The Heptadecagon Grant Franks October 8, 2019

Review

We’ve looked at the first encounters with the -1 , the development of complex arithmetic, the roots of unity, constructibility of points on the complex plane (considered algebraically) and followed the construction of the pentagon. These points will be reviewed a final time next Wednesday in the final lecture. (Remember: Wednesday, October 16, Junior Common Room, 3:15 pm.)

Let’s move on to the Heptadecagon.

Remember the Pentagon?

We’ve seen how to approach the algebraic construction of the pentagon. The process begins with the equation:

x5 - 1 = 0

One factors out the one real solution that all “roots of unity” equations share, namely (x - 1):

1 + x + x2 + x3 + x4 = 0

This equation is irreducible so long as one allows only rational numbers, Q, as solutions. (In math- jargon, it is “irreducible over the rationals.”) But if one constructs a finite quadratic field extension Q( 5 ), it can be factored into two quadratics. A second finite quadratic field extension allows it to be factored fully into four linear factors from which you can read oﬀ the solutions readily. 2 �� 5 The Heptadecagon.nb

0.31 + 0.95 ⅈ

-0.81 + 0.59 ⅈ

-0.81 - 0.59 ⅈ

0.31 - 0.95 ⅈ

x4 + x3 + x2 + x + 1 = 0

Irreducible over Q Then adjoin 5

2 2 x - η2 x +1 x - η1 x +1

Irreducible over F1 Irreducible over F1

Then adjoin ζ1 Then adjoin ζ2

(x - ζ ) (x - ζ ) (x - ζ1) 4 (x - ζ2) 3

Do the Same Thing, But More O�en 5 The Heptadecagon.nb ��3

We’ll follow the same basic plan to construct the heptadecagon. However, the procedure has a few additional complications due to the greater number of steps.

The Equation For the heptadecagon, we start with the equation:

x17 - 1 = 0.

Again we factor out the one real solution (x - 1) to obtain:

1 + x + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14 + x15 + x16 = 0

This 16th degree equation has sixteen roots, all complex, that we designate:

ζ1, ζ2, ζ3, ζ4, ζ5, ζ6, ζ7, ζ8, ζ9, ζ10, ζ11, ζ12, ζ13, ζ14, ζ15, ζ16

Graphically, these roots appear on the complex plane as vertices of a regular 17-gon, making equal angles at the center:

ζ 4 ζ 5 0 + 1 i ζ 3 ζ 6

ζ 2

ζ 7

ζ 8

-1 + 0 i 1 + 0 i

ζ 9

ζ 16

ζ 10

ζ 15

ζ 11 ζ 14 ζ 12 0 - 1 i ζ 13

Unsurprisingly, these occur in eight pairs of complex conjugates. (In the diagram above, complex conjugates are joined by orange dotted lines).

Arrangement of the Sixteen Roots: the Eight-Periods 4 �� 5 The Heptadecagon.nb

Following the general procedure seen with the pentagon, we are going to split these roots up into --

two groups of eight, then four groups of four, then eight groups of two, then sixteen individuals.

At each stage, we will make numbers by taking the sums of the members in each group. With the pentagon, we found that even though we didn’t know the values of any of the roots (the ζ’s), we could figure out a quadratic formula for the values of the two intermediate sums:

η1 = ζ1 + ζ4 η2 = ζ2 + ζ3

because we could figure out their sum and the product, η1 + η2 and η1 ×η2.

When working on the pentagon, the way in which to subdivide the four roots presented little trouble. We had reason to believe that the complex conjugates had to stay together, so there was only one possible subdivision of the four roots into two pairs. The second division separated the two pairs roots from their conjugate mates. Now, however, we have sixteen roots and eight pairs of conjugates. For the first subdivision, there are 8 x 7 x 6 x 5 = 1,680 possible ways to separate the eight pairs into two groups of four. We don’t know a priori whether some or all, or not all or possibly only one will work. On the surface, it seems that trial and error might not work.

Sorting out the roots properly is more than half the battle in doing this construction. It will require a small detour.

Half the Problem is Pretty Easy First, some good news. We’re looking for a sorting of the roots that will allow us to find the sum and the product of the two groups. In that quest, the sum of the two groups will pose no problem. No matter how we divide the sixteen roots into two bunches, we will be able to get their total sum. Say we just sort out the first eight and the last eight:

A = ζ1 + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8 B = ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16

Now, when we add A + B, we get the sum of all sixteen roots. And we know that to be equal to negative one. In fact, no matter what subdivision we make, the sum of the two divisions will be negative one. Remember: 5 The Heptadecagon.nb ��5

1 + x + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14 + x15 + x16 = 0

which is to say:

x + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14 + x15 + x16 = -1

Just put in ζ for x:

ζ + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 = -1

So the issue that has to be addressed is η1 × η2, the product of two eight-term sums.

