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I remember a survey about the Pell equation. At the start it was claimed that it can be proven with the pigeonhole principle that the Pell-equation $$p^2-Nq^2=1$$ ($N$ is a positive integer not being a perfect square) always has a non-trivial solution ($p,q>0$)

Unfortunately, I do not remember whether the proof used the Dirichlet-approximation-theorem or modulo-arithmetic (and didn't understand, how it worked) , and I do not have access to the source anymore.

Can the pigeonhole-principle actually prove that the Pell-equation is always solveable without using the continued-fraction-expansion of $\sqrt{N}$ ? And if yes, how ?

Rodrigo de Azevedo
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Peter
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  • Take a search engine you trust and feed it with pell and pigeonhole. –  Feb 01 '17 at 12:57
  • Does this help http://math.stackexchange.com/a/207775/186189 – kingW3 Feb 01 '17 at 13:04
  • @kingW3 Seems to be exactly what I want, thanks! – Peter Feb 01 '17 at 13:42
  • Looking closer at the proof , I am stuck : I do not understand why we can conclude $|p^2-Dq^2|<\frac{(q+1)\sqrt{D}}{q}$. It is clear that $|q\sqrt{D}-p|\le \frac{1}{q}$, but what about $|p+q\sqrt{D}|$ ? – Peter Feb 01 '17 at 13:54
  • I hope that the proof does not contain a serious error... – Peter Feb 01 '17 at 14:06
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    If it was possible to prove that $$ x^2-Dy^2 = 1 $$ always has a solution by the pigeonhole principle only, it would be probably possible to show that $$ x^2-Dy^2 = -1$$ always has a solution by the same lines. But that is simply wrong. Arithmetic has to play a major role at some point. – Jack D'Aurizio Feb 01 '17 at 19:29

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