6
$\begingroup$

Prove that whenever the equation $x^2 - dy^2 = c$ is solvable, then it has infinitely many solutions.

I consider that, if $u$ and $v$ satisfy $x^2 -dy^2 = c$ and then $r$ and $s$ satisfy $x^2 -cy^2 = 1$, then $$(ur \pm dvs)^2 - d(us \pm vr)^2 = (u^2 - dv^2)(r^2 - ds^2) = c\;.$$ But, still I failed to complete the proof. I am requesting members to spare some time for this. Thanks in advnace.

  • 0
    @Brain M. Scott! thank you so much for editing2012-04-07
  • 0
    Shouldn't it be $(u^2-dv^2)\color{red}{(r^2-cs^2)}=c$?2012-04-07
  • 2
    Where is the difficulty? You seem to have done most of the work. You know there are infinitely many solutions to Pell's equation right? There's infinitely many $r,s$ you can use, so infinitely many pairs $ (ur + dvs, us + vr) $ that solve $x^2-dy^2=c $.2012-04-07
  • 0
    @gandhi : I agree with the above comment, it seems you have everything done, but before that please read FAQ in asking questions and TeX your questions properly. Some papers that can add something to it are [here](http://nyjm.albany.edu/j/2001/7-7.pdf) and [here](http://www.maths.tcd.ie/pub/ims/bull58/R5801.pdf) . Thank you. Try googling your answers before putting them here. It will help you and others in case of saving time.2012-04-07
  • 0
    @Ragib Zaman! I got it sir, where I was stacked. Thank you.2012-04-08
  • 0
    What if $d = 1$? Then $x^2-y^2=2n+1$ has only a finite number of solutions for any integer $n$ and it always has the solution $x = n+1$, $y = n$.2014-06-12

3 Answers 3