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Does every soluble negative pell equation, $a^2-Db^2=-1$, have infinitely many integer solutions $(a,b)$ where $a,b$ are both positive integers?

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Assume that $D$ is an integer greater than $1$. From a solution $(a,b)$ of the equation $a^2-Db^2=-1$, we can obtain infinitely many solutions $(a_n,b_n)$ by setting $$a_n+b_n\sqrt{D}=(a+b\sqrt{D})^{2n+1}.$$

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  • $\begingroup$ On the other hand, for $D=1$ the only solutions are $a = 0$, $b = \pm 1$. $\endgroup$ – Robert Israel Feb 10 '15 at 19:16
  • $\begingroup$ Hello, thanks for your answer. Is it possible to change a negative pell equation into a positive one? I imagine it might be possible through a linear substitution, e.g. to change $a^2-5b^2=-1$ into $x^2-5y^2=1$. EDIT: I tried doing it by putting $a=x+5y$ and $b=x+y$, but this gave the contradictory $(2x)^2-5(2y)^2=1$. (contradictory since 4 doesn't divide 1) $\endgroup$ – user45220 Feb 10 '15 at 20:17
  • $\begingroup$ In my opinion there is no simple way. We can obtain the solutions of $x^2-Dy^2=-1$ by using the fundamental solution, and then "multiplying" by solutions of $x^2-Dy^2=1$. However, for the negative Pell, there is no useful condition for existence of a solution. If a simple substitution did it, there would be no mysteries about negative Pell equations. $\endgroup$ – André Nicolas Feb 10 '15 at 20:25
  • $\begingroup$ @AndréNicolas: I found that the substitution $x=2a+5b$ and $y=a+2b$ works. $\endgroup$ – user45220 Feb 10 '15 at 20:26
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    $\begingroup$ @user, no matter what, the set of numbers that can be represented as $x^2 - D y^2$ is closed under multiplication. Meanwhile, if $a^2 - D b^2 = D,$ then $D|a,$ let $a = Dt,$ then $D^2 t^2 - D b^2 = D,$ then $b^2 - D t^2 = -1.$ For a very complete treatment using quadratic forms language rather than number field language, see math.stackexchange.com/questions/742181/… At this level, the two types of language are equivalent. Number fields take longer to learn, later you can do more. $\endgroup$ – Will Jagy Feb 10 '15 at 20:40

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