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 $z=a+bi$.]

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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.

enter image description here

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.

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  • $\begingroup$ Thanks! Can you elaborate on your last statement? $\endgroup$ – rae306 May 16 '19 at 13:44
  • $\begingroup$ 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 $b$. $\endgroup$ – Oscar Lanzi May 16 '19 at 14:01
  • $\begingroup$ @RybinDmitry your equation is not correct $\endgroup$ – rae306 May 17 '19 at 6:37

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