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I want to find all integer solutions of the equation $$x^2-7y^2=-3$$ I don't really know where to start... I tried the one trick I know which is to factor in some quadratic ring: $$(x+\sqrt{-3})(x-\sqrt{-3})=7y^2$$ But I don't think that this tells us much since $\mathbb{Z}[\sqrt{-3}]$ is not a UFD. Any help would be appreciated.

I will add that I'll have to solve this type of problem on an exam, hence I want a solution that it quick and suited for use on exams.

user2520938
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    Possible duplicate of How to solve inhomogeneous quadratic forms in integers? – Dietrich Burde May 24 '16 at 18:57
  • @DietrichBurde thank you, I think that does indeed solve this problem. – user2520938 May 24 '16 at 18:59
  • There is an online "Pell's equation solver", which gives you the recursion how to obtain all integer solutions. See also here. – Dietrich Burde May 24 '16 at 19:00
  • @DietrichBurde Hm actually I'm not sure this solves my problem. I would have to be able to solve this without a computer, just using paper, and in a reasonable time. Does this method allow for that? – user2520938 May 24 '16 at 19:05
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    Yes, it solves your problem, giving you a very easy recursion for all solutions. See also this question. Pell's equation is the standard here, also for your exam. – Dietrich Burde May 24 '16 at 19:13
  • @DietrichBurde Not convinced I'd want to do this from the top of my head on an exam... I'd prefer a more 'ad hoc' method, there must be something clever we can do in this specific case. – user2520938 May 24 '16 at 19:16
  • The problem is the all solutions. It is easy to get one solution and a recurrence which gives you infinitely many. – almagest May 24 '16 at 19:18
  • All solutions are given by recursion form either $(x,y)=(2,1)$ or $(x,y)=(-2,1)$. For the first one, the recursion is $x_{n+1}=8x_n+21y_n$ and $y_{n+1}=3x_n+8y_n$. – Dietrich Burde May 24 '16 at 19:21
  • @DietrichBurde Yes, but proving that is a little harder. Whereas it is easy to show that (2,1) and the recurrence generate solutions. – almagest May 24 '16 at 19:23
  • @almagest I agree with you. I do not agree with the OP, that we can avoid generalised Pell's equation by "some trick". – Dietrich Burde May 24 '16 at 19:25
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    @almagest there is a trick in the style of an infinite descent: for each solution $(x,y),$ both positive, we may find an earlier solution by inverting the action Dietrich gives. That is, we back up with $$ (x,y) \mapsto (8x - 21 y, -3x + 8y). $$ A "seed" solution is when either $8x - 21 y \leq 0$ or $-3x + 8y \leq 0.$ These conditions give one representative for each orbit of solutions under Dietrich's action. – Will Jagy May 24 '16 at 19:30

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As Dietrich is saying:

there is a trick in the style of an infinite descent: for each "non-seed" solution $(x,y),$ both positive, we may find an earlier positive solution by inverting the action Dietrich gives. That is, we back up with $$ (x,y) \mapsto (8x - 21 y, -3x + 8y). $$ A "seed" solution is when either $8x - 21 y \leq 0$ or $-3x + 8y \leq 0.$

I should add that, as $|-3|$ is prime, we get at most two "seed" solutions. I wrote this program to emphasize positive $x,y,$ however, note $ (5,2) \mapsto (-2, 1). $ There is a 2016 article by Brillhart that gives detail on why more than two such seed points would cause the target number to be composite. So, being able to guess the solutions $(\pm 2,1),$ we know we have found all the orbits of solutions. 

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental

  8^2 - 7 3^2 = 1

 x^2 - 7 y^2 = -3

Tue May 24 12:20:40 PDT 2016


 Pell automorph 
8  21
3  8


x:  2  y:  1 ratio: 2  SEED 
x:  5  y:  2 ratio: 2.5  SEED 
x:  37  y:  14 ratio: 2.642857142857143
x:  82  y:  31 ratio: 2.645161290322581
x:  590  y:  223 ratio: 2.645739910313901
x:  1307  y:  494 ratio: 2.645748987854251
x:  9403  y:  3554 ratio: 2.645751266178953
x:  20830  y:  7873 ratio: 2.645751301917947
x:  149858  y:  56641 ratio: 2.645751310887873
x:  331973  y:  125474 ratio: 2.64575131102858
x:  2388325  y:  902702 ratio: 2.645751311063895
x:  5290738  y:  1999711 ratio: 2.645751311064449

Tue May 24 12:21:00 PDT 2016

 x^2 - 7 y^2 = -3

jagy@phobeusjunior:~$

In addition, since the trace of the "Automorph" matrix is $16,$ but there are two seeds so we alternate,

$$ x_{n+4} = 16 x_{n+2} - x_n, $$ $$ y_{n+4} = 16 y_{n+2} - y_n. $$

Will Jagy
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  • Thank you for your answer. I'm a bit disappointed that I'll have to do this on an exam, but it is what it is I guess. – user2520938 May 24 '16 at 19:37
  • @user2520938 note that with $|-3|$ prime, we get at most two seed values. Furthermore we can use Dietrich's pairs $(\pm 2,1)$ as the descent gives $(5,2) \mapsto (-2,1).$ I wrote the program this way for a certain type of convenience. Put it together, you can use the obvious solution and be done. – Will Jagy May 24 '16 at 19:42