How Can I Solve $2x^2+3xy+8y^2\equiv0\pmod{13}$

Question

This problem is on Pg 135 of Adams and Goldstein's "Introduction To Number Theory". They give the answer but they don't show how to solve it. They also treat it as a Diophantine equation,so how can one get from a listing of congruence pairs i.e., $(x,y) =(0,0)$, etc., for all 169 possibilities to a set of parametric equations (which they do give)?

Answer 1

HINT:

If $13|x\iff13|y$

Else $(x,13)=1\iff(y,13)=1$

$$0\equiv2x^2+3xy+8y^2\equiv2(x^2+8xy+4y^2)\iff(x+4y)^2\equiv12y^2\equiv-y^2$$

$$\iff\left(\dfrac{x+4y}y\right)^2\equiv-1$$

Now $(\pm5)^2\equiv-1\pmod{13}\implies\dfrac{x+4y}y\equiv\pm5\pmod{13}$

$\implies x\equiv y$ or $-9y\equiv 4y\equiv x\pmod{13}$

Answer 2

By completing the square, just as you would over the real numbers. Computing modulo $13$, $$\begin{align} 2x^2+3xy+8y^2&\equiv 2(x^2+8xy+4y^2)\equiv2((x+4y)^2+y^2)\\ &\equiv2((x+4y)^2-25y^2)\equiv 2((x+4y)^2-(5y)^2)\\ &\equiv 2(x-y)(x+9y)\equiv2(x-y)(x-4y). \end{align} $$ So the solutions are the $(x,y)$ with $x\equiv y$ or $x\equiv 4y\pmod{13}$.