Construction of Regular 17-gon: Gauss

June 1, 1796 Jenenser Intelligenzblatt : “Every beginner in geometry knows that it is possible to construct different regular polygons [with compass and straightedge], for example triangles, pentagons, 15-gons, and those regular polygons that result from doubling the number of sides of these figures. One had already come this far in Euclid’s time, and it seems that since then one has generally believed that the field for elementary geometry ended at that point, and in any case I do not know of any successful attempt to extend the boundaries beyond that line. Therefore it seems to me that this discovery possesses special interest, that besides these regular polygons, a number of others are geometrically constructible, for example the 17-gon.” -Gauss

Gauss proved this result Disquistiones Arithmeticae, and in general, he proves the constructibility of the $\displaystyle n$ -gon for any $\displaystyle n$ that is a prime of the form $\displaystyle 2^{2^k} + 1$ . (called Fermat’s primes).

Geometry and Arithmetic! : What all lengths can be constructed from a straight edge is precisely described by addition, subtraction, multiplication, division, and extracting square roots starting from unit length. Geometry to Algebra –This is easy to see from coordinate viewpoint, all the constructions involve intersecting circles and lines described by equations of degree at most 2. Algebra to Geometry — The fact that all of these operations can be constructed can be seen by explicit constructions. For instance to construct square root of $\displaystyle x$ is equivalent to constructing geometric mean of lengths $\displaystyle 1$ and $\displaystyle x$ .

Construction of Regular n-gons: To construct a regular n-gon, we need to be able to split a circle into n parts and thus construct the angle $\displaystyle \frac{2\pi}{n}.$ This in terms is equivalent to construction of the coordinates $\displaystyle \cos (\frac{2\pi}{n})$ and $\displaystyle \sin (\frac{2\pi}{n})$ (In fact, it’s enough to construct one of the coordinates)

$n=3$ : Equilateral triangle is easy to construct- it’s easy to see this is equivalent to construction of $\displaystyle \cos (\frac{2\pi}{3}) =-1/2.$ . (So we just need to construct midpoints).

$n=4$ : Square is easy. We need to draw perpendiculars.

$n=5$ : Regular Pentagon requires the constructions of the angle $\displaystyle \frac{2\pi}{5}. .$
$\displaystyle \frac{\pi}{5}$ is easy to construct (hence regular 10-gon!)– Observe that in an isosceles triangle on unit length base and base angles &fg=000000 $\displaystyle\frac{2\pi}{5})$ , the sides are of length $\displaystyle x=\frac{\sqrt 5 - 1}{2}$ . By constructing this length $\displaystyle x$ , we can construct the triangle and hence the angle $\displaystyle \frac{\pi}{5}$ . (we can construct $\displaystyle x$ because you need basic arithmetic and just square roots).
Once you have $\displaystyle \frac{\pi}{5}$ , doubling it we get $\displaystyle \frac{2\pi}{5}$ !

$n=6$ : Regular Hexagon is similar to $\displaystyle n=3$ and constructing $\displaystyle \frac{\pi}{3} =60^\circ.$

$n=7$ is impossible.

$n=8$ is obtained by doubling (bisecting) the square.

$n=9$ is impossible. (How do we prove that it’s impossible to construct? We switch to algebra/arithmetic and show that such angles/the coordinates cannot be obtained by basic algebra and square-root extractions – and we achieve that by studying the symmetries of quantities we construct! Galois theory)

Roots of unity: Constructing the coordinates $\displaystyle \cos (\frac{2\pi}{n})$ and $\displaystyle \sin (\frac{2\pi}{n})$ is equivalent to construction of the root of unity $\displaystyle \zeta_n =e^{\frac{2\pi i}{n}},$ which satisfies $\displaystyle z^n=1.$ Thus in terms of algebra, we should be able to solve this equation $\displaystyle z^n=1$ by a sequences of operations just involving square root extractions (In fancy modern terms, the roots should lie in the “field” obtained by a sequence of “quadratic extensions”)

Example:

For $\displaystyle n=3$ we obtain the equation

$\displaystyle z^3-1= (z-1)z^{2}+z+1=0$ , with roots
$\displaystyle 1, \zeta_3=-\frac{1}{2}+i \frac{\sqrt{3}}{2}, \quad \zeta_3^2=-\frac{1}{2}-i \frac{\sqrt{3}}{2}$

For $\displaystyle n=5$ , we need to solve

$\displaystyle z^5-1=(z-1) (z^4+z^3+z^2+z+1).$

Let’s focus on $\displaystyle z^4+z^3+z^2+z+1=0.$ Dividing by $z^2$ , we get

$\displaystyle z^{2}+\frac{1}{z^{2}}+z+\frac{1}{z}+1=0.$

$\displaystyle \left(z+\frac{1}{z}\right)^{2}+\left(z+\frac{1}{z}\right)-1=0$

Makign the substitution $\displaystyle w=z+\frac{1}{z},$ we have

$\displaystyle w^{2}+w-1=0,$

with roots

$\displaystyle w=\frac{-1 \pm \sqrt{5}}{2}$

Thus we see that $\displaystyle w$ is a quadratic irrational (thus constructible).

