Find more integral points on a hyperbola

Question

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.

Answer 1

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.

Answer 2

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 $$ x_{n+2} = 3 x_{n+1} - x_n,$$ $$ y_{n+2} = 3 y_{n+1} - y_n.$$

Answer 3

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