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I want to show that if $(x,y)$ is a solution to the negative pell equation ($x^2-dy^2=-1)$, then $\frac{x}{y}$ is a convergent of the continued fraction expansion of $\sqrt{d}.$

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  • $\begingroup$ One reference is this $\endgroup$ – rogerl Oct 13 '18 at 21:08
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I think it's easier to see the connection in the other direction. Here's a slightly imprecise way to see this. Let $[a_0; a_1, a_2, \dots]$ be the regular continued fraction of $\sqrt{d}$. Cutting this infinite expression off at $a_m$ gives the convergent $$\frac{h_m}{k_m} \approx \sqrt{d}$$ which gives the best approximation by any rational with denominator less than or equal to $k_m$. So $$h_m \approx \sqrt{d}k_m$$ $$h_m^2 \approx dk_m^2$$ $$h_m^2 - dk_m^2 \approx 0.$$

But this is an integer expression, so being as close to $0$ as possible without actually equaling $0$ means $|h_m^2 - dk_m^2| = 1.$ So $(h_m, k_m)$ satisfies the Pell equation.

The fact that the convergents give the "best possible rational approximation" correspond to the minimality condition on $x$ and $y$ in the Pell equation! This is also the fact that allows us to (sometimes) use the Pell equation to find the fundamental unit of the real quadratic number field $\mathbb{Q}[\sqrt{d}]$.

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