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I want to show that $x^2-6y^2=523$ has infinitely many solutions. For the special case $x^2-dy^2=1$, I know what I need to do. I can get the result by using continued fractions. Also, in the kinds of $x^2-dy^2=m$ for some examples, I can say that there is no solution using modulo prime $p$. But in general, I'm not sure how to find the solution set for $ax^{2}+by^{2}+c=0$ where $a,b,c\in \mathbf{Z}$.

I would appreciate if you can help me with this question or direct it to a resource that can help.

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  • $\begingroup$ The general question you ask at the end turns out to be quite a difficult problem in general. In the most general case whether a prime $p$ may be written as $ax^2 + by^2 = p$ boils down to asking about the splitting behaviour of a prime $p$ in the Hilbert Class Field of $\mathbb{Q}(\sqrt{D})$ where $D$ is the discriminant of $ax^2 + by^2$. $\endgroup$ – Mummy the turkey Jul 3 '20 at 2:13
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You can use a twist on Pell's equation in this case. The fundamental Pell solution for $D=6$ is $(x,y)=(5,2).$ So all positive solutions to $$a^2-6b^2=1$$ are given by $$a_k+b_k \sqrt{6}=(5+2\sqrt{6})^k$$ for $k\in \mathbb{Z}_{+}.$ Here's the fun part: since $(x,y)=(23,1)$ is a solution to $$x^2-6y^2 = 523$$ (found by trying out smaller values of $x$ or $y$ and seeing if the other variable turns out to be an integer), a solution is given by $$x_k + y_k\sqrt{6}=(23+\sqrt{6})(a_k+b_k \sqrt{6})$$ for each positive integer $k.$ It is easy to prove that these are all solutions because if $a^2-6b^2=1$ then $$(23+\sqrt{6})(a+b\sqrt{6}) = (23a+6b)+(23b+a)\sqrt{6}$$ and \begin{align*} (23a+6b)^2-6(23b+a)^2 &= 23^2 (a^2-6b^2) - 6(a^2-6b^2)\\ &= 23^2-6\\ &=523. \end{align*}

This does not necessarily find all solutions, but you do get infinitely many since they are monotonically increasing in some sense. You can read what Keith Conrad has written about finding all solutions.

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  • $\begingroup$ I understood the idea very well. I will read the paper you said to see more details. Thank you for clarifying. $\endgroup$ – Tuğba Yesin Jul 2 '20 at 19:59
  • $\begingroup$ @TuğbaYesin here Favst displays the process beginning with $(23,1).$ You get every solution if you do the same process beginning with $(23,-1).$ The reason that just two beginning "seed" points suffice is that 523 is prime. I recommend Weissman, bookstore.ams.org/mbk-105 $\endgroup$ – Will Jagy Jul 3 '20 at 17:29
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$(23, 1)$ is a solution.

If $(x, y)$ is a solution, then $(5x + 12y, 2x + 5y)$ is also a solution.

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    $\begingroup$ it’s better if you show your analysis ( some key details ) that leads to the answer. $\endgroup$ – DeepSea Jul 2 '20 at 18:23
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    $\begingroup$ @DeepSea I would like to, but without any context I'm not sure whether the OP is supposed to know the ring $\Bbb Z[\sqrt 6]$ or any of the algebraic number theory backgrounds. $\endgroup$ – WhatsUp Jul 2 '20 at 18:25
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    $\begingroup$ Then I would suppose that the preferred method is to leave a comment inquiring that instead of writing an answer. $\endgroup$ – Andrew Chin Jul 2 '20 at 18:27
  • $\begingroup$ Thank you for the answer, but actually I am curious about the process that continues after finding a solution. Also, finding the first solution isn't always that easy, right? $\endgroup$ – Tuğba Yesin Jul 2 '20 at 18:41
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    $\begingroup$ @TuğbaYesin The first question is easier to answer: simply multiply $x + y\sqrt 6$ by $5 + 2\sqrt 6$. The second question (i.e. finding the first solution) is more complicated. In this simple case, just try every value of $y$, starting from $0$. $\endgroup$ – WhatsUp Jul 2 '20 at 18:47

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