Gauss and the Regular Heptadecagon

How many regular polygons are constructible? In other words, how many regular polygons can be drawn in a finite number of steps using just a compass and an unmarked straightedge?

Since the time of Euclid, geometers knew how to construct the equilateral triangle, the square, and the regular pentagon.

Construction of an equilateral triangle inscribed in a circle.
Construction of a square inscribed in a circle.
Construction of a regular pentagon inscribed in a circle attributed to H. W. Richmond (1893).

They also knew how to construct the regular fifteen-sided polygon, or pentadecagon, by constructing an equilateral triangle and a regular pentagon inscribed in the same circle and sharing a vertex, and they knew how to double the number of sides of a regular polygon inscribed in a circle by bisecting the arcs between its vertices.

Construction of a regular pentadecagon inscribed in a circle.
Doubling the number of sides of an inscribed square to construct a regular octagon.

This doubling technique can be used any number of times on the triangle, square, pentagon, and pentadecagon to generate an infinite number of regular polygons with nothing but a compass and straightedge. But this still leaves an infinite number of regular polygons that mathematicians couldn’t construct like the heptagon and the enneagon, and for over 2,000 years, it wasn’t known if any of these left-out polygons were constructible at all. That all changed in 1796 when a teenage Carl Friedrich Gauss proved the constructibility of the regular seventeen-sided polygon, or heptadecagon. But how did he do this?

The Constructible Numbers

Before I get into Gauss’s proof, I must first explain what it means for numbers to be constructible. A real number, let’s call it c, is considered constructible if and only if, given just a line segment with length 1, a line segment with length |c| is constructible. For example, \sqrt{2} is a constructible number because a right triangle whose legs are of unit length can be constructed, and its hypotenuse will have a length of \sqrt{2}.

Construction of a line segment with length \sqrt{2}.

It’ll be helpful to define what it means for a number to be constructible algebraically. Take two line segments with lengths a and b. One can then easily construct a line segment with length a+b.

Construction of a line segment with length a+b.

It’s also straightforward to construct a line segment with length |a-b|.

Construction of a line segment with length a-b.

We’ve just translated addition and subtraction into geometric construction techniques. It’s also possible to translate multiplication and division into geometric constructions using the intercept theorem.

Construction of a line segment with length ab.
Construction of a line segment with length a/b.

Starting with a line segment of unit length, one can use addition any number of times to construct a line segment of any integer length. Furthermore, it’s possible to divide two line segments of integer length using the intercept theorem to construct a line segment of any rational length. This means all rational numbers are constructible.

Construction of line segments with integral lengths.

There’s one more operation that I’ve left out so far which can be translated into a geometric construction, and that’s the square root. Given some line segment of length a, it’s possible to construct a line segment of length \sqrt{a} using the geometric mean theorem.

Construction of a line segment with length \sqrt{a}.

Hence, we can say that a number is constructible algebraically if it can be expressed as a finite combination of integers using addition, subtraction, multiplication, division, and square roots (of positive numbers, let’s keep it real), and because all these operations have analogous geometric constructions, every algebraically constructible is also geometrically constructible.

The converse is also true. Points can be uniquely defined by coordinates of two numbers ((x_0,y_0)), lines by slope and y-intercept (y=mx+b), and circles by center and radius ((x-x_0)^2+(y-y_0)^2=r^2). There are five ways a geometric object can be added in a construction: by drawing a line through two points, by drawing a circle with center one point through another point, or by creating a point of intersection from two lines, a line and a circle, or two circles.

The five ‘elementary’ constructions. New objects are colored red.

Finding the values uniquely describing each additional object is just a matter of plugging values uniquely describing existing objects into formulas with nothing but the four basic arithmetic operations and square roots. Thus, if all existing geometric objects in a construction are uniquely described by algebraically constructible numbers, as is true of the base case where there is only a line segment of unit length, then any additional geometric object must also be uniquely described by algebraic numbers.

The length of a constructible line segment must be algebraically constructible for the same reason, and recalling the geometric definition of constructible numbers, all geometrically constructible numbers are lengths of constructible line segments.

Therefore, every geometrically constructible number is also algebraically constructible. The two definitions are equivalent.

