Conversion of $x^2-2xy-y^2-x+y=0$ to $Ax^2-BY^2=C$ to solve diophantine equation.

Question

Given $x^2-2xy-y^2-x+y$ we see that $\Delta=(-2)^2-4(1)(-1)=4+4=8>0$, therefore it is a hyperbola, I was required to convert it to $AX^2-BY^2=C$, so first I removed the $xy$ term by noticing that rotation of $-\pi/8$ converted it to $$\sqrt2x^2-\left(\frac{\sqrt{2+\sqrt2}+\sqrt{2-\sqrt2}}{2}\right)x-\sqrt2y^2+\left(\frac{\sqrt{2+\sqrt2}-\sqrt{2-\sqrt2}}{2}\right)y=0$$ Also now I found the center of the conic by using $\partial F/\partial x=\partial F/\partial y =0$ which gave me $$(x_0,y_0)=\left(\frac{\sqrt{2+\sqrt2}+\sqrt{2-\sqrt2}}{4\sqrt2},\frac{\sqrt{2+\sqrt2}+\sqrt{2-\sqrt2}}{4\sqrt2}\right)$$ Now I get by translation to origin by using $(x_0,y_0)$ $$x^2-y^2=\frac{\sqrt{2+\sqrt2}\sqrt{2-\sqrt2}}{8}=\frac{\sqrt2}{8}$$ But the RHS is not an integer, therefore how should we solve the diophantine equation $x^2-2xy-y^2-x+y$, since I saw that we needed to convert it to Pell type equation $X^2-DY^2=N$. Should I square both sides?

Answer 1

Discriminant of the quadratic (in terms of $x$ for example) must be a perfect square, thus $$(2y+1)^2-4(-y^2+y)=z^2$$or $$8y^2+1=z^2$$Its a Pell equation now.

Answer 2

$$x^2-2xy-y^2-x+y=0$$

We use the solutions of the equation Pell. $p^2-2s^2=\pm1$

Decisions can be recorded.

$$x=\pm{p(p+s)}$$

$$y=\pm{ps}$$

Solutions can be found. For $+1$ use the first solution $(p_1;s_1)-(3;2)$

For $-1$ use the first solution $(p_1;s_1)-(1;1)$

And the formula we find the following.

$$p_2=3p_1+4s_1$$

$$s_2=2p_1+3s_1$$

$p;s - $ can have any sign.