Modular Arithmetic Sorting out the other half of the problem will involve us in modular arithmetic. It’s not at all diﬀicult for anyone who has read clock.

�

ζ 6

ζ ζ 7

ζ 2

ζ 5

ζ 3

ζ 8

ζ 4

Remember that multiplying roots of unity by themselves -- that is, raising them to powers -- will move the solution around the unit circle in the complex plane like a clock hand. (In this case the clock hand goes counterclockwise; all analogies have problems!). For example, there are five fi�h roots of unity. If I square the first one, then cube it and so forth, the result moves around the unit circle. Also, when the 6 �� 5 The Heptadecagon.nb

hand has gone completely around, all later solutions are equivalent to one or another of the first five solutions. Thus, ζ6 is equivalent to ζ1.

6 1 The technical term for this sort of equivalence is “congruence”; we write ζ ≡ (ζ )Mod 5, “zeta to the sixth is congruent with zeta one, modulo 5.”

Congruence of this sort will be very useful for us, for Gauss’s solution to the sorting problem involves some very high powers of the 17th roots of unity.

Primitive Roots One observation about modular arithmetic before we go on. Suppose we are working in modulo 17 -- as we will be doing. Take some number, a, and raise it to successive powers. In ordinary arithmetic, it will grow continually. In modular arithmetic, it will go around the cycle of available numbers. Some numbers in doing to touch all the values available; some do not.

Take 2, for instance:

TableFormTablen, 2n, Mod2n, 17, {n, 1, 16}, n n TableHeadings → None, "n", "2 ", "(2 )mod 17" n n n 2 (2 )mod 17 1 2 2 2 4 4 3 8 8 4 16 16 5 32 15 6 64 13 7 128 9 8 256 1 9 512 2 10 1024 4 11 2048 8 12 4096 16 13 8192 15 14 16 384 13 15 32 768 9 16 65 536 1

If I use “n” to designate the power to which one raises the root, notice that 2 cycles through seven values before coming to n=1 and repeating itself.

On the other hand, 3 cycles through all the possible values before repeating itself 5 The Heptadecagon.nb ��7

TableFormTablen, 3n, Mod3n, 17, {n, 1, 16}, n n TableHeadings → None, "n", "3 ", "(3 )mod 17" n n n 3 (3 )mod 17 1 3 3 2 9 9 3 27 10 4 81 13 5 243 5 6 729 15 7 2187 11 8 6561 16 9 19 683 14 10 59 049 8 11 177 147 7 12 531 441 4 13 1 594 323 12 14 4 782 969 2 15 14 348 907 6 16 43 046 721 1

It can be shown that, for prime numbers, there is always at least one such value. For our purposes, with the 17-gon, we only need one. The number three will work for us. (There are others; 2, 4, 8, 9, 13 and 15 don’t work; 3, 5, 6, 7, 10, 11, 12, and 14 do.)

Ordering of the Sixteen Roots Gauss ordered the sixteen roots in accordance with the expression:

ζ(3n)

Since 3n modulo 17 cycles through all values from 1 to 16 before repeating, this ordering will encom- pass all sixteen complex roots. The ordering looks like this: n n 3n n 3 (3 )mod 17 ζ mod 17 0 1 1 ζ 1 3 3 ζ3 2 9 9 ζ9 3 27 10 ζ10 4 81 13 ζ13 5 243 5 ζ5 6 729 15 ζ15 7 2187 11 ζ11 8 6561 16 ζ16 9 19 683 14 ζ14 10 59 049 8 ζ8 11 177 147 7 ζ7 12 531 441 4 ζ4 13 1 594 323 12 ζ12 14 4 782 969 2 ζ2 15 14 348 907 6 ζ6 16 43 046 721 1 ζ 8 �� 5 The Heptadecagon.nb

Notice that each term is obtained from the previous one by successive powers of three. Notice also, that if one takes every other term, one has a succession by powers of nine:

1, 9, 81, 729 … or 3, 27 = 3 x9, 243 = 3 x 81, 2187 = 3 x 729 …

This will be useful in what follows.

Two Eight Periods The 16-period is divided into two 8-periods by taking alternate members of the series and summing the.

9 13 15 16 8 4 2 η1 = ζ + ζ + ζ + ζ + ζ + ζ + ζ + ζ 3 10 5 11 14 7 12 6 η2 = ζ + ζ + ζ + ζ + ζ + ζ + ζ + ζ

Notice that each period contains four pairs of complex conjugates.