Now once we have $\displaystyle w$ , we can solve the equation $\displaystyle z+\frac{1}{z} =w$ . Each values of $\displaystyle w$ gives a quadratic equation in $\displaystyle z,$ so $\displaystyle z$ itself is constructible!

Regular $\displaystyle 17$ -gon: Gauss shows once can construct regular 17-gon by the same logic as above by solving $\displaystyle z^{16}+z^{15}+z^{14}+\cdots+z+1=0.$ But what substitution do we need to make sure that we always get quadratic equations? This is where the beauty lies, and Gauss introduces Gauss Sums to do the computations. The roots of this equation are $\displaystyle \zeta, \zeta^2, \cdots \zeta^{16}$ . The quantities in modern terms will be sums over “subgroups” of the Galois group $\displaystyle (\mathbb Z/17\mathbb Z)^{*}$ , but Gauss does it by using primitive roots to identify $\displaystyle (\mathbb/\mathbb 17Z)^{*}$ with $\displaystyle \mathbb Z/16\mathbb Z.$ Concretely for each $k =1, 2, \cdots 16$ consider $\displaystyle l \in \{0, 1, 2, \cdot 15\} \text{such that} k =3^l \mod 17.$ (3 is a primitive element modulo 1, any primitive element works)

Define the quantities $\displaystyle r_l =\zeta^{3^l}$ . That is we have

$\displaystyle \begin{array}{|l|l|l|l|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline \ell &0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \hline k & 1 & 3 & 9 & 10 & 13 & 5 & 15 & 11 & 16 & 14 & 8 & 7 & 4 & 12 & 2 & 6 \\ \hline \end{array}$

$\displaystyle k\to l: (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16) \to (0, 14, 1, 12, 5, 15, 11, 10, 2, 3, 7, 13, 4, 9, 6, 8)$

So $\displaystyle r_0 = \zeta, r_1 = \zeta^3, r_2=\zeta^9, \cdots r_15 =\zeta^3{^15}=\zeta^6$

Now consider the sums even and odd sums (with respect to the labelling in $\displaystyle r_l$ )

$\displaystyle \sigma_{2,0}=r_{0}+r_{2}+r_{4}+\cdots+r_{14} = \zeta +\zeta^9+\zeta^{13}+\zeta^{15}+\zeta^{16}+\zeta^8+ \zeta^4+\zeta^2$
$\displaystyle \sigma_{2,1}=r_{1}+r_{3}+r_{5}+\cdots+r_{15} = \zeta^3 +\zeta^{10}+\zeta^{5}+\zeta^{11}+\zeta^{14}+\zeta^7+ \zeta^{12}+\zeta^6$

Observe that

$\displaystyle \sigma_{2,0}+\sigma_{2,1}=r_{0}+r_{1}+\cdots+r_{15}= \zeta+\zeta^2+\cdots+\zeta^{16}=-1 .$

$\displaystyle \sigma_{2,0}\cdot \sigma_{2,1}=4(r_{0}+r_{1}+\cdots+r_{15})= 4(\zeta+\zeta^2+\cdots+\zeta^{16})=-4 .$

Hence we can see that $\displaystyle \sigma_{2,0}, \sigma_{2,1}$ satisfy $X^{2}+X-4=0$ and deduce that

$\displaystyle \sigma_{2,0}=\frac{-1 + \sqrt{17}}{2}$
$\displaystyle \sigma_{2,1}=\frac{-1 -\sqrt{17}}{2}$

are both quadratic irrationals (constructible). (Note that we need to chose the signs properly by comparing $\sigma_{2,0}, \sigma_{2,1}$ )

