034792701, 27514517394837012211)    j = 6
(826261040744272502059539, 86615692018629198015779)    j = 7
(2601069493791202326448636071, 272666170960127320516660081)    j = 8


Norm: -x - 9


(9, 1)    j = 0
(29181, 3059)    j = 1
(91861779, 9629731)    j = 2
(289180851111, 30314390129)    j = 3
(910341227435649, 95429690496361)    j = 4
(2865753894786571941, 300412635368154299)    j = 5
(9021392350446901034619, 945698880709259236891)    j = 6
(28399340253452949670408671, 2977059776060112709578569)    j = 7
(89401114096477535115545461689, 9371783229338354100494098321)    j = 8
Dmitry Ezhov
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  • My issue with computing the continued fraction for large D was speed. bnfinit runs in sub-exponential time if I am correct. I am looking for a polynomial time algorithm. – ender Jul 18 '23 at 22:19
  • Currently is no polynomial time algorithm, otherwise the factorization problem and RSA would have already been breacked. – Dmitry Ezhov Jul 19 '23 at 05:35
1

here is a method for standard Pell

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

$$ \sqrt { 91} = 9 + \frac{ \sqrt {91} - 9 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {91} - 9 } = \frac{ \sqrt {91} + 9 }{10 } = 1 + \frac{ \sqrt {91} - 1 }{10 } $$ $$ \frac{ 10 }{ \sqrt {91} - 1 } = \frac{ \sqrt {91} + 1 }{9 } = 1 + \frac{ \sqrt {91} - 8 }{9 } $$ $$ \frac{ 9 }{ \sqrt {91} - 8 } = \frac{ \sqrt {91} + 8 }{3 } = 5 + \frac{ \sqrt {91} - 7 }{3 } $$ $$ \frac{ 3 }{ \sqrt {91} - 7 } = \frac{ \sqrt {91} + 7 }{14 } = 1 + \frac{ \sqrt {91} - 7 }{14 } $$ $$ \frac{ 14 }{ \sqrt {91} - 7 } = \frac{ \sqrt {91} + 7 }{3 } = 5 + \frac{ \sqrt {91} - 8 }{3 } $$ $$ \frac{ 3 }{ \sqrt {91} - 8 } = \frac{ \sqrt {91} + 8 }{9 } = 1 + \frac{ \sqrt {91} - 1 }{9 } $$ $$ \frac{ 9 }{ \sqrt {91} - 1 } = \frac{ \sqrt {91} + 1 }{10 } = 1 + \frac{ \sqrt {91} - 9 }{10 } $$ $$ \frac{ 10 }{ \sqrt {91} - 9 } = \frac{ \sqrt {91} + 9 }{1 } = 18 + \frac{ \sqrt {91} - 9 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccccccc} & & 9 & & 1 & & 1 & & 5 & & 1 & & 5 & & 1 & & 1 & & 18 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 9 }{ 1 } & & \frac{ 10 }{ 1 } & & \frac{ 19 }{ 2 } & & \frac{ 105 }{ 11 } & & \frac{ 124 }{ 13 } & & \frac{ 725 }{ 76 } & & \frac{ 849 }{ 89 } & & \frac{ 1574 }{ 165 } \\ \\ & 1 & & -10 & & 9 & & -3 & & 14 & & -3 & & 9 & & -10 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 91 \cdot 0^2 = 1 & \mbox{digit} & 9 \\ \frac{ 9 }{ 1 } & 9^2 - 91 \cdot 1^2 = -10 & \mbox{digit} & 1 \\ \frac{ 10 }{ 1 } & 10^2 - 91 \cdot 1^2 = 9 & \mbox{digit} & 1 \\ \frac{ 19 }{ 2 } & 19^2 - 91 \cdot 2^2 = -3 & \mbox{digit} & 5 \\ \frac{ 105 }{ 11 } & 105^2 - 91 \cdot 11^2 = 14 & \mbox{digit} & 1 \\ \frac{ 124 }{ 13 } & 124^2 - 91 \cdot 13^2 = -3 & \mbox{digit} & 5 \\ \frac{ 725 }{ 76 } & 725^2 - 91 \cdot 76^2 = 9 & \mbox{digit} & 1 \\ \frac{ 849 }{ 89 } & 849^2 - 91 \cdot 89^2 = -10 & \mbox{digit} & 1 \\ \frac{ 1574 }{ 165 } & 1574^2 - 91 \cdot 165^2 = 1 & \mbox{digit} & 18 \\ \end{array} $$

Putting the $(x_n, y_n)$ pairs that solve $x_n^2 - 91 y_n^2 = -10$ in increasing order, we get recurrences $$ x_{n+4} = 3148 x_{n+2} - x_n,$$

$$ y_{n+4} = 3148 y_{n+2} - y_n.$$

The pairs are $$ (-849, 89) $$ $$ (-9, 1) $$ $$ ( 9, 1) $$ $$ (849, 89) $$ $$ (29181, 3059) $$ $$ (2672661, 280171) $$

J. W. Tanner
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Will Jagy
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