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I have to find all integral solutions of $3x^2 - 4y^2 = 11$.

I looked at the equation $\text{mod}\ 3$, and $\text{mod}\ 4$ to see that $x^2 \equiv 1\ \text{mod}\ 4$ and $y^2 \equiv 1\ \text{mod}\ 3$

How do I proceed from here? Or is there a better way to approach this?

Naweed G. Seldon
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1 Answers1

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Two good books. I have given more detailed answers for many of this type of problem.

http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf

https://www.math.cornell.edu/~hatcher/TN/TNbook.pdf

Here is a good one http://math.stackexchange.com/questions/739752/how-to-solve-binary-form-ax2bxycy2-m-for-integer-and-rational-x-y/739765#739765

As $11$ is prime, there are two orbits; one sequence of $x$ values is $$ 3, \; \; 37, \; \; 515, \; \; 7173, \ldots $$ with $$ x_{n+2} = 14 x_{n+1} - x_n, $$ with matching $y$ values $$ 2, \; \; 32, \; \; 446, \; \; 6212, \ldots $$ $$ y_{n+2} = 14 y_{n+1} - y_n. $$

Here is the second orbit: sequence of $x$ values is $$ 5, \; \; 67, \; \; 933, \; \; 12995, \ldots $$ with $$ x_{n+2} = 14 x_{n+1} - x_n, $$ with matching $y$ values $$ 4, \; \; 58, \; \; 808, \; \; 11254, \ldots $$ $$ y_{n+2} = 14 y_{n+1} - y_n. $$

The linear recurrences above come from Cayley-Hamilton for the matrix $$ \left( \begin{array}{cc} 7 & 8 \\ 6 & 7 \\ \end{array} \right) $$

enter image description here

Will Jagy
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  • Can you give more context? Orbit is from group theory? Is this some generalization of Pell's equation? – mvw Dec 25 '17 at 01:05
  • I'm interested. How do you know that there are two orbits? – Kenny Lau Dec 25 '17 at 01:05
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    @mvw see references I put in the answer now. Worth the effort to draw the topograph for $3x^2 - 4 y^2,$ out enough layers to find all occurrences of $11.$ – Will Jagy Dec 25 '17 at 01:17
  • @KennyLau see references I put in the answer now. Worth the effort to draw the topograph for $3x^2 - 4 y^2,$ out enough layers to find all occurrences of $11.$ – Will Jagy Dec 25 '17 at 01:17
  • Can it be one orbit if we propagate the recursion in both directions? – Oscar Lanzi Dec 25 '17 at 01:27
  • That is great stuff. Thank you very much. – mvw Dec 25 '17 at 01:37
  • @OscarLanzi yes, exactly. I have posted the diagram, enough for you to run the automorphism (inverse matrix) backwards and get negative $y,$ repeating the other orbit of positive $x,y$ – Will Jagy Dec 25 '17 at 02:07
  • @mvw I put in part of the diagram. The linear recursion is Cayley-Hamilton for the "automorphism" matrix – Will Jagy Dec 25 '17 at 02:09
  • This is a very interesting approach. Thank you! Especially for the resources – Naweed G. Seldon Dec 25 '17 at 12:54