ζ 4 ζ 5 ζ 3 ζ 6

ζ 2

ζ 7

ζ 8

ζ 9

ζ 16

ζ 10

ζ 15

ζ 11 ζ 14 ζ 12 ζ 13

Complex 17th roots of unity - Two Eight-Periods

Sum of the 8-periods

η1 + η2 = -1, as explained above.

Product of the 8-periods

The product η1 η2 requires some calculation. We have the multiplication of two eight-term sums, which will yield sixty four terms: 5 The Heptadecagon.nb ��9

ζ + ζ9 + ζ13 + ζ15 + ζ16 + ζ8 + ζ4 + ζ2 ζ3 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6 = …

exponent1 exponent 2 sum summod 17 1 1 + 3 = 4 ≡ 4 2 1 + 10 = 11 ≡ 11 3 1 + 5 = 6 ≡ 6 4 1 + 11 = 12 ≡ 12 5 1 + 14 = 15 ≡ 15 6 1 + 7 = 8 ≡ 8 7 1 + 12 = 13 ≡ 13 8 1 + 6 = 7 ≡ 7 9 9 + 3 = 12 ≡ 12 10 9 + 10 = 19 ≡ 2 11 9 + 5 = 14 ≡ 14 12 9 + 11 = 20 ≡ 3 13 9 + 14 = 23 ≡ 6 14 9 + 7 = 16 ≡ 16 15 9 + 12 = 21 ≡ 4 16 9 + 6 = 15 ≡ 15 17 13 + 3 = 16 ≡ 16 18 13 + 10 = 23 ≡ 6 19 13 + 5 = 18 ≡ 1 20 13 + 11 = 24 ≡ 7 21 13 + 14 = 27 ≡ 10 22 13 + 7 = 20 ≡ 3 23 13 + 12 = 25 ≡ 8 24 13 + 6 = 19 ≡ 2 25 15 + 3 = 18 ≡ 1 26 15 + 10 = 25 ≡ 8 27 15 + 5 = 20 ≡ 3 28 15 + 11 = 26 ≡ 9 29 15 + 14 = 29 ≡ 12 30 15 + 7 = 22 ≡ 5 31 15 + 12 = 27 ≡ 10 32 15 + 6 = 21 ≡ 4 33 16 + 3 = 19 ≡ 2 34 16 + 10 = 26 ≡ 9 35 16 + 5 = 21 ≡ 4 36 16 + 11 = 27 ≡ 10 37 16 + 14 = 30 ≡ 13 38 16 + 7 = 23 ≡ 6 39 16 + 12 = 28 ≡ 11 40 16 + 6 = 22 ≡ 5 41 8 + 3 = 11 ≡ 11 42 8 + 10 = 18 ≡ 1 43 8 + 5 = 13 ≡ 13 44 8 + 11 = 19 ≡ 2 45 8 + 14 = 22 ≡ 5 46 8 + 7 = 15 ≡ 15 47 8 + 12 = 20 ≡ 3 48 8 + 6 = 14 ≡ 14 49 4 + 3 = 7 ≡ 7 50 4 + 10 = 14 ≡ 14 51 4 + 5 = 9 ≡ 9 52 4 + 11 = 15 ≡ 15 10 �� 5 The Heptadecagon.nb

53 4 + 14 = 18 ≡ 1 54 4 + 7 = 11 ≡ 11 55 4 + 12 = 16 ≡ 16 56 4 + 6 = 10 ≡ 10 57 2 + 3 = 5 ≡ 5 58 2 + 10 = 12 ≡ 12 59 2 + 5 = 7 ≡ 7 60 2 + 11 = 13 ≡ 13 61 2 + 14 = 16 ≡ 16 62 2 + 7 = 9 ≡ 9 63 2 + 12 = 14 ≡ 14 64 2 + 6 = 8 ≡ 8

This is a little hard to digest. However, there are some patterns and repetitions. Looking just at the exponents of the terms: 1 + 3 = 4 ; 9 + 10 = 19 ; 13 + 5 = 18 ; 15 + 11 = 26 ; 16 + 14 = 30 ; 8 + 7 = 15 ; 4 + 12 = 16 ; 2 + 6 = 8 ;

1 + 10 = 11 ; 9 + 5 = 14 ; 13 + 11 = 24 ; 15 + 14 = 29 ; 16 + 7 = 23 ; 8 + 12 = 20 ; 4 + 6 = 10 ; 2 + 3 = 5 ;

1 + 5 = 6 ; 9 + 11 = 20 ; 13 + 14 = 27 ; 15 + 7 = 22 ; 16 + 12 = 28 ; 8 + 6 = 14 ; 4 + 3 = 7 ; 2 + 10 = 12 ;