The Trigonometric Functions

A regular polygon always has the same number of radii as it does sides. The radii of a regular n-gon divide it into n congruent pieces, so the angles between each consecutive radii must all be one n-th of a revolution. For the heptadecagon, this angle is one seventeenth of a revolution, or 2\pi/17 radians.

The radii of the regular heptadecagon.

Assume \cos(2\pi/17) is constructible. Given a line segment of length \cos(2\pi/17), extend the line segment and draw a unit circle with center one of the endpoints of the line segment. Now it’s possible to construct a perpendicular line segment of length \sin(2\pi/17) which meets the other endpoint. This allows for the construction of the rest of the heptadecagon as shown bellow. In other words, if \cos(2\pi/17) is a constructible number, then the regular heptadecagon is constructible.

A construction of the regular heptadecagon is possible given two line segments of length 1 and \cos(2\pi/17).

Roots of Unity

Let’s look at the equation x^{17}-1=0. You might think that there’s only one solution of this equation: x=1. After all, it’s the only real solution of our equation. But in the complex plane, there’s more. Seventeen of them, to be exact. For example, e^{2\pi i/17}, which I’ll denote with \zeta, yields (e^{2\pi i/17})^{17}=e^{2\pi i}=1 when raised to the power of 17.

Another solution is e^{4\pi i/17}=(e^{2\pi i/17})^2=\zeta^2, which yields (e^{4\pi i/17})^{17}=e^{4\pi i}=(e^{2\pi i})^2=1^2=1 when raised to the power of 17.

In fact, all powers of \zeta are solutions of our equation since

(\zeta^n)^{17}=\zeta^{17n}=(e^{2\pi i/17})^{17n}=e^{2n\pi i}=(e^{2\pi i})^n=1^n=1

for all n\in\mathbb{Z}, and they are known as the seventeenth roots of unity.

The seventeenth roots of unity.

This means for all k=0,1,2,\dots,16,


Equivalently, for all a\in\mathbb{Z} and for all b=0,1,2,\dots,16, \zeta^a can be simplified to \zeta^b if a\equiv b\ (\mathrm{mod}\ 17). Therefore, there are only seventeen unique powers of \zeta, which are all the seventeenth roots of unity.

This brings us to primitive roots. An integer g is said to be a primitive root modulo n if, for all a relatively prime to n, there exists some integer k such that a\equiv g^k\ (\mathrm{mod}\ n). All non-negative integers less than 17 except for 0 are relatively prime to 17, so if there exists some primitive root modulo 17 g, then all integers between 1 and 16 inclusive should be congruent to some power of g modulo 17. In fact, there are eight primitive roots modulo 17, with 3 being the smallest.

3^0=1\equiv1\ (\mathrm{mod}\ 17)

3^1=3^0\cdot3\equiv1\cdot3=3\equiv3\ (\mathrm{mod}\ 17)

3^2=3^1\cdot3\equiv3\cdot3=9\equiv9\ (\mathrm{mod}\ 17)

3^3=3^2\cdot3\equiv9\cdot3=27\equiv10\ (\mathrm{mod}\ 17)

3^4=3^3\cdot3\equiv10\cdot3=30\equiv13\ (\mathrm{mod}\ 17)

3^5=3^4\cdot3\equiv13\cdot3=39\equiv5\ (\mathrm{mod}\ 17)

3^6=3^5\cdot3\equiv5\cdot3=15\equiv15\ (\mathrm{mod}\ 17)

3^7=3^6\cdot3\equiv15\cdot3=45\equiv11\ (\mathrm{mod}\ 17)

3^8=3^7\cdot3\equiv11\cdot3=33\equiv16\ (\mathrm{mod}\ 17)

3^9=3^8\cdot3\equiv16\cdot3=48\equiv14\ (\mathrm{mod}\ 17)

3^{10}=3^9\cdot3\equiv14\cdot3=42\equiv8\ (\mathrm{mod}\ 17)

3^{11}=3^{10}\cdot3\equiv8\cdot3=24\equiv7\ (\mathrm{mod}\ 17)

