How to find integral solutions of a integral binary quadratic form?

Question

How to find all integral solutions of $f(x, y) = 2x^2-xy-3y^2=8$? I know that the solutions of this equation are in a bijection with $\alpha \in M_f$ where $N(\alpha) = 16$. But I don't know how to use this fact to solve the problem. Can anyone help me?

Answer 1

$$ (x+y)(2x-3y) = 8 $$

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Answer 2

$$2x^2-xy-3y^2-8=0 \implies 3y^2+xy-2x^2+8=0 \implies y=\frac{-x\pm \sqrt{25x^2-96}}{6}$$ Let $$25x^2-96=m^2 \implies 25x^2-m^2=96 \implies (5x+m)(5x-m)=96$$ Factorise $$96=48\times 2 \implies 5x+m=48, 5x-m=2 \implies x=5, m=23 \implies y= 3$$ Simlarly, $$96=12\times 8 \implies 5x+m=12, 5x-m=8\implies x=2, m=2 \implies y=0.$$

Answer 3

Above equation shown below:

$ 2x^2-xy-3y^2=8$ -------$(1)$

Knowing a numerical solution (p,q) to equation (1)

another numerical solution can be arrived at by

using the formula below:

$x=(1/2)*(5p-7q)$

$y=(1/2)*(3p-5q)$

for known, $(p,q)=[(14/4),(11/6)]$ we get another

solution after substituting in the above:

$(x,y)=[(7/3),(4/3)]$