# Calculating the continued fraction of $\sqrt{47}$ using a different result

I have calculated the continued fraction of $\alpha=\frac{6+\sqrt{47}}{11}$ which equals $\overline{[1,5,1,12]}$. Now I am asked to calculated the cont. fraction of $\sqrt{47}$ using this result. I am not sure whether there is a simple formula to calculate the continued fraction of $\sqrt{47}=11\alpha-6$.

I know the answer to be $\sqrt{47}=[6,\overline{1,5,1,12}]$ (checked by Mathematica) but it's not clear how to arrive at this result using our previous answer.

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I think I found the answer: Those two cont'd fractions are nearly identical because $47=6^2+11$, and thus when calculating the continued fraction the second step in calculating $\sqrt{47}$ coincides with the first step of $\alpha$, thus those two expressions are the same from that point onwards. I.e., $(\sqrt{47}-6)^{-1}=\alpha$ – ClausW Mar 4 '12 at 13:52
Hint: $6=\sqrt{36}$ and $11=47-36$ (consider the conjugate!). (seeing your answer : yes you are right!) – Raymond Manzoni Mar 4 '12 at 13:54

$(\sqrt{47}-6)(\sqrt{47}+6)=47-36=11$, so $$(\sqrt{47}-6)\alpha=(\sqrt{47}-6)\left(\frac{\sqrt{47}+6}{11}\right)=1\;,$$ and $$\sqrt{47}-6=\frac1{\alpha}\;.$$

Clearly $\lfloor\sqrt{47}\rfloor=6$, so you know that $$\sqrt{47}=6+\frac1{\left(\frac1{\sqrt{47}-6}\right)}=6+\frac1\alpha=[6,\overline{1,5,1,12}]\;.$$

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Just my comment of a minute ago, only you explained it better. Thanks. :) – ClausW Mar 4 '12 at 13:54
@Claus: I saw your comment just after I posted; good job. – Brian M. Scott Mar 4 '12 at 13:55

I know the answer to be $\sqrt{47}=[6,\overline{1,5,1,12}]$ (checked by Mathematica) but it's not clear how to arrive at this result using our previous answer.

No ingenuity is needed. The above observation makes the proof mechanical. The above is true

$$\iff\ \sqrt{47}\: =\: 6 + \dfrac{1}{\overline{1,5,1,12}}\: =\: 6 + \dfrac{1}\alpha\ \iff\ \alpha \:=\: \dfrac{1}{\sqrt{47}-6}\: =\: \dfrac{\sqrt{47}+6}{11}$$

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