# Constructibility of an inscribed square using Galois Theory

A problem in my Galois theory syllabus in the chapter on constructible numbers is as follows:

Check if the small square if constructible from the big square. [Hint: Choose coordinates $$0,1\in\mathbf{C}$$ as in the picture, and find $$z=a+bi$$.]

I found two equations from which I can solve $$a$$ and $$b$$. Adding up the area's of the four congruent triangles and the small square, we get $$1=4\cdot \frac{1}{2}b+(1-a)^2$$. Using similar triangles, I can extract a second relation, which is $$\frac{1}{\sqrt{b^2+(1-a)^2}}=\frac{\sqrt{b^2+(1-a)^2}}{1-a}$$ or $$b^2=a(1-a)$$.

This gives the following system of equations to solve:

$$\begin{cases} 2b+(1-a)^2=1 \\ b^2=a(1-a) \end{cases}$$

I tried finding the real solution of this system of equations, in vain, which after seeing the solution from Mathematica is not such a surprise.

The minimal polynomial of $$z=a+bi$$ is, according to Mathematica, $$X^3 - 4 X^2 + 6 X - 2$$. This polynomial is Eisenstein with $$p=2$$, so irreducible. By Galois theory, we can then conclude that $$z$$ is not constructible.

My question is: how can I find either a closed expression for $$z$$ or it's minimal polynomial without a ton of calculations or use of Mathematica? The author of this problem in my syllabus obviously intended an approach by hand.

We can get the polynomial which $$a$$ satisfies: $$4a(1 - a) = 4b^2 = (1 - (1 - a)^2)^2$$ and thus $$4(1 - a) = a(2 - a)^2$$, equivalently: $$a^3 - 4a^2 + 8a - 4 = 0$$ which is irreducible $$\mathrm{mod}$$ $$3$$ (because it got no root $$\mathrm{mod}$$ $$3$$ ). So we can't construct $$a$$. But $$z$$ is constructible if and only if both $$a, b$$ are.
• Suppose $a$ and $b$ are constructible. Then $bi$ is a $90°$ counterclockwise rotation of $b$ and adding that vectorially to $a$ gives $a+bi$. Now suppose $a+bi$ is constructive. Drop a perpendicular from there to the real axis giving $a$ as the intersection. The vector from $a$ to $a+bi$ is $bi$ which is rotated $90°$ clockwise to resolve the real quantity $b$. – Oscar Lanzi May 16 '19 at 14:01