1 + 11 = 12 ; 9 + 14 = 23 ; 13 + 7 = 20 ; 15 + 12 = 27 ; 16 + 6 = 22 ; 8 + 3 = 11 ; 4 + 10 = 14 ; 2 + 5 = 7 ;

1 + 14 = 15 ; 9 + 7 = 16 ; 13 + 12 = 25 ; 15 + 6 = 21 ; 16 + 3 = 19 ; 8 + 10 = 18 ; 4 + 5 = 9 ; 2 + 11 = 13 ;

1 + 7 = 8 ; 9 + 12 = 21 ; 13 + 6 = 19 ; 15 + 3 = 18 ; 16 + 10 = 26 ; 8 + 5 = 13 ; 4 + 11 = 15 ; 2 + 14 = 16 ;

1 + 12 = 13 ; 9 + 6 = 15 ; 13 + 3 = 16 ; 15 + 10 = 25 ; 16 + 5 = 21 ; 8 + 11 = 19 ; 4 + 14 = 18 ; 2 + 7 = 9 ;

1 + 6 = 7 ; 9 + 3 = 12 ; 13 + 10 = 23 ; 15 + 5 = 20 ; 16 + 11 = 27 ; 8 + 14 = 22 ; 4 + 7 = 11 ; 2 + 12 = 14 ;

Or, with the sums reduced modulo 17: 1 + 3 = 4 ; 9 + 10 = 2 ; 13 + 5 = 1 ; 15 + 11 = 9 ; 16 + 14 = 13 ; 8 + 7 = 15 ; 4 + 12 = 16 ; 2 + 6 = 8 ;

1 + 10 = 11 ; 9 + 5 = 14 ; 13 + 11 = 7 ; 15 + 14 = 12 ; 16 + 7 = 6 ; 8 + 12 = 3 ; 4 + 6 = 10 ; 2 + 3 = 5 ;

1 + 5 = 6 ; 9 + 11 = 3 ; 13 + 14 = 10 ; 15 + 7 = 5 ; 16 + 12 = 11 ; 8 + 6 = 14 ; 4 + 3 = 7 ; 2 + 10 = 12 ;

1 + 11 = 12 ; 9 + 14 = 6 ; 13 + 7 = 3 ; 15 + 12 = 10 ; 16 + 6 = 5 ; 8 + 3 = 11 ; 4 + 10 = 14 ; 2 + 5 = 7 ;

1 + 14 = 15 ; 9 + 7 = 16 ; 13 + 12 = 8 ; 15 + 6 = 4 ; 16 + 3 = 2 ; 8 + 10 = 1 ; 4 + 5 = 9 ; 2 + 11 = 13 ;

1 + 7 = 8 ; 9 + 12 = 4 ; 13 + 6 = 2 ; 15 + 3 = 1 ; 16 + 10 = 9 ; 8 + 5 = 13 ; 4 + 11 = 15 ; 2 + 14 = 16 ;

1 + 12 = 13 ; 9 + 6 = 15 ; 13 + 3 = 16 ; 15 + 10 = 8 ; 16 + 5 = 4 ; 8 + 11 = 2 ; 4 + 14 = 1 ; 2 + 7 = 9 ;

1 + 6 = 7 ; 9 + 3 = 12 ; 13 + 10 = 6 ; 15 + 5 = 3 ; 16 + 11 = 10 ; 8 + 14 = 5 ; 4 + 7 = 11 ; 2 + 12 = 14 ;

Or, again, more graphically: 5 The Heptadecagon.nb ��11

ζ4 ζ2 ζ ζ9 ζ13 ζ15 ζ16 ζ8 ζ11 ζ14 ζ7 ζ12 ζ6 ζ3 ζ10 ζ5 ζ6 ζ3 ζ10 ζ5 ζ11 ζ14 ζ7 ζ12 ζ12 ζ6 ζ3 ζ10 ζ5 ζ11 ζ14 ζ7 ζ15 ζ16 ζ8 ζ4 ζ2 ζ ζ9 ζ13 ζ8 ζ4 ζ2 ζ ζ9 ζ13 ζ15 ζ16 ζ13 ζ15 ζ16 ζ8 ζ4 ζ2 ζ ζ9 ζ7 ζ12 ζ6 ζ3 ζ10 ζ5 ζ11 ζ14

Notice that the yellow rows have all the same members; so do the green rows. Also, that the yellow rows and the green rows have all diﬀerent members, so that a yellow row and a green row together have all sixteen elements. And all sixteen elements together equal negative one, so that the total of all sixty- four terms is … (drum roll, please) … negative four.

