Use continued fractions to find the minimal solution to $x^2-11y^2=1$.

I know that $\sqrt{11}=3+\frac{1}{3+\frac{1}{6+\frac{1}{3+...}}}$

I took $\sqrt{11}=3+\frac{1}{3+\frac{1}{6+\sqrt{11}}}$ and I got $$\frac{63+10\sqrt{11}}{19+3\sqrt{11}}.$$ This gave me $(19)(10)-(63)(3)=1$ but I wasn't sure what to do from there.

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    Find the first few convergents of the continued fraction of $\sqrt{11}$. Soon you will bump into a solution. – André Nicolas Nov 7 '13 at 4:11

Your expansion of $\sqrt n$ will always look like $m+\frac1{a_1+}\cdots\frac1{2m+\cdots}\,$. Here $m=\lfloor\sqrt n\rfloor$. You cut off the expansion just before the first appearance of $2m$, and use the numerator and denominator for the values of $X$ and $Y$, to get the first nontrivial solution of $X^2-nY^2=\pm1$. If you got minus instead of plus, go to the next appearance of $2m$. (I’ve left off lots of details).

In your case, $n=11$ and $m=3$, so you evaluate $3+\frac13$ to get $X=10$, $Y=3$, voilà. Want more solutions? Go to the next appearance of $2m$, etc.

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