3^{12}=3^{11}\cdot3\equiv7\cdot3=21\equiv4\ (\mathrm{mod}\ 17)

3^{13}=3^{12}\cdot3\equiv4\cdot3=12\equiv12\ (\mathrm{mod}\ 17)

3^{14}=3^{13}\cdot3\equiv12\cdot3=36\equiv2\ (\mathrm{mod}\ 17)

3^{15}=3^{14}\cdot3\equiv2\cdot3=6\equiv6\ (\mathrm{mod}\ 17)

3^{16}=3^{15}\cdot3\equiv6\cdot3=18\equiv1\ (\mathrm{mod}\ 17)


The integers 1 to 16 form a cyclic group under multiplication modulo 17, denoted by (\mathbb{Z}/17\mathbb{Z})^{\times}, and 3 is said to be a generator of (\mathbb{Z}/17\mathbb{Z})^{\times}.

Because two powers of \zeta are equal if their exponents are congruent modulo 17, and because every integer from 1 to 16 is congruent to some power of 3 modulo 17, all seventeenth roots of unity except for 1 can be written as \zeta raised to the power of some power of 3.



















These roots are said to be primitive because they cannot be raised to the power of a positive integer less than 17 to yield 1.

Finding a polynomial whose roots are all the seventeenth roots of unity except for 1 is a simple matter of dividing x^{17}-1 by x-1. Remembering the finite geometric series formula

\displaystyle a+ar+ar^2+\cdots+ar^n=\frac{a(r^{n+1}-1)}{r-1},\ r\neq1,

we obtain

\displaystyle \frac{x^{17}-1}{x-1}=1+x+x^2+\cdots+x^{16},\ x\neq1.

This is known as the seventeenth cyclotomic polynomial, and it’s easy to see that \zeta is one of its roots by viewing the complex numbers as vectors. The sum of all seventeen powers of \zeta should be equal to 0 due to their rotational symmetry.

The seventeenth roots of unity all add up to zero.

Hence, the sum of all seventeenth roots of unity is equal to 0. Subtracting 1 from our sum gives us


which we can rewrite as


Gaussian Periods

If there exists an integer k such that k^2\equiv q\ (\mathrm{mod}\ n), then q is said to be a quadratic residue modulo n. Otherwise, it is a quadratic nonresidue modulo n. We can split the sum of primitive seventeenth roots of unity into two parts: the sum of all powers of \zeta sans 1 whose exponents are quadratic residues, and the sum of all powers of \zeta whose exponents are quadratic nonresidues. I’ll denote the former sum with x_1 and the latter sum with x_2.



It’s obvious that x_1+x_2=-1 since adding the two sums together gives us the sum of all primitive seventeenth roots of unity again. Solving for x_1x_2 is a bit trickier.

Because each term of x_1 and x_2 is a seventeenth root of unity, each term of x_1x_2 should be a product of two seventeenth roots of unity. There are eight terms each in x_1 and x_2, so x_1x_2 must have sixty-four terms.

\displaystyle x_1x_2=\sum_{k=0}^7\sum_{n=0}^7\zeta^{3^{2n}}\zeta^{3^{2k+1}}

The product of two seventeenth roots of unity is, itself, a seventeenth root of unity, so our expression can be simplified as


Where A,a_0,a_1,\dots,a_{15} are yet-to-be-determined integers. Replacing \zeta with \zeta^3, we obtain two new sums, which I’ll call x_1^{\prime} and x_2^{\prime}. As it turns out, replacing \zeta with \zeta^3 simply swaps x_1 and x_2 (Remember that \zeta^{3^{16}}=\zeta^{3^0}).



This means





Comparing the coefficients of the latter two expressions, we see that


Comparing the coefficients of x_1x_2 and x_1^{\prime}x_2^{\prime}.

They’re all equal to the same integer value, which I’ll call a. We can now write x_1x_2 as


Because a and A are integers, x_1x_2 must be an integer.

-1 is congruent to 16 modulo 17, and since 16 is a perfect square, -1 is a quadratic residue modulo 17. Let q be a quadratic residue modulo 17.

q\equiv k^2\ (\mathrm{mod}\ 17)

Then -q must also be a quadratic residue modulo 17.

