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How can one find a quadratic irrational when knowing its periodic continued fraction?

For example(using Wikipedia notion), how can one find the quadratic irrational that its continued fraction is $[0; \overline{1,4,1}]$ ?

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  • $\begingroup$ It is not a "fraction," it is a quadratic irrationality. $\endgroup$ – André Nicolas Feb 23 '14 at 20:18
  • $\begingroup$ If periodic (therefore infinite) it is a "quadratic irrational" $\endgroup$ – Will Jagy Feb 23 '14 at 20:19
  • $\begingroup$ Sorry about that, changed it. $\endgroup$ – Eliran Koren Feb 23 '14 at 20:32
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Let $a$ be a quadratic irrational with periodic continued fraction of the form $[0; \overline{1,4,1}]$.

$a=\cfrac{1}{1+\cfrac{1}{4+\cfrac{1}{1+a}}} =\cfrac{1}{1+\cfrac{1}{\cfrac{5+4a}{1+a}}} = \cfrac{1}{1+\cfrac{1+a}{5+4a}}=\cfrac{1}{\cfrac{6+5a}{5+4a}}=\cfrac{5+4a}{6+5a} \Rightarrow$

$6a+5a^2=5+4a \Rightarrow$ $5a^2+2a-5=0 \Rightarrow$ $a=\frac{-1 \pm \sqrt{26}}{5} $
We know that $a>0$ ,thus:

$a=\frac{ \sqrt{26}-1}{5}$

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You might consider taking it over to Wolfram and seeing what it says

$\frac{\sqrt{26}-1}{5} $

Now you can try to prove it.

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