This is not an accident. It was carefully orchestrated by Gauss’s arrangement of the two eight-groups. In fact, the necessity that leads to this arrangement can be seen through examination and analysis of the patterns of the terms entering into the multiplication together with about a week of practice with modular multiplication. I can’t undertake that dissection in more detail here; any one interested can pursue a more complete presentation in the texts referred to in the handout.

For our purposes, what is essential is that we know the sum and the product of the eight-periods,

η1 and η2.

Constructing and Solving an Appropriate Quadratic

We now know that the sum η1 + η2 = -1 , and the product η1 η2 = -4. We can therefore make a quadratic equation with these two numbers as its solutions. Using y as a variable, we have:

2 y - (-1) y + ( -4) = 0 η1 + η2 η1 η2

whose solutions will be η1 and η2 . This equation can be solved with the quadratic formula.

-1 ± 1 + 16 = -1 ± 17 y = 2 2

Beautiful. So I know values of the η’s, which are the sums of the 8-periods. The two values are: 12 �� 5 The Heptadecagon.nb

η = -1 ± 17 1, 2 2

The approximate values for η1 and η2 are 1.56155 and -2.56155.

The Four Periods Next we make four periods of four by taking every fourth root from the original series, starting with the first, second, third and fourth, respectively:

ζ, ζ3, ζ9, ζ10, ζ13, ζ5, ζ15, ζ11, ζ16, ζ14, ζ8, ζ7, ζ4, ζ12, ζ2, ζ6 Period 1 = ζ, ζ13, ζ16, ζ4, Period 2 = ζ3, ζ5, ζ14, ζ12 Period 3 = ζ9, ζ15, ζ8, ζ2 Period 4 = ζ10, ζ11, ζ7, ζ6

Sums of the Four Periods We make the sums of each of the four-periods:

4 13 16 μ1 = ζ + ζ + ζ + ζ 3 5 12 14 μ2 = ζ + ζ + ζ + ζ 2 8 9 15 μ3 = ζ + ζ + ζ + ζ 6 7 10 11 μ4 = ζ + ζ + ζ + ζ

Again, the sums present no diﬀiculy: μ1 + μ3 = η1 and μ2 + μ4 = η2 , since the the four-periods are gotten by segregating elements of the two eight-periods.

Graphically, the four periods are pictured below. Notice that the 8-periods have been subdivided: the red 8-period into red and green 4-periods; the blue 8-period into blue and orange 4-periods. Once again, notice that each four-period includes two pairs of complex conjugates: 5 The Heptadecagon.nb ��13

ζ 4 ζ 5 ζ 3 ζ 6

ζ 2

ζ 7

ζ 8

ζ 9

ζ 16

ζ 10

ζ 15

ζ 11 ζ 14 ζ 12 ζ 13

Inner dots = 8 periods; Outer dots = 4 periods

Products of the 4-Periods

As for the products of the μ’s, we can work out the terms directly. First, take μ1 times μ3 :

Expand[μ1 μ3] ζ3 + ζ6 + ζ9 + ζ10 + ζ12 + ζ13 + ζ15 + ζ16 + ζ18 + ζ19 + ζ21 + ζ22 + ζ24 + ζ25 + ζ28 + ζ31

Which, when simplified by re-expressing the exponents modulo 17: ζ3 + ζ6 + ζ9 + ζ10 + ζ12 + ζ13 + ζ15 + ζ16 + ζ1 + ζ2 + ζ4 + ζ5 + ζ7 + ζ8 + ζ11 + ζ14

Put in numerical order of the exponents:

ζ1 + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 = -1

And for μ2 times μ4:

Expand[μ2 μ4] ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 + ζ18 + ζ19 + ζ20 + ζ21 + ζ22 + ζ23 + ζ24 + ζ25

Again, reduced by re-expressing the exponents modulo 17: ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 + ζ1 + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8

Put in numerical order of the exponents

ζ1 + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 = -1 14 �� 5 The Heptadecagon.nb

Cool.

