So i have the curve:

$\displaystyle \gamma(t) =(t^2-t+1, t^2-t)$

And i need to show that it has one self-intersection point.

To me, it is obviously at 0 and 1, as at these points $\displaystyle \gamma$ is equal to zero.

Now, the thing is, i think i must have missed this lecture. I can't find any reference in my differential geometry book, and my google fu is failing me.

Can anyone show me the method on how to show self-intersection points?

I'll solve the question once i know how.

Thus far, i've been unable to figure out my own way. These damn parametric equations upset me so.

so far i've though maybe if

$\displaystyle \gamma(t) = \gamma(u)$

where

$\displaystyle (t^2-t+1, t^3-t) = (u^2-u+1, u^3-u)$

but yeh.

Halp?