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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?

user45220
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1 Answers1

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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 $$a_n+b_n\sqrt{D}=(a+b\sqrt{D})^{2n+1}.$$

André Nicolas
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  • On the other hand, for $D=1$ the only solutions are $a = 0$, $b = \pm 1$. – Robert Israel Feb 10 '15 at 19:16
  • 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) – user45220 Feb 10 '15 at 20:17
  • 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. – André Nicolas Feb 10 '15 at 20:25
  • @AndréNicolas: I found that the substitution $x=2a+5b$ and $y=a+2b$ works. – user45220 Feb 10 '15 at 20:26
  • @AndréNicolas: Maybe we cause the solutions of the equation itself to find a suitable transformation. We need to force the $x$ coefficient to equal 1 in the transformed version. – user45220 Feb 10 '15 at 20:28
  • @AndréNicolas: At least for linear substitutions, it appears to rest on the following statement: If $a^2-Db^2=-1$ is solvable, then $a^2-Db^2=D$ is solvable. If this statement is true, we can always find such a transformation. – user45220 Feb 10 '15 at 20:33
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    @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 http://math.stackexchange.com/questions/742181/find-all-integer-solutions-for-the-equation-5x2-y2-4/756972#756972 At this level, the two types of language are equivalent. Number fields take longer to learn, later you can do more. – Will Jagy Feb 10 '15 at 20:40
  • I'm probably assuming $D$ squarefree in the previous comment. That does not matter as much as one would expect, but more proof is required. – Will Jagy Feb 10 '15 at 20:44
  • @WillJagy: Thanks for the simple proof. So it is always possible to transform negative pell into the corresponding positive one. Your post in the link looks interesting, I am reading it now. – user45220 Feb 10 '15 at 20:44
  • @WillJagy: Ah, yes, the boundary case rearing its ugly head in. – user45220 Feb 10 '15 at 20:44
  • About your substitution, it is equivalent to finding the primitive solution. We can do that by a continued fraction calculation, but there is no simple criterion for existence. – André Nicolas Feb 10 '15 at 23:57