Solving for the μ’s

Once again, we know the sums and products of pairs of variables, in this case μ1 and μ3 and also

μ2 and μ4 :

μ1 + μ3 = η1 μ2 + μ4 = η2

μ1 × μ3 = -1 μ2 × μ4 = -1

With these sums-and-products, we can make two quadratic equations; we use v and w as variables:

2 v - η1 v - 1 = 0 whose solutions are μ1 and μ3

2 μ μ = η1 ± η1 + 4 1, 3 2

2 And w - η2 w - 1 = 0 whose solutions are μ2 and μ4

2 μ μ = η2 ± η2 + 4 2, 4 2 Just to show where we are at this point, we can identify the values of the μ’s:

2 μ = η1 + η1 + 4 = 1 1 - +  + + 1 - + 2 ≈ 1 2 2 2 1 17 4 4 1 17 2.04948

2 μ = η2 + η2 + 4 = 1 1 - -  + + 1 - - 2 ≈ 2 2 2 2 1 17 4 4 1 17 0.344151

2 μ = η1 - η1 + 4 = 1 1 - +  - + 1 - + 2 ≈ 3 2 2 2 1 17 4 4 1 17 0.487928

2 μ = η2 - η2 + 4 = 1 1 - -  - + 1 - - 2 ≈ - 1 2 2 2 1 17 4 4 1 17 2.9057

The Two Periods With this in hand, we look at the two-periods, obtained as before but this time taking every eighth root from the original list:

16 β1 = ζ + ζ 3 14 β2 = ζ + ζ 8 9 β3 = ζ + ζ 5 The Heptadecagon.nb ��15

7 10 β4 = ζ + ζ 4 13 β5 = ζ + ζ 5 12 β6 = ζ + ζ 2 15 β7 = ζ + ζ 6 11 β8 = ζ + ζ It may be worth noting that each pair of roots that make up a β is a complex conjugate pair. This is importanT, although its special importance won’t appear until the next stage.

As in previous steps, these two-periods come about by separating elements of the four-periods. Their sums thus lead us back to the variables of the previous step:

16 4 13 β1 + β5 = ζ + ζ + ζ + ζ = μ1 3 14 5 12 β2 + β6 = ζ + ζ + ζ + ζ = μ2 8 9 2 15 β3 + β7 = ζ + ζ + ζ + ζ = μ3 7 10 6 11 β4 + β8 = ζ + ζ + ζ + ζ = μ4

We thus have sums of pairs of the β’s. It remains to figure out the products of the same pairs.

Products of the 2-Periods With a little labor, we can figure out the products of the 2-periods paired in way given above. The products are given below, including the reduction of the exponents modulo 17:

16 4 13 5 14 20 29 5 14 3 12 β1 β5 = ζ + ζ  ζ + ζ  = ζ + ζ + ζ + ζ = ζ + ζ + ζ + ζ = μ2

5 12 3 14 8 15 19 26 8 15 2 9 β2 β6 = ζ + ζ  ζ + ζ  = ζ + ζ + ζ + ζ = ζ + ζ + ζ + ζ = μ3

2 15 8 9 10 11 23 24 10 11 6 7 β3 β7 = ζ + ζ  ζ + ζ  = ζ + ζ + ζ + ζ = ζ + ζ + ζ + ζ = μ4

6 11 7 10 13 16 18 21 13 16 1 4 β4 β8 = ζ + ζ  ζ + ζ  = ζ + ζ + ζ + ζ = ζ + ζ + ζ + ζ = μ1 Now we have defined the products as well as the sums of the four pairs of 2-periods. Again, the way in which this multiplication works out is not an accident; it follows from Gauss’s original ordering of the roots that combinations taken by twos, by fours, and so forth will always multiply so as to produce these intermediate periods.

Constructing Four Quadratic Equations With that in mind, we can construct four quadratic equations with the β’s as roots:

2 q - μ1 q + μ2 = 0 whose roots are β1 and β5 16 �� 5 The Heptadecagon.nb

2 r - μ2 r + μ3 = 0 whose roots are β2 and β6

2 s - μ3 s + μ4 = 0 whose roots are β3 and β7

2 t - μ4 t + μ1 = 0 whose roots are β4 and β8

Their solutions can be obtained with the quadratic formula. Since we know the values of the μ’s, we can calculate the values of the β’s:

μ + μ 2 - μ μ - μ 2 - μ β = 1 1 4 2 = 1.86494 β = 1 1 4 2 = 0.184537 1 2 5 2 μ + μ 2 - μ μ - μ 2 - μ β = 2 2 4 3 = 0.891477 β = 2 2 4 3 = -0.547326 2 2 6 2 μ + μ 2 - μ μ - μ 2 - μ β = 3 3 4 4 = 1.47802 β = 3 3 4 4 = -1.96595 3 2 7 2 μ + μ 2 - μ μ - μ 2 - μ β = 4 4 4 1 = -1.20527 β = 4 4 4 1 = -1.70043 4 2 8 2

The Singletons

One more step remains: dividing the 2-periods into individual roots. This is in some ways the easiest step of all.

