<s>[INST] Continued fraction of $ \sqrt{3} $ How can I solve the continued fraction of $\sqrt{3}$?
I know that a continued fraction is [1;1,2,1,2,1,2,....] -> this is the continued fraction of $\sqrt{3}$, but how can I solve? What are the steps? 
I don't understand this proof.

Thank you so much.
 [/INST] Method described  by Prof. Lubin  at Continued fraction of $\sqrt{67} - 4$ 
$$  \sqrt { 3} = 1 +     \frac{  \sqrt {3} - 1 }{ 1 }  $$ 
 $$    \frac{ 1 }{   \sqrt {3} - 1 }  =  \frac{    \sqrt {3} + 1 }{2 } = 1 +  \frac{    \sqrt {3} - 1 }{2 } $$ 
 $$    \frac{ 2 }{   \sqrt {3} - 1 }  =  \frac{    \sqrt {3} + 1 }{1 } = 2 +  \frac{    \sqrt {3} - 1 }{1 } $$ 
Simple continued fraction tableau:
 $$ 
 \begin{array}{cccccccccc}
 & & 1 & & 1 & & 2 & \\ 
 \\ 
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 1 }{ 1 }   & &   \frac{ 2 }{ 1 }    \\ 
  \\ 
 & 1 & & -2 & & 1
 \end{array}
 $$ 
$$ 
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 3 \cdot 0^2 = 1 &     \mbox{digit}  &  1  \\  
  \frac{ 1 }{ 1 }   & 1^2 - 3 \cdot 1^2 = -2 &     \mbox{digit}  &  1  \\  
  \frac{ 2 }{ 1 }   & 2^2 - 3 \cdot 1^2 = 1 &     \mbox{digit}  &  2  \\  
 \end{array}
 $$ 
........................
</s>