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Let $d\in\mathbb{N}$ be square-free and $(a,b)$ the fundamental solution to the negative Pell equation $x^2-dy^2=-1$. Prove that $(\alpha,\beta)$, defined by $\alpha+\beta\sqrt{d}=(a+b\sqrt{d})^2$, is the fundamental solution of the positive Pell equation $x^2-dy^2=1$.

I'm sure this is supposed to be elementary, but I still can't figure it out.

My first attempt was to consider $(\gamma,\delta)$ a solution to the positive equation such that $\gamma+\delta\sqrt{d}<\alpha+\beta\sqrt{d}$ and find a contradition. I've tried to cook up some solution $(a',b')$ to the negative equation such that $\gamma+\delta\sqrt{d}=(a'+b'\sqrt{d})^2$ and $a'+b'\sqrt{d}<a+b\sqrt{d}$ but that didn't work out.

What's the idea?

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    $\begingroup$ You may find details on this in Elementary Number Theory D.M.Burton $\endgroup$ – Shubhrajit Bhattachrya Sep 12 '17 at 19:43
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Let $c$ be the fundamental solution to the improper Pell equation. Then it's easy to see that $g = c^2$ is a solution to the proper Pell equation. To show it's the fundamental solution, suppose not. Then there's a solution $h$ that's smaller in absolute value than $g$. Then $h/c$ is a solution to the improper equation that's smaller than $c$, contradicting the fact that $c$ is fundamental.

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    $\begingroup$ $h/c$ is of the form $u+v\sqrt{d}$, so don't we have to prove that both $u$ and $v$ are positive? $\endgroup$ – rmdmc89 Sep 12 '17 at 14:30
  • $\begingroup$ I don't think so. It's a unit, and it's between $1$ and $|c|$ in absolute value. I might be wrong - it's been a long time since I worked on number theory. $\endgroup$ – Ethan Bolker Sep 12 '17 at 17:09

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