Their Sums There are sixteen individual roots:

ζ, ζ2, ζ3, ζ4, ζ5, ζ6, ζ7, ζ8, ζ9, ζ10, ζ11, ζ12, ζ13, ζ14, ζ15, ζ16

These, take pairwise in a particular order, constitute the β’s:

16 β1 = ζ + ζ 3 14 β2 = ζ + ζ 8 9 β3 = ζ + ζ 7 10 β4 = ζ + ζ 4 13 β5 = ζ + ζ 5 12 β6 = ζ + ζ 2 15 β7 = ζ + ζ 6 11 β8 = ζ + ζ

Here we see that we already have the sums of the sixteen ζ’s, taken pairwise.

Their Products 5 The Heptadecagon.nb ��17

This time, the product of the roots just as simple as the sums. Since, as already noted above, each β pair constitutes a pair of complex conjugates, their product -- obtained by adding their exponents -- is always seventeen or, on the unit circle in the complex plane, +1.

ζ ζ16 = ζ17 = 1 ζ3 ζ14 = ζ17 = 1 ζ8 ζ9 = ζ17 = 1 ζ7 ζ10 = ζ17 = 1 ζ4 ζ13 = ζ17 = 1 ζ5 ζ12 = ζ17 = 1 ζ2 ζ15 = ζ17 = 1 ζ6 ζ11 = ζ17 = 1 WIth this information, we have the sum and the products of the roots (taken in this special order) we can construct eight quadratic equations whose solutions are the ζ’s.

2 16 r - β1 r + 1 = 0 whose solutions are ζ and ζ 2 3 14 s - β2 s + 1 = 0 whose solutions are ζ and ζ 2 8 9 t - β3 t + 1 = 0 whose solutions are ζ and ζ 2 7 10 v - β4 v + 1 = 0 whose solutions are ζ and ζ 2 4 13 w - β5 w + 1 = 0 whose solutions are ζ and ζ 2 5 12 x - β6 x + 1 = 0 whose solutions are ζ and ζ 2 2 15 y - β7 y + 1 = 0 whose solutions are ζ and ζ 2 6 11 z - β8 z + 1 = 0 whose solutions are ζ and ζ We can apply the quadratic formula to find the solutions. Since we have the values for the β’s, we can obtain values for the ζ’s:

β ± β 2 - ζ and ζ16 = 1 1 4 = 0.932472 ± 0.361242 ⅈ 2 β ± β 2 - ζ3 and ζ14 = 2 2 4 = 0.445738 ± 0.895163 ⅈ 2 β ± β 2 - ζ8 and ζ9 = 3 3 4 = -0.982973 ± 0.18375 ⅈ 2 β ± β 2 - ζ7 and ζ10 = 4 4 4 = -0.850217 ± 0.526432 ⅈ 2 β ± β 2 - ζ4 and ζ13 = 5 5 4 = 0.0922684 ± 0.995734 ⅈ 2 β ± β 2 - ζ5 and ζ12 = 6 6 4 = -0.273663 ± 0.961826 ⅈ 2 β ± β 2 - ζ2 and ζ15 = 7 7 4 = 0.739009 ± 0.673696 ⅈ 2 β ± β 2 - ζ6 and ζ11 = 8 8 4 = -0.602635 ± 0.798017 ⅈ 2 Shown graphically: 18 �� 5 The Heptadecagon.nb

ζ 4 ζ 5 ζ 3 ζ 6

ζ 2

ζ 7

ζ 8

ζ 9

ζ 16

ζ 10

ζ 15

ζ 11 ζ 14 ζ 12 ζ 13

As advertised.

Full Algebraic Presentation of the Sixteen Complex Roots of Unity

The stack of quadratic equations involved in calculating the heptadecagon vertices is diﬀicult to grasp when they are all assembled into a single formula. One of the roots is represented as:

1 1 1 17 - 17  + 1 -1 + 17  + 1 + 2 4 2 8 4 2

34+6 17 + 578-34 17 - 34-2 17 -8 2 17+ 17

 -4 + 1 -1 + 17 + 34 - 2 17 +  2 34 + 6 17 + 64

2 578 - 34 17 - 34 - 2 17 - 8 2 17 + 17 

It’s a challenge to grasp such a thing, but even a casual inspection shows that it consists exclusively of stacks of rational numbers and square roots combined with rational functions (addition, subtraction, multiplication and division). That alone is enough to guarantee its constructibility. 5 The Heptadecagon.nb ��19

An abbreviated graphic representation of the process of constructing the heptadecagon might look like this:

2 βn ± βn - 4 ζ1 … 16= 2

2 μ1, 2, 3, 4 ± μ1, 2, 3, 4 - 4 μ2, 3, 4, 1 β1, 2, 3, 4, 5, 6, 7, 8 = 2

2 η1,2 ± η1,2 + 4 μ1, 3, 2, 4 = 2

-1 ± 17 η1,2 = 2

The sixteen roots were subdivided successively as follows.

