TITLE: Continued fraction of $\sqrt{67} - 4$
QUESTION [7 upvotes]: Find the continued fraction of $ \sqrt{67}-4 $ . $$ $$ We Know that if $ N $ is not a perfect square  and if continued fraction of $ \sqrt N $  is $ \sqrt N  =  [a_{1} , \overline {a_{2},a_{3} , \ldots , 2a_{1}} ]$  , then the continued fraction of $ \sqrt N-a_{1}$ is  $ \sqrt N-a_{1}=[\overline {0,a_{2},a_{3},\ldots,a_{n}, 2a_{1}}] $ . But I can't find the continued fraction of $\sqrt {67} $. Please someone help me

REPLY [4 votes]: Typeset the hand calculation method Prof. Lubin indicated. I would say that the reason I did not know this was that I had never calculated one of these by hand. There is a lesson in that.
$$  \sqrt { 67} = 8 +     \frac{  \sqrt {67} - 8 }{ 1 }  $$
$$    \frac{ 1 }{   \sqrt {67} - 8 }  =  \frac{    \sqrt {67} + 8 }{3 } = 5 +  \frac{    \sqrt {67} - 7 }{3 } $$
$$    \frac{ 3 }{   \sqrt {67} - 7 }  =  \frac{    \sqrt {67} + 7 }{6 } = 2 +  \frac{    \sqrt {67} - 5 }{6 } $$
$$    \frac{ 6 }{   \sqrt {67} - 5 }  =  \frac{    \sqrt {67} + 5 }{7 } = 1 +  \frac{    \sqrt {67} - 2 }{7 } $$
$$    \frac{ 7 }{   \sqrt {67} - 2 }  =  \frac{    \sqrt {67} + 2 }{9 } = 1 +  \frac{    \sqrt {67} - 7 }{9 } $$
$$    \frac{ 9 }{   \sqrt {67} - 7 }  =  \frac{    \sqrt {67} + 7 }{2 } = 7 +  \frac{    \sqrt {67} - 7 }{2 } $$
$$    \frac{ 2 }{   \sqrt {67} - 7 }  =  \frac{    \sqrt {67} + 7 }{9 } = 1 +  \frac{    \sqrt {67} - 2 }{9 } $$
$$    \frac{ 9 }{   \sqrt {67} - 2 }  =  \frac{    \sqrt {67} + 2 }{7 } = 1 +  \frac{    \sqrt {67} - 5 }{7 } $$
$$    \frac{ 7 }{   \sqrt {67} - 5 }  =  \frac{    \sqrt {67} + 5 }{6 } = 2 +  \frac{    \sqrt {67} - 7 }{6 } $$
$$    \frac{ 6 }{   \sqrt {67} - 7 }  =  \frac{    \sqrt {67} + 7 }{3 } = 5 +  \frac{    \sqrt {67} - 8 }{3 } $$
$$    \frac{ 3 }{   \sqrt {67} - 8 }  =  \frac{    \sqrt {67} + 8 }{1 } = 16 +  \frac{    \sqrt {67} - 8 }{1 } $$
My favorite tableau for a simple continued fraction:
$$ 
\small
 \begin{array}{cccccccccccccccccccccccccc}
 & & 8 & & 5 & & 2 & & 1 & & 1 & & 7 & & 1 & & 1 & & 2 & & 5 & & 16 & \\ 
 \\ 
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 8 }{ 1 }   & &   \frac{ 41 }{ 5 }   & &   \frac{ 90 }{ 11 }   & &   \frac{ 131 }{ 16 }   & &   \frac{ 221 }{ 27 }   & &   \frac{ 1678 }{ 205 }   & &   \frac{ 1899 }{ 232 }   & &   \frac{ 3577 }{ 437 }   & &   \frac{ 9053 }{ 1106 }   & &   \frac{ 48842 }{ 5967 }    \\ 
  \\ 
 & 1 & & -3 & & 6 & & -7 & & 9 & & -2 & & 9 & & -7 & & 6 & & -3 & & 1
 \end{array}
 $$
$$ 
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 67 \cdot 0^2 = 1 &     \mbox{digit}  &  8  \\  
  \frac{ 8 }{ 1 }   & 8^2 - 67 \cdot 1^2 = -3 &     \mbox{digit}  &  5  \\  
  \frac{ 41 }{ 5 }   & 41^2 - 67 \cdot 5^2 = 6 &     \mbox{digit}  &  2  \\  
  \frac{ 90 }{ 11 }   & 90^2 - 67 \cdot 11^2 = -7 &     \mbox{digit}  &  1  \\  
  \frac{ 131 }{ 16 }   & 131^2 - 67 \cdot 16^2 = 9 &     \mbox{digit}  &  1  \\  
  \frac{ 221 }{ 27 }   & 221^2 - 67 \cdot 27^2 = -2 &     \mbox{digit}  &  7  \\  
  \frac{ 1678 }{ 205 }   & 1678^2 - 67 \cdot 205^2 = 9 &     \mbox{digit}  &  1  \\  
  \frac{ 1899 }{ 232 }   & 1899^2 - 67 \cdot 232^2 = -7 &     \mbox{digit}  &  1  \\  
  \frac{ 3577 }{ 437 }   & 3577^2 - 67 \cdot 437^2 = 6 &     \mbox{digit}  &  2  \\  
  \frac{ 9053 }{ 1106 }   & 9053^2 - 67 \cdot 1106^2 = -3 &     \mbox{digit}  &  5  \\  
  \frac{ 48842 }{ 5967 }   & 48842^2 - 67 \cdot 5967^2 = 1 &     \mbox{digit}  &  16  \\  
