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Let $\mathcal H$ be a hyperbola (in the affine plane) whose defining equation has integers coefficients. Assume that one knows 2 points of $\mathcal H$ with integral coordinates. Is there a way to build a third one? For the hyperbola (Pell-Fermat equation) $x^2-Ny^2=1$ ($N\in\mathbb N$ not a square), there is Euler formula: if $(h,g)$ and $(h_0,g_0)$ are two not trivial points of the Pell-Fermat curve then $(hh_0−Ngg_0,hg_0−gh_0)$ is an another one. But for the general hyperbola, is it still possible to do the same? Thanks in advance for any answer.

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  • $\begingroup$ give an example. Every indefinite form, discriminant not a square, has an infinite (oriented) automorphism group. If the center is not at the origin, absorbing the linear terms into $q(x,y) = C$ with $q$ homogeneous quadratic gives, in effect, linear congruences on top of the automorphism group of $q.$ $\endgroup$ – Will Jagy Jul 25 '15 at 20:40
  • $\begingroup$ memories are coming back....given homogeneous quadratic $q(x,y) = C,$ with non-square discriminant, all solutions are given by the (infinite) automorphism group acting on a finite number of "seed" solutions. Each seed account for an "orbit" under the group action. Now, given a generator matrix $A$ of the (oriented) group, we get a degree two recurrence using Cayley-Hamilton on $A.$ Plenty of those among my answers. However, if there were linear terms that got absorbed, the degree two recurrence has also constant terms added in, annoying but not fatal. $\endgroup$ – Will Jagy Jul 25 '15 at 20:54
  • $\begingroup$ math.stackexchange.com/questions/739752/… $\endgroup$ – Will Jagy Jul 25 '15 at 20:58
  • $\begingroup$ math.stackexchange.com/questions/181758/… $\endgroup$ – Will Jagy Jul 25 '15 at 21:01
  • $\begingroup$ Here is an example: $-x^2+y^2+16x-14y=-20$ where $(14,8)$ and $(2,6)$ are integral points on the hyperbola. $\endgroup$ – joaopa Jul 25 '15 at 21:04
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given $$ x^2 - 15 xy + y^2 - 7 x - 8 y = 2,$$ multiply through by $4$ and complete the first square, for $$ u = 2x - 15 y - 7. $$ The remaining nonconstant terms are $-221 y^2 - 242y,$ which is kind of bad. $221 = 13 \cdot 17$ is squarefree, so we multiply through by $221$ and use $v = 221 y + 121.$ So far, we have $$ \frac{221u^2 - v^2 + 3812}{884} = 2. $$ This becomes $$ v^2 - 221 u^2 = 2044. $$

Find several solutions, including imprimitive (both u,v even). After that, the successor to a $(v,u)$ solution is $$ v' = 1665 v + 24752 u, \; \; \; \; u' = 112 v + 1665 u. $$

A portion of these will be solvable for integers $x,y.$

I have done enough. Buy the books.

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Note that $x^2 - 3 x y + y^2$ has the improper automorphism $(x,y) \mapsto (y,x).$

Given a discriminant $\Delta=5$ positive and not a square, solve $\tau^2 - \Delta \sigma^2 = 4.$ The one with smallest variables is the fundamental solution, in this case $3^2 - 5 \cdot 1^2 = 4.$

The generator of the oriented automorphism group of $A x^2 + B xy + C y^2$ is $$ \left( \begin{array}{cc} \frac{\tau - B \sigma}{2} & - C \sigma \\ A \sigma & \frac{\tau + B \sigma}{2} \end{array} \right) = \left( \begin{array}{cc} 3 & - 1 \\ 1 & 0 \end{array} \right) $$ and $$ \left( \begin{array}{cc} 3 & 1 \\ -1 & 0 \end{array} \right) \left( \begin{array}{cc} 2 & - 3 \\ -3 & 2 \end{array} \right) \left( \begin{array}{cc} 3 & - 1 \\ 1 & 0 \end{array} \right) = \left( \begin{array}{cc} 2 & - 3 \\ -3 & 2 \end{array} \right) $$

Given a solution $(x,y)$ we get the next solution in the forward direction as $$ (3x-y,x) $$ so $$ (1,1) $$ $$ (2,1) $$ $$ (5,2) $$ $$ (13,5) $$ $$ (34,13) $$ $$ (89,34) $$ $$ (233,89) $$ and so on, with evident Fibonacci numbers. If we call the generator matrix $G,$ Cayley-Hamilton says $G^2 - 3 G + I + 0.$ Which says that the $(x,y)$ coordinates in these solutions satisfy $$ x_{n+2} = 3 x_{n+1} - x_n,$$ $$ y_{n+2} = 3 y_{n+1} - y_n.$$

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  • $\begingroup$ Thanks a lot!! Just a last question. With this method, I will obtain new solutions, but will I obtain ALL the solutions? $\endgroup$ – joaopa Jul 25 '15 at 21:33
  • $\begingroup$ @joaopa You gave target $-1,$ so this time it is simple, and yes. If you gave a target with several prime factors, all represented primes, so $1,4 \pmod 5,$ you need more "seeds," more orbits. Why don't you run a simple computer program, find integer solutions to $x^2 - 3xy+y^2 = 209 = 11 \cdot 19,$ with $|x|, |y| < 1000?$ You will find more that one orbit of solutions. Given your questions, I recommend springer.com/us/book/9780387970370 and maa.org/publications/maa-reviews/the-sensual-quadratic-form Neither one tells you everything I wrote here, but..... $\endgroup$ – Will Jagy Jul 25 '15 at 21:53
  • $\begingroup$ Thanks. But I do not have these books unfortunately. Can you explain the method when there are non-quadratic terms as in the equation: $x^2+y^2-15 xy-7 x-8 y=2$ which has integer solutions $(-82,-6)$ and $(-7,-1)$? $\endgroup$ – joaopa Jul 26 '15 at 9:03
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For generalization of the Pell equation question arises constantly.

So writes the Pell equation in General form.

$$Ap^2-Bs^2=1$$

If we know any solution of this equation. $( p ; s)$

If we use any solutions of the following equation Pell.

$$x^2-ABy^2=1$$

Then the following solution of the desired equation can be found by the formula.

$$p_2=xp+Bys$$

$$s_2=xs+Ayp$$

For a more General equation, one can use the formula. http://www.artofproblemsolving.com/community/c3046h1048219

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