3 9 10 13 5 15 11 16 14 8 7 4 12 2 6 The whole ζ, ζ , ζ , ζ , ζ , ζ , ζ , ζ , ζ , ζ , ζ , ζ , ζ , ζ , ζ , ζ t

9 13 15 16 8 4 2 ζ 3, ζ 10, ζ 5, ζ 11, ζ 14, ζ 7, ζ 12, ζ 6 The η's ζ, ζ , ζ , ζ , ζ , ζ , ζ , ζ

3 5 14 12 13 16 4 9 15 8 2 ζ , ζ , ζ , ζ 10 11 7 6 The μ's ζ, ζ , ζ , ζ ζ , ζ , ζ , ζ ζ , ζ , ζ , ζ

13 4 9 8 5 12 16 ζ , ζ ζ , ζ ζ 15, ζ 2 ζ 3, ζ 14 ζ , ζ ζ 10, ζ 7 ζ 11, ζ 6 The β's ζ, ζ

3 14 7 6 16 13 9 8 15 2 ζ ζ 5 12 10 ζ 11 ζ The ζ's ζ ζ ζ ζ 4 ζ ζ ζ ζ ζ ζ ζ ζ

The sum of the whole set was -1. The sum of the η’s required a finite quadratic field extension to -1 ± 17 include 2 . Each successive subdivision required another finite quadratic field extension of what went before. But (and?) that is the sine qua non of constructibility: a point is constructible if (and only 20 �� 5 The Heptadecagon.nb

if) it is defined by numbers that are either rational or the result of a succession of finite, quadratic field extensions from the rationals.

Extension

This is a pretty remarkable result: the 17-gon is constructible using the techniques available in Euclid’s Elements. In fact, Gauss’s result is even more remarkable than that. It has both a positive and a negative side, which I can report although we haven’t done quite enough work to demonstrate both sides fully.

The Positive Side The positive side is that any figure is constructible if it the number of its sides is either,

(a) a prime number equal to 2n + 1, or (b) some multiple of 2p times such a number, or (c) the sum of two of the primes in (a), or 2p times that sum.

So, 20 = 1, 1 + 1 is 2, which is prime. You can’t make a 2-gon, really, but you can make polygons that are multiples of 2 times it: a 4-gon (square), and 8-gon (octagon), etc.

Next, 21 = 2. 2 + 1 is 3, which is prime. You can make a 3-gon (triangle), or a 6-gon (hexagon), etc.

A�er that, 22 = 4. 4 + 1 is 5, which is prime. You can make a 5-gon (pentagon). Euclid could do this.

These are all the constructible prime n-gons that Euclid knew. He doesn’t say so much, but if he had known the construction of another it is hard to believe he would not have given it.

A�er that, 24 = 16. 16 + 1 is 17, which is prime. You can make a 17-gon (heptadecagon). This was Gauss’s great discovery. I cannot believe that Euclid knew that the 17-gon was constructible.

But there are more!

Consider: 28 = 256. 256 + 1 = 257, which is prime. The 257-gon is constructible.

Consider: 216 = 64536. 64,536 + 1 = 64,537, which is prime. The 64,537-gon is constructible.

In fact, if you can find another number of the form 2n + 1 which is prime, it too will be constructible. These are the so-called “Fermat numbers,” named for Pierre Fermat who conjectured that all numbers of the form 2n + 1 were prime, provided that n itself is a power of 2. Such numbers are: 5 The Heptadecagon.nb ��21

3, 5, 17, 257, 64537, …

The next one would be 4,294,967,297 … but this on turns out not to be prime. (It is 641 times 6,700,417). As of last year, only the first eleven such numbers have been fully tested. The last one, 2211 + 1, has 617 digits; it has two prime factors. The next one has 1,234 digits; whether it is prime or not is still undetermined. Las Vegas is not giving odds. In fact, now it is conjectured that apart from the first five, no other Fermat numbers are prime, although as far as I know that guess hasn’t been proven or disproven.

The Negative Side The negative claim is that only these polygons are constructible. Gauss did not demonstrate that his construction method was the only one possible. The negative claim was demonstrated later by Pierre Wantzel in 1837. The seven-gon, which Euclid just skips, cannot be constructed. Neither can the 9-gon, the 11-gon or the 13-gon. Euclid constructs the (non-prime) 15-gon by the combination of the triangle and the pentagon. His leap from the hexagon to the 15-gon is completely unexplained in the Elements, and to my knowledge few students remark on it. (They should.)

So What?

What follows from this exercise? That is the subject of next week’s talk.