Find where a curve crosses itself?

Question

I have the curve $x=t^2,y=t^3-4t$. I made a $t_2$ such that $t_2>t_1$ and from $x$ found that $0=t_1^2 -t_2^2$, from here I solved for values by basic guess and test and then subbed them into the y equation to see which values worked for both. While it did give me the right answer ($t_1 = -2$, $t_2 =2$) this way seems like it would be very difficult given more complex equations. Is there a more systematic approach?

Answer 1

Suppose $t^2 = s^2$ and $t^3 - 4 t = s^3 - 4 s$. Since $t^2 - s^2 = (t+s)(t-s) = 0$ we have (if $t \ne s$) $t = -s$. Then $-s^3 + 4 s = s^3 - 4 s$ so $0 = s^3 - 4 s = s (s^2 - 4) = s (s-2)(s+2)$, so $s = 0$ or $\pm 2$.

EDIT: More generally, suppose you want to know where the curve $x = f(t)$, $y = g(t)$ intersects itself, where $f$ and $g$ are polynomials. Thus you want to solve $f(t) - f(s) = 0$, $g(t) - g(s) = 0$. As polynomials in $s$ and $t$, $f(t) - f(s)$ and $g(t) - g(s)$ are divisible by $t-s$, so let $f(t) - f(s) = (t-s) h(s,t)$ and $g(t) - g(s) = (t-s) k(s,t)$. The resultant of $h(s,t)$ and $k(s,t)$ with respect to $t$ is a polynomial $r(s)$. The values of $s$ at which the polynomial intersects itself will be found among the roots of $r(s)$.