Question: Find a rational parametrization of the conic whose equation in homogeneous coordinates is:
$x^2+y^2-xy-z^2=0$
Hence find all rational numbers $x, y$ such that:
$x^2+y^2-xy=1$
It seems I don't quite understand the rational parametrization lecture notes. Can someone show me how to deal with this example so that I see how to approach those type of problems? I think I am familiar with the basic concepts
Basically what I've done is to reduce it to $-z_1^2+\frac{1}{2}z_2^2+\frac{3}{2}z_3^2=0$
For a parametrization of the affine curve you need a rational point, say $(-1,0)$. Then a line through $(-1,0)$ will intersect the conic in another point. For the line through $(-1,0)$ take $y=t(x+1)$ as the equation. Substitute the equation of the line into the equation of the affine conic. so $$x^2 + (t(x+1))^2-x(t(x+1))=1$$ After expanding and collecting terms we find $$x^2(1-t+t^2)+(2t^2-t)x+(t^2-1)=0$$ Divide through by $1-t+t^2$ to finally get $$x^2+\frac{(2t^2-t)}{(1-t+t^2)}x+\frac{(t^2-1)}{(1-t+t^2)}=0$$ We know one root of the quadratic is $-1$, and the product of the roots must be $\frac{(t^2-1)}{(1-t+t^2)}$, so we conclude $$x= \frac{(1-t^2)}{(1-t+t^2)}$$ Returning to the equation of the line $y=t(x+1)$ to find $y$ gives $$y = \frac{t(2-t)}{(1-t+t^2)}$$ So we have a rational parametrization of the conic in terms of the rational parameter $t$.