TITLE: Show that $x^2 - 79y^2 = \pm 3$ has no solutions in $\Bbb Z$
QUESTION [5 upvotes]: I am trying to compute the ideal class group of $\Bbb Q(\sqrt{79})$ and this came up:
"Show that $x^2 - 79y^2 = \pm 3$ has no solutions in $\Bbb Z$ using congruences."
I've tried using quadratic reciprocity up to $p=11$ but I've not had any success. Hints are welcome.

REPLY [4 votes]: Note: there is no proof using only congruences; $-3$ is represented by a form in the principal genus. The forms below are reduced in the Gauss-Lagrange sense, $ax^2 + bxy +cy^2$ coefficients $\langle a,b,c \rangle$ such that $ac<0$ and $b > |a+c|$
316    factored   2^2 *  79

    1.             1          16         -15   cycle length             4
    2.            -1          16          15   cycle length             4
    3.             3          16          -5   cycle length             6
    4.            -3          16           5   cycle length             6
    5.             5          16          -3   cycle length             6
    6.            -5          16           3   cycle length             6

  form class number is   6

Any squarefree number $n$ with $|n| < \sqrt {79}$ is represented by $x^2 - 79 y^2$ if and only if it appears with $x/y$ as a convergent of the continued fraction for $\sqrt {79}.$ The numbers that qualify are $1,2.$ The slightly larger $-15$ also comes up
$$  \sqrt { 79} = 8 +     \frac{  \sqrt {79} - 8 }{ 1 }  $$
$$    \frac{ 1 }{   \sqrt {79} - 8 }  =  \frac{    \sqrt {79} + 8 }{15 } = 1 +  \frac{    \sqrt {79} - 7 }{15 } $$
$$    \frac{ 15 }{   \sqrt {79} - 7 }  =  \frac{    \sqrt {79} + 7 }{2 } = 7 +  \frac{    \sqrt {79} - 7 }{2 } $$
$$    \frac{ 2 }{   \sqrt {79} - 7 }  =  \frac{    \sqrt {79} + 7 }{15 } = 1 +  \frac{    \sqrt {79} - 8 }{15 } $$
$$    \frac{ 15 }{   \sqrt {79} - 8 }  =  \frac{    \sqrt {79} + 8 }{1 } = 16 +  \frac{    \sqrt {79} - 8 }{1 } $$ 
Simple continued fraction tableau:
$$ 
 \begin{array}{cccccccccccccc}
 & & 8 & & 1 & & 7 & & 1 & & 16 & \\ 
 \\ 
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 8 }{ 1 }   & &   \frac{ 9 }{ 1 }   & &   \frac{ 71 }{ 8 }   & &   \frac{ 80 }{ 9 }    \\ 
  \\ 
 & 1 & & -15 & & 2 & & -15 & & 1
 \end{array}
 $$ 
$$ 
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 79 \cdot 0^2 = 1 &     \mbox{digit}  &  8  \\  
  \frac{ 8 }{ 1 }   & 8^2 - 79 \cdot 1^2 = -15 &     \mbox{digit}  &  1  \\  
  \frac{ 9 }{ 1 }   & 9^2 - 79 \cdot 1^2 = 2 &     \mbox{digit}  &  7  \\  
  \frac{ 71 }{ 8 }   & 71^2 - 79 \cdot 8^2 = -15 &     \mbox{digit}  &  1  \\  
  \frac{ 80 }{ 9 }   & 80^2 - 79 \cdot 9^2 = 1 &     \mbox{digit}  &  16  \\  
 \end{array}
 $$