-q\equiv-k^2\equiv16k^2=(4k)^2\ (\mathrm{mod}\ 17)

It follows that if \zeta^q is a term of x_1, then \zeta^{-q} is also a term of x_1, and thus, not a term of x_2. Therefore, no term in x_1 is the reciprocal of a term in x_2 and vice versa. Equivalently, any term of x_1 times any term of x_2 never equals 1. Therefore, A=0.

Remembering that x_1x_2 has sixty-four terms, all of which are seventeenth roots of unity, and that x_1x_2=A+a(\zeta^{3^0}+\zeta^{3^1}+\cdots+\zeta^{3^{15}})=a(\zeta^{3^0}+\zeta^{3^1}+\cdots+\zeta^{3^{15}}) (a sixteen-term expression), one can conclude that a must be equal to 4, and because x_1x_2=A-a=-a, x_1x_2=-4.

x_1 and x_2 are examples of Gaussian periods, and they will be of great use when we get to Gauss’s proof of the constructibility of the regular heptadecagon.

The Proof

Imagine x_1 and x_2 are the solutions of some quadratic equation ax^2+bx+c=0. Then by Vieta’s formulas, x_1+x_2=-b/a and x_1x_2=c/a. To simplify things, we’ll let a=1. Now x_1+x_2=-b and x_1x_2=c. We know that x_1+x_2=-1 and x_1x_2=-4, so b=1 and c=-4. This means x_1 and x_2 are solutions of the quadratic equation


x_1 only has two terms in its sum whose real parts are negative, so it will be the positive solution of our quadratic equation.

The terms of x_1 and x_2.

We can then solve for x_1 and x_2 using the quadratic formula.

\displaystyle x_1=\frac{-1+\sqrt{17}}{2}

\displaystyle x_2=\frac{-1-\sqrt{17}}{2}

x_1 and x_2 as roots of the quadratic equation x^2+x-4=0.

Define y_1 and y_2 from the terms of x_1 such that y_1=\zeta^{3^0}+\zeta^{3^4}+\zeta^{3^8}+\zeta^{3^{12}}=\zeta+\zeta^{13}+\zeta^{16}+\zeta^4 and y_2=\zeta^{3^2}+\zeta^{3^6}+\zeta^{3^{10}}+\zeta^{3^{14}}=\zeta^9+\zeta^{15}+\zeta^8+\zeta^2.

The terms of y_1 and y_2.

It’s clear that

\displaystyle y_1+y_2=x_1=\frac{-1+\sqrt{17}}{2}.

What’s interesting is that


Like x_1x_2, y_1y_2 is an simple integer value. (Exercise for the reader: show that y_1y_2=-1 by employing a similar method used to find x_1x_2. Don’t use brute force!)

Once again, we can use Vieta’s formulas to find a quadratic equation whose solutions are y_1 and y_2.

\displaystyle y^2+\frac{1-\sqrt{17}}{2}y-1=0

We can then use the quadratic formula to solve for y_1 and y_2.

\displaystyle y_1=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}}{4}

\displaystyle y_2=\frac{-1+\sqrt{17}-\sqrt{34-2\sqrt{17}}}{4}.

Define y_3 and y_4 from x_2 such that y_3=\zeta^{3^1}+\zeta^{3^5}+\zeta^{3^9}+\zeta^{3^{13}}=\zeta^3+\zeta^5+\zeta^{14}+\zeta^{12} and y_4=\zeta^{3^3}+\zeta^{3^7}+\zeta^{3^{11}}+\zeta^{3^{15}}=\zeta^{10}+\zeta^{11}+\zeta^7+\zeta^6.

The terms of y_3 and y_4.

Just as before, we find y_3+y_4 and y_3y_4.

\displaystyle y_3+y_4=x_2=\frac{-1-\sqrt{17}}{2}

\displaystyle y_3y_4=-1.

We can then solve for y_3 and y_4 in the same fashion as we did for y_1 and y_2.

