Show that $x^2 - 79y^2 = \pm 3$ has no solutions in $\Bbb Z$

Question

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.

Answer 1

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:
$$ \begin{array}{cccccccccccccc} & & 8 & & 1 & & 7 & & 1 & & 16 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 8 }{ 1 } & & \frac{ 9 }{ 1 } & & \frac{ 71 }{ 8 } & & \frac{ 80 }{ 9 } \\ \\ & 1 & & -15 & & 2 & & -15 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 79 \cdot 0^2 = 1 & \mbox{digit} & 8 \\ \frac{ 8 }{ 1 } & 8^2 - 79 \cdot 1^2 = -15 & \mbox{digit} & 1 \\ \frac{ 9 }{ 1 } & 9^2 - 79 \cdot 1^2 = 2 & \mbox{digit} & 7 \\ \frac{ 71 }{ 8 } & 71^2 - 79 \cdot 8^2 = -15 & \mbox{digit} & 1 \\ \frac{ 80 }{ 9 } & 80^2 - 79 \cdot 9^2 = 1 & \mbox{digit} & 16 \\ \end{array} $$