 \end{array}
 $$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
I also did 73 which seems to be the only example in Perron(1913)
$$  \sqrt { 73} = 8 +     \frac{  \sqrt {73} - 8 }{ 1 }  $$
$$    \frac{ 1 }{   \sqrt {73} - 8 }  =  \frac{    \sqrt {73} + 8 }{9 } = 1 +  \frac{    \sqrt {73} - 1 }{9 } $$
$$    \frac{ 9 }{   \sqrt {73} - 1 }  =  \frac{    \sqrt {73} + 1 }{8 } = 1 +  \frac{    \sqrt {73} - 7 }{8 } $$
$$    \frac{ 8 }{   \sqrt {73} - 7 }  =  \frac{    \sqrt {73} + 7 }{3 } = 5 +  \frac{    \sqrt {73} - 8 }{3 } $$
$$    \frac{ 3 }{   \sqrt {73} - 8 }  =  \frac{    \sqrt {73} + 8 }{3 } = 5 +  \frac{    \sqrt {73} - 7 }{3 } $$
$$    \frac{ 3 }{   \sqrt {73} - 7 }  =  \frac{    \sqrt {73} + 7 }{8 } = 1 +  \frac{    \sqrt {73} - 1 }{8 } $$
$$    \frac{ 8 }{   \sqrt {73} - 1 }  =  \frac{    \sqrt {73} + 1 }{9 } = 1 +  \frac{    \sqrt {73} - 8 }{9 } $$
$$    \frac{ 9 }{   \sqrt {73} - 8 }  =  \frac{    \sqrt {73} + 8 }{1 } = 16 +  \frac{    \sqrt {73} - 8 }{1 } $$
Tableau:
$$ 
\tiny
 \begin{array}{cccccccccccccccccccccccccccccccc}
 & & 8 & & 1 & & 1 & & 5 & & 5 & & 1 & & 1 & & 16 & & 1 & & 1 & & 5 & & 5 & & 1 & & 1 & & 16 & \\ 
 \\ 
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 8 }{ 1 }   & &   \frac{ 9 }{ 1 }   & &   \frac{ 17 }{ 2 }   & &   \frac{ 94 }{ 11 }   & &   \frac{ 487 }{ 57 }   & &   \frac{ 581 }{ 68 }   & &   \frac{ 1068 }{ 125 }   & &   \frac{ 17669 }{ 2068 }   & &   \frac{ 18737 }{ 2193 }   & &   \frac{ 36406 }{ 4261 }   & &   \frac{ 200767 }{ 23498 }   & &   \frac{ 1040241 }{ 121751 }   & &   \frac{ 1241008 }{ 145249 }   & &   \frac{ 2281249 }{ 267000 }    \\ 
  \\ 
 & 1 & & -9 & & 8 & & -3 & & 3 & & -8 & & 9 & & -1 & & 9 & & -8 & & 3 & & -3 & & 8 & & -9 & & 1
 \end{array}
 $$
$$ 
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 73 \cdot 0^2 = 1 &     \mbox{digit}  &  8  \\  
  \frac{ 8 }{ 1 }   & 8^2 - 73 \cdot 1^2 = -9 &     \mbox{digit}  &  1  \\  
  \frac{ 9 }{ 1 }   & 9^2 - 73 \cdot 1^2 = 8 &     \mbox{digit}  &  1  \\  
  \frac{ 17 }{ 2 }   & 17^2 - 73 \cdot 2^2 = -3 &     \mbox{digit}  &  5  \\  
  \frac{ 94 }{ 11 }   & 94^2 - 73 \cdot 11^2 = 3 &     \mbox{digit}  &  5  \\  
  \frac{ 487 }{ 57 }   & 487^2 - 73 \cdot 57^2 = -8 &     \mbox{digit}  &  1  \\  
  \frac{ 581 }{ 68 }   & 581^2 - 73 \cdot 68^2 = 9 &     \mbox{digit}  &  1  \\  
  \frac{ 1068 }{ 125 }   & 1068^2 - 73 \cdot 125^2 = -1 &     \mbox{digit}  &  16  \\  
  \frac{ 17669 }{ 2068 }   & 17669^2 - 73 \cdot 2068^2 = 9 &     \mbox{digit}  &  1  \\  
  \frac{ 18737 }{ 2193 }   & 18737^2 - 73 \cdot 2193^2 = -8 &     \mbox{digit}  &  1  \\  
  \frac{ 36406 }{ 4261 }   & 36406^2 - 73 \cdot 4261^2 = 3 &     \mbox{digit}  &  5  \\  
  \frac{ 200767 }{ 23498 }   & 200767^2 - 73 \cdot 23498^2 = -3 &     \mbox{digit}  &  5  \\  
  \frac{ 1040241 }{ 121751 }   & 1040241^2 - 73 \cdot 121751^2 = 8 &     \mbox{digit}  &  1  \\  
  \frac{ 1241008 }{ 145249 }   & 1241008^2 - 73 \cdot 145249^2 = -9 &     \mbox{digit}  &  1  \\  
  \frac{ 2281249 }{ 267000 }   & 2281249^2 - 73 \cdot 267000^2 = 1 &     \mbox{digit}  &  16  \\  
 \end{array}
 $$
Note how we need to go twice through the "digits" to get to $1,$ as
$$      1068^2 - 73 \cdot 125^2 = -1    $$
See how Perron (1913) displays the beginning of the calculation for $\sqrt {73}$

Perron does (on just one page) show the diagram I have been calling a tableau. He calls it a Schema: