# Gauss and Regular Polygons: Conclusion

### The Central Result

This is the concluding post in this series and we aim to prove the following result (proved in part by Gauss and finally the converse by Wantzel):
A regular polygon of $n, n > 2$ sides can be constructed by Euclidean tools if and only if $\phi(n) = 2^{k}$.

Here $\phi(n)$ denotes the Euler's Totient function. It is clear that a regular polygon of $n$ sides can be constructed if and only if $\cos(2\pi / n)$ is constructible. To do so we find the minimal polynomial of $\xi = 2\cos(2\pi / n)$. We have already established that the minimal polynomial of $\zeta = \cos(2\pi / n) + i\sin(2\pi /n)$ is the $n^{th}$ cyclotomic polynomial $\Phi_{n}(z)$.

We also have $\xi = \zeta + \zeta^{-1}$ and we need to note that $\zeta^{-1}$ is a root of $\Phi_{n}(z) = 0$ so that the polynomial $z^{\phi(n)}\Phi_{n}(1 / z)$ is the same as $\Phi_{n}(z)$. Since $n > 2$ we see that $\phi(n)$ is even and we can write $\phi(n) = 2m$.

It therefore follows we can express $z^{-m}\Phi_{n}(z)$ can be expressed in the form $$(z^{m} + z^{-m}) + a_{1}(z^{m - 1} + z^{-(m - 1)}) + \cdots + a_{m - 1}(z + z^{-1}) + a_{m}$$ where the $a_{i}$ are integers. Since $$z^{k} + z^{-k} = (z + z^{-1})(z^{k - 1} + z^{-(k - 1)}) - (z^{k - 2} + z^{-(k - 2)})$$ we can express $z^{-m}\Phi_{n}(z)$ as a polynomial in $(z + z^{-1})$ of degree $m$ with integral coefficients only and moreover this polynomial will be monic. Let this polynomial be called $\Psi_{m}(z)$. Then $\Phi_{n}(z) = z^{m}\Psi_{m}(z + z^{-1})$.

From this relation it follows that $\Psi_{m}(z)$ is irreducible over rationals. For if it were not, we could write $\Psi_{m}(z) = f(z)g(z)$ where $f(z)$ is of degree $r < m$. Therefore \begin{align} \Phi_{n}(z) &= z^{m}\Psi_{m}(z + z^{-1})\notag\\ &= z^{m}f(z + z^{-1})g(z + z^{-1})\notag\\ &= \{z^{r}f(z + z^{-1})\}\{z^{m - r}g(z + z^{-1})\}\notag \end{align} which leads to a contradiction as $\Phi_{n}(z)$ is irreducible.

It is thus established that $\Psi_{m}(z)$ is the minimal polynomial of degree $m = \phi(n) / 2$ for $\xi$. And therefore $\xi = 2\cos(2\pi / n)$ is constructible only if $m = \phi(n) / 2$ is a power of $2$ i.e. when $\phi(n)$ is a power of $2$.

Therefore we have established the "only if" part of the main result. We now would like to establish the "if" part. To proceed we need to observe that $\phi(n) = 2^{k}$ if and only if $$n = 2^{l}p_{1}p_{2}\ldots p_{m}$$ where $p_{i}$ are distinct Fermat primes. If $n$ if of the form above then it is easily seen that $\phi(n)$ is a power of $2$. On the other hand if $n$ has the prime factorization $$n = p_{0}^{k_{0}}p_{1}^{k_{1}}p_{2}^{k_{2}}\ldots p_{m}^{k_{m}}$$ then $$\phi(n) = p_{0}^{k_{0} - 1}p_{1}^{k_{1} - 1}\ldots p_{m}^{k_{m} - 1}(p_{0} - 1)(p_{1} - 1) \ldots (p_{m} - 1)$$ and it is clear that $\phi(n)$ will be a power of $2$ only when one of the primes $p_{i}$ is $2$ and the exponents of other primes is $1$ and for those primes $(p_{i} - 1)$ is a power of $2$. Thus we must have $n$ in the desired form to have $\phi(n)$ a power of $2$.

In the expression of $n$ we can obviously ignore the factor $2^{l}$ because as we have seen, if a polygon of $n$ sides can be constructed by Euclidean tools then a polygon of $2n$ sides can also be so constructed. So we can assume that $n$ is a product of distinct Fermat primes. We now prove the following result:

If a regular polygon of $m$ sides and a regular polygon of $n$ sides can be constructed by Euclidean tools, and $m$ is coprime to $n$ then a regular polygon of $mn$ sides can also be constructed using Euclidean tools.
Proof: Since $m$ is coprime to $n$, we have integers $a, b$ such that $am + bn = 1$ and then we get \begin{align} \frac{a}{n} + \frac{b}{m} &= \frac{1}{mn}\notag\\ \Rightarrow a \cdot \frac{2\pi}{n} + b \cdot \frac{2\pi}{m} &= \frac{2\pi}{mn}\notag \end{align} Since we can construct angles $2\pi / m$ and $2\pi / n$ using Euclidean tools, we can therefore construct angle $2\pi / (mn)$ in this fashion and therefore a regular polygon of $mn$ sides is constructible using Euclidean tools only.

In the previous post we had established that if $p$ is a Fermat prime then a regular polygon of $p$ sides is constructible using Euclidean tools. It now follows that if $n$ is a product of distinct Fermat primes then a regular polygon of $n$ sides is constructible using Euclidean tools.

Therefore we have established the main result concerning the construction of regular polygons. Its now time to show the construction of a regular polygon of $17$ sides.

### Construction of Regular Polygon of $17$ Sides

We will solve the equation $$\Phi_{17}(z) = 1 + z + z^{2} + \cdots + z^{16} = 0$$ using the concept of Gaussian periods. Let $\alpha = 2\pi / 17$ so that $\zeta = \cos\alpha + i\sin\alpha$ is a root of the above cyclotomic equation. Since $3$ is a primitive root of $17$, we arrange $\zeta^{k}$ in the form $\zeta_{m} = \zeta^{3^{m}}$. The correspondence between $m$ and $k$ is as follows: \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline\\ m & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15\\ \hline\\ k & 1 & 3 & 9 & 10 & 13 & 5 & 15 & 11 & 16 & 14 & 8 & 7 & 4 & 12 & 2 & 6\\ \hline \end{array} The periods of $8$ terms are given by \begin{align} x_{1} &= \sum_{m\,\, \text{even}}\zeta_{m} = \zeta^{1} + \zeta^{9} + \zeta^{13} + \zeta^{15} + \zeta^{16} + \zeta^{8} + \zeta^{4} + \zeta^{2}\notag\\ x_{2} &= \sum_{m\,\, \text{odd}}\zeta_{m} = \zeta^{3} + \zeta^{10} + \zeta^{5} + \zeta^{11} + \zeta^{14} + \zeta^{7} + \zeta^{12} + \zeta^{6}\notag \end{align} and periods of $4$ terms are given by \begin{align} y_{1} &= \sum_{m \equiv 0\pmod 4}\zeta_{m} = \zeta^{1} + \zeta^{13} + \zeta^{16} + \zeta^{4}\notag\\ y_{2} &= \sum_{m \equiv 2 \pmod 4}\zeta_{m} = \zeta^{9} + \zeta^{15} + \zeta^{8} + \zeta^{2}\notag\\ y_{3} &= \sum_{m \equiv 1\pmod 4}\zeta_{m} = \zeta^{3} + \zeta^{5} + \zeta^{14} + \zeta^{12}\notag\\ y_{4} &= \sum_{m \equiv 3 \pmod 4}\zeta_{m} = \zeta^{10} + \zeta^{11} + \zeta^{7} + \zeta^{6}\notag \end{align} Since $\zeta^{k} + \zeta^{17 - k} = 2\cos k\alpha$, we get \begin{align} x_{1} &= 2(\cos\alpha + \cos 8\alpha + \cos 4\alpha + \cos 2\alpha)\notag\\ x_{2} &= 2(\cos 3\alpha + \cos 7\alpha + \cos 5\alpha + \cos 6\alpha)\notag\\ y_{1} &= 2(\cos\alpha + \cos 4\alpha)\notag\\ y_{2} &= 2(\cos 8\alpha + \cos 2\alpha)\notag\\ y_{3} &= 2(\cos 3\alpha + \cos 5\alpha)\notag\\ y_{4} &= 2(\cos 7\alpha + \cos 6\alpha)\notag \end{align} Clearly $x_{1} + x_{2} = -1$ (sum of roots) and if we multiply we will find that $x_{1}x_{2} = 4(x_{1} + x_{2}) = -4$ so that $x_{1}, x_{2}$ are roots of the equation $$x^{2} + x - 4 = 0\tag{1}$$ where $x_{1} > x_{2}$ (check this yourself). Again by multiplying we get $y_{1}y_{2} = -1$ and since $y_{1} + y_{2} = x_{1}$, so that $y_{1}, y_{2}$ are roots of $$y^{2} - x_{1}y - 1 = 0\tag{2}$$ where $y_{1} > y_{2}$. Similarly $y_{3}, y_{4}$ are roots of $$y^{2} - x_{2}y - 1= 0\tag{3}$$ and $y_{3} > y_{4}$.

Two periods of $2$ terms are $z_{1} = 2\cos\alpha$ and $z_{2} = 2\cos 4\alpha$ and then we have $$z_{1} + z_{2} = y_{1},\,\, z_{1}z_{2} = y_{3}$$ so that $z_{1}, z_{2}$ are roots of equation $$z^{2} - y_{1}z + y_{3} = 0\tag{4}$$ and $z_{1} > z_{2}$.

Solving the above equations and taking care of the inequalities involved we can see that $z_{1}$ is given by the somewhat cumbersome expression \begin{align} &2\cos\left(\frac{2\pi}{17}\right)\notag\\ &\,\,\,\,= \frac{-1 + \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2\sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2\sqrt{34 + 2 \sqrt{17}}}}{8}\notag \end{align} We should also note that the inequalities which differentiate the above roots are obtained using non-algebraic means. Therefore obtaining an expression of $2 \cos (2 \pi / 17)$ is not entirely an algebraic problem. In fact, from the view point of algebra, all the roots of $\Psi_{8}(z) = 0$ are interchangeable.

We now show the construction of a regular polygon of $17$ sides graphically and then explain the steps.

Fig 1. Construction of Regular 17-gon

We start with a circle and two perpendicular radii $OA$ and $OB$. Then a point $C$ is found on $OB$ such that $OC = OB / 4$. Another point $D$ on $OA$ is found such that $\angle OCA = 4 \angle OCD$. A point $E$ is located on $OA$ extended towards $O$ such that $\angle DCE = \pi / 4$. A semicircle with $AE$ as diameter is drawn which intersects $OB$ in $F$. A circle is drawn with center $D$ and radius $DF$ which intersects $OA$ in $G$ and $H$. Perpendiculars to $OA$ are drawn from points $G$ and $H$ which meet the initial circle (with center $O$ and radius $OA$) at points $P_{3}$ and $P_{5}$. Bisecting the arc $P_{3}P_{5}$ we get $P_{4}$. Now $P_{3}P_{4}$ is the side of the desired regular polygon and the construction can be completed easily by cutting arcs of that length starting from point $A = P_{0}$.

The reason the above construction works is not difficult to explain. Let us then assume that $\beta$ is an acute angle such that $\tan 4\beta = 4$, so that $4\cot 4\beta = 1$. Thus equation (1) for $x_{1}, x_{2}$ can be written as $$x^{2} + 4x\cot 4\beta - 4 = 0$$ so that the roots are $x_{1} = 2\tan 2\beta$ and $x_{2} = - 2\cot 2\beta$. Putting these in equations $(2)$ and $(3)$ we get \begin{align} y_{1} &= \tan(\beta + \pi / 4)\notag\\ y_{2} &= \tan(\beta - \pi / 4)\notag\\ y_{3} &= \tan\beta\notag\\ y_{4} &= - \cot\beta\notag \end{align} and therefore we have \begin{align} 2\cos 3\alpha + 2\cos 5\alpha &= y_{3} = \tan\beta\notag\\ 2\cos 3\alpha \cdot 2\cos 5\alpha &= 2\cos 2\alpha + 2\cos 8\alpha\notag\\ &= y_{2} = \tan(\beta - \pi / 4)\notag \end{align} In the given construction we have $\angle OCA = 4\beta, \angle OCD = \beta$ and $\angle OCE = \pi / 4 - \beta$. Let us assume $OA = 1$ so that we have \begin{align} 2\cos \angle AOP_{3} + 2\cos \angle AOP_{5} &= 2 (OG - OH)\notag\\ &= 2(OD + DG) - 2(DH - OD)\notag\\ &= 4OD\text{ (since } DH = DG)\notag\\ &= OD / (1 / 4) = OD / OC\notag\\ &= \tan\beta\notag \end{align} and \begin{align} 2\cos \angle AOP_{3} \cdot 2\cos \angle AOP_{5} &= -4 OG \cdot OH\notag\\ &= -4 OF^{2}\notag\\ &= -4 OE \cdot OA\notag\\ &= -OE / (1 / 4) = -OE/OC\notag\\ &= \tan(\beta - \pi / 4)\notag \end{align} It follows that $\angle AOP_{3} = 3\alpha$ and $\angle AOP_{5} = 5\alpha$.

Note: The construction of regular 17-gon as well as its proof is taken from "An Introduction to the Theory of Numbers" by G. H. Hardy and E. M. Wright. The construction is originally by Richmond and till date is the simplest one available. See MathWorld page for details. Also the wikipedia entry for heptadecagon gives another construction (steps are shown in an animated GIF figure), but it has relatively greater number of steps.

#### 2 comments :: Gauss and Regular Polygons: Conclusion

1. Interesting you didnt mention the term “Heptadecagon”. http://en.wikipedia.org/wiki/Heptadecagon has a step by step animation of how you can actually draw the Heptadecagon.

Here is another interesting link: http://mathworld.wolfram.com/GeometricConstruction.html

2. SD, the post has been updated and references to Wikipedia and MathWorld have been provided. Thanks for providing the references.