So, I was doing some exercises and one of them was about a regular triangle inscribed in a parabola given by, naturally:
$f(x)=ax^2+bx+c$, with $a > 0$ and $\Delta>0$.
This question later got me thinking of how I could possibly solve it if, instead of a triangle, I had a square, let's say, or a regular pentagon. I went to Geogebra, started plotting some stuff and realized it seems to be impossible to inscribe any regular polygon with number of sides n > 3 to a given parabola of the mentioned form.
I looked it up on the internet, but couldn't find anything satisfactory. How can I prove that it actually is impossible? I mean, if it is so.
$n=4$ is not possible. For any square (not aligned with the axes), one vertex has its $y$ coordinate greater than those of the other vertices, and its $x$ coordinate between those of the other vertices. If those three vertices are on the parabola, then the fourth cannot be.
Specifically, if $(x_L,y_L)$, $(x_R,y_R)$, and $(x_U,y_U)$ are the leftmost, rightmost, and uppermost vertices (respectively), then
$$x_L<x_U<x_R,\quad y_L<y_U>y_R.$$
We assume that $y_L=x_L^2$ and $y_R=x_R^2$. Then there are two cases to consider. The first case has $x_U\geq0$, so we can square the inequality $x_R>x_U$:
$$y_U>y_R=x_R^2>x_U^2$$
The second case has $x_U\leq0$, so we can square the inequality $-x_L>-x_U$:
$$y_U>y_L=x_L^2>x_U^2$$
In either case we get $y_U>x_U^2$, so the uppermost vertex is not on the parabola.
The same reasoning works for $n\geq5$: Consider the uppermost vertex, and the two adjacent vertices to its left and right. (Since the polygon's angle is greater than $90^\circ$, we do actually have $x_L<x_U<x_R$. This wouldn't be true for $n=3$.) Or you could use Calvin Lin's hint.