Given parabola $x^2$ on plane, how can I construct 7th roots of unity?
I was straying with my only idea, that sum of squares of real and imaginary part of roots of 1 equals 1 and belongs to $\mathbb{Q}$, but it leads me nowhere.
Also I know that group of 7th roots of unity is cyclic and isomorphic to $\mathbb{Z}_7$, but I think it won't be useful here.
Start with the fact that the algebraic integers $2\cos(2k\pi/7), k\in\{1,2,3\}$ solve the equation $x^3+x^2-2x-1=0$ (see here, for instance).
Begin by multiplying this equation by $x-1$, turning it into a depressed quartic:
$x^4-3x^2+x+1=0$
Now from the given parabola, $x^4=y^2$. With this $y^2$ term we will begin the process of defining a circle by "borrowing" $x^2$ from the quadratic term in $x$:
$y^2-3x^2+x+1=0$
$(x^2+y^2)-4x^2+x+1=0$
Next eliminate the $4x^2$ term in favor of $4y$, and now you can use completing the square to define the circle:
$(x^2+y^2)-4y+x+1=0$
$(x^2+x)+(y^2-4y)=-1$
$\color{blue}{(x+\frac12)^2+(y-2)^2=\frac{13}4}$
Thus, identify the point $(-\frac12,2)$ by usual means and draw a circle centered there through the point $(1,1)$ on the parabola (corresponding to the auxiliary root $x=1$). The circle will intersect the parabola at three other points corresponding to $x=2\cos(2k\pi/7), k\in\{1,2,3\}$. From there the construction of the actual seventh roots as points on the coordinate plane (now reinterpreted as the complex number plane) is straightforward.
This method can be used to solve any cubic equation. For example, if we start with $x^3=2$, the steps above lead to $(x-1)^2+(y-\frac12)^2=\frac54$, so you draw the circle centered at $(1,\frac12)$ and passing through the origin to get $x^3=2$ at the other intersection. Ultimately, because quartic equations may be rendered from resolvent cubics, this shows that all solid constructions are derivable using the parabola $y=x^2$ as the only noncircular conic section.