\displaystyle y^2+\frac{1+\sqrt{17}}{2}y-1

\displaystyle y_3=\frac{-1-\sqrt{17}+\sqrt{34+2\sqrt{17}}}{4}

\displaystyle y_4=\frac{-1-\sqrt{17}-\sqrt{34+2\sqrt{17}}}{4}

At this point, you probably already know what we’re going to do next. Define z_1 and z_2 from y_1 such that z_1=\zeta+\zeta^{16} and z_2=\zeta^{13}+\zeta^4.

The terms of z_1 and z_2.

Solving z_1+z_2 and z_1z_2 gives us

\displaystyle z_1+z_2=y_1=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}}{4}


\displaystyle z_1z_2=(\zeta+\zeta^{16})(\zeta^{13}+\zeta^4)=\zeta^{14}+\zeta^5+\zeta^{29}+\zeta^{20}

\displaystyle =\zeta^3+\zeta^5+\zeta^{14}+\zeta^{12}=y_3=\frac{-1-\sqrt{17}+\sqrt{34+2\sqrt{17}}}{4}.

We then solve for z_1 and z_2.

\displaystyle z^2+\frac{1-\sqrt{17}-\sqrt{34-2\sqrt{17}}}{4}z-\frac{1+\sqrt{17}-\sqrt{34+2\sqrt{17}}}{4}=0

\displaystyle z_1=\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}

\displaystyle z_2=\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}

Now rewrite z_1 as

z_1=\zeta+\zeta^{16}=\zeta+\zeta^{-1}=e^{2\pi i/17}+e^{-2\pi i/17}



A vector diagram demonstrating the equality z_1=\zeta+\zeta^{16}=2\cos(2\pi/17).

It follows that

\displaystyle \cos(2\pi/17)=\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}}}}{16}.

Consequentially, it can be said that \cos(2\pi/17) is expressible as a finite number of additions, subtractions, multiplications, divisions, and square roots of integers. In other words, \cos(2\pi/17) is a constructible number. Hence, the regular heptadecagon can be constructed in a finite number of steps with just an unmarked straightedge and compass.

Analytic Geometry

Notice how Gauss never actually provided a construction of the regular heptadecagon in order to prove its constructibility. In fact, despite the problem being fundamentally geometric, his proof relied mostly on algebraic methods and reasoning. This is just one example of many which demonstrates the power of analytic geometry, the study of geometry using algebraic methods.

(A side note: Eventually, in 1893, H. W. Richmond provided a truly marvelous construction of the regular heptadecagon based off of Gauss’s result which, unfortunately, I’m too lazy to animate the margin of this blog post is too narrow to contain.)

Analytic geometry is useful not just in pure mathematical settings, but also in more applied settings like physics, engineering, aviation, and architecture, where it’s important to represent and describe objects with numerical values and coordinates.

The Gauss-Wantzel Theorem

In 1798, when he was just 21, Gauss wrote Disquisitiones Arithmeticae, one of the most influential texts in the field of number theory. In its seventh and final section, Gauss gives the criteria that determine the constructibility of all regular polygons. A regular n-gon is constructible if n is a power of 2 greater than 2, a Fermat prime, a power of 2 times a Fermat prime, a product of distinct Fermat primes, or a power of 2 times a product of distinct Fermat primes. A Fermat prime, by the way, is a prime number of the form 2^{2^m}+1 where m is a non-negative integer. Gauss proved that this was a sufficient condition for the constructibility of regular polygons, and in 1837, Pierre Wantzel proved that it was also necessary.

There’s a question that remains unanswered as of 2021: how many Fermat primes are there? So far, we only know of five Fermat primes: 3, 5, 17, 257, and 65537. However, we don’t know if these are the only Fermat primes. They appear to be, given that 2^{2^5}+1,2^{2^6}+1,\dots,2^{2^{32}}+1 are composite, but we can’t say for sure until we prove or disprove the existence of more Fermat primes.

Consequentially, we only know of thirty-one constructible polygons with an odd number of sides. If we find another Fermat prime, however, this number would jump up to sixty-three, and if we prove that there are an infinite number of Fermat primes, then we could conclude that there are an infinite number of constructible polygons with an odd number of sides. If, however, there are only five Fermat primes, then there are only thirty-one constructible polygons with an odd number of sides.

Sources and Additional Reading

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