Next, we consider progressions which increase by 4 instead of 2. That is

$\displaystyle \sigma_{4,0}=r_{0}+r_{4}+r_{8}+r_{12}$
$\displaystyle \sigma_{4,1}=r_{1}+r_{5}+r_{9}+r_{13}$
$\displaystyle \sigma_{4,2}=r_{2}+r_{6}+r_{10}+r_{14}$
$\displaystyle \sigma_{4,3}=r_{3}+r_{7}+r_{11}+r_{15}$

$\displaystyle \sigma_{4,0}+\sigma_{4,2}=\sigma_{2,0}$
$\displaystyle \sigma_{4,1}+\sigma_{4,3}=\sigma_{2,1}$
$\displaystyle \sigma_{4,0} \cdot \sigma_{4,2}=-1$
$\displaystyle \sigma_{2,0}+\sigma_{2,1}=-1$

$\displaystyle \sigma_{4,0}$ and $\displaystyle \sigma_{4,2}$ are roots of $\displaystyle X^{2}-\sigma_{2,0} X-1=0 ,$ hence by solving, noting that $\displaystyle \sigma_{4,0}>\sigma_{4,2}$ , we get

$\displaystyle \sigma_{4,0}=\frac{1}{4}(\sqrt{17}-1+\sqrt{34-2 \sqrt{17}})$
$\displaystyle \sigma_{4,2}=\frac{1}{4}(\sqrt{17}-1-\sqrt{34-2 \sqrt{17}}) .$

Similarly $\displaystyle \sigma_{4,1}$ and $\displaystyle \sigma_{4,3}$ are roots of $\displaystyle X^{2}-\sigma_{2,1} X-1=0 ,$ hence by solving, noting that $\displaystyle \sigma_{4,1}>\sigma_{4,3}$ , we get

$\displaystyle \sigma_{4,1}=\frac{1}{4}(-\sqrt{17}-1+\sqrt{34+2 \sqrt{17}})$ ,
$\displaystyle \sigma_{4,3}=\frac{1}{4}(-\sqrt{17}-1-\sqrt{34+2 \sqrt{17}}) .$

We move to larger progressions:

$\displaystyle \sigma_{8,0}=r_{0}+r_{8}=\zeta+\zeta^{16}$
$\displaystyle \sigma_{8,4}=r_{4}+r_{12}=\zeta^4+\zeta^{13}$

satisfy

$\displaystyle X^{2}-\sigma_{4,0} X+\sigma_{4,1}=0$

and $\displaystyle \sigma_{8,0} = \zeta+\zeta^{16} =2\cos{\frac{2pi}{17}}$ is the larger root of the equation given by

$\displaystyle \frac{1}{8}(\sqrt{17}-1+\sqrt{34-2 \sqrt{17}})+\frac{1}{4} \sqrt{17+3 \sqrt{17}-\sqrt{170+38 \sqrt{17}}}$

We are done here because we expressed the cosine of the angle as nested expressions in terms of square roots!

Remarks:

Once can generalize the argument from $\displaystyle 17$ to primes of the form $\displaystyle p=2^{2^k}+1.$ We just need $\displaystyle 17-1=2^{2^2}.$
General $\displaystyle n$ : Similar arguments dealing with the roots of unity show that regular $\displaystyle n-$ gon is constructible iff and only if $\displaystyle n$ is of the form $\displaystyle n=2^{k} p_{1} p_{2} \ldots p_{l}$ , where $\displaystyle p_{1}, p_{2}, \ldots, p_{l}$ are distinct Fermat primes that is $\displaystyle p=2^{2^k}+1$
Note that the identification with $\displaystyle r_l$ using primitive elements is crucial! How would you dream of considering those special sums in terms of $\displaystyle \zeta^k$ . In modern terms, we are considering sums over subgroups (and their cosets)

Actual constructions: We showed that it’s possible to construct a regular heptadecagon, but how would you construct it? We can look at the proof and geometrize it to construct the angles and the coordinates, but that would involve a lot of constructions! Is there a simpler way?

Richmond’s construction (1893): “Let $OA$ , $OB$ be two perpendicular radii of a circle. Make $Ol$ one-fourth of $OB$ , and the angle $OIE$ one-fourth of $OIA$ ; also find in $OA$ produced a point $F$ such that $EIF$ is $45^{\circ}$ . Let the circle on $AF$ as diameter cut $OB$ in $K$ , and let the circle whose centre is $E$ and radius $EK$ cut $OA$ in $N_{3}$ and $N_{5}$ ; then if ordinates $N_3P_3, N_5P_5$ are drawn to the circle, the $\text{arcs} AP_3, AP_5$ will be $3 / 17$ and $5 / 17$ of the circumference