## Sunday, January 20, 2008

### Gauss: Seventeenth Root of Unity Expressed As Radicals

Carl Friedrich Gauss was 18 when he was able to solve the seventeenth root of unity in terms of radicals. He not only made this discovery but also realized that this implied that it would be possible