<s>[INST] continued fraction of $\sqrt{41}$ 
Show that $\sqrt{41} = [6;\overline {2,2,12}]$

here's my try:
$$\sqrt{36}<\sqrt{41}<\sqrt{49}\implies6<\sqrt{41}<7\implies\lfloor\sqrt{41}\rfloor=6$$
$$\sqrt{41}=6+\sqrt{41}-6=6+\frac{1}{\frac{1}{\sqrt{41}-6}}$$
$$\frac{1}{\sqrt{41}-6}=\frac{\sqrt{41}+6}{41-36}=\frac{\sqrt{41}+6}{5}=\frac{12+\sqrt{41}-6}{5}=2+\frac{\sqrt{41}-4}{5}$$
So far,
$$\sqrt{41}=6+\frac{1}{2+\frac{\sqrt{41}-4}{5}}=6+\frac{1}{2+\frac{1}{\frac{5}{\sqrt{41}-4}}}$$
But, $$\frac{5}{\sqrt{41}-4}=\frac{5(\sqrt{41}+4)}{41-16}=\frac{\sqrt{41}+4}{5}=\frac{6+\sqrt{41}-2}{5}=\color{red}{1}+\frac{\sqrt{41}-1}{5}$$
It suppose to be $2$ and not $1$. 
Where is the mistake? (I triple-checked and it seems fine to me)
 [/INST] Method described  by Prof. Lubin  at Continued fraction of $\sqrt{67} - 4$ 
$$  \sqrt { 41} = 6 +     \frac{  \sqrt {41} - 6 }{ 1 }  $$ 
 $$    \frac{ 1 }{   \sqrt {41} - 6 }  =  \frac{    \sqrt {41} + 6 }{5 } = 2 +  \frac{    \sqrt {41} - 4 }{5 } $$ 
 $$    \frac{ 5 }{   \sqrt {41} - 4 }  =  \frac{    \sqrt {41} + 4 }{5 } = 2 +  \frac{    \sqrt {41} - 6 }{5 } $$ 
 $$    \frac{ 5 }{   \sqrt {41} - 6 }  =  \frac{    \sqrt {41} + 6 }{1 } = 12 +  \frac{    \sqrt {41} - 6 }{1 } $$ 
Simple continued fraction tableau:
 $$ 
 \begin{array}{cccccccccccccccc}
 & & 6 & & 2 & & 2 & & 12 & & 2 & & 2 & & 12 & \\ 
 \\ 
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 6 }{ 1 }   & &   \frac{ 13 }{ 2 }   & &   \frac{ 32 }{ 5 }   & &   \frac{ 397 }{ 62 }   & &   \frac{ 826 }{ 129 }   & &   \frac{ 2049 }{ 320 }    \\ 
  \\ 
 & 1 & & -5 & & 5 & & -1 & & 5 & & -5 & & 1
 \end{array}
 $$ 
$$ 
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 41 \cdot 0^2 = 1 &     \mbox{digit}  &  6  \\  
  \frac{ 6 }{ 1 }   & 6^2 - 41 \cdot 1^2 = -5 &     \mbox{digit}  &  2  \\  
  \frac{ 13 }{ 2 }   & 13^2 - 41 \cdot 2^2 = 5 &     \mbox{digit}  &  2  \\  
  \frac{ 32 }{ 5 }   & 32^2 - 41 \cdot 5^2 = -1 &     \mbox{digit}  &  12  \\  
  \frac{ 397 }{ 62 }   & 397^2 - 41 \cdot 62^2 = 5 &     \mbox{digit}  &  2  \\  
  \frac{ 826 }{ 129 }   & 826^2 - 41 \cdot 129^2 = -5 &     \mbox{digit}  &  2  \\  
  \frac{ 2049 }{ 320 }   & 2049^2 - 41 \cdot 320^2 = 1 &     \mbox{digit}  &  12  \\  
 \end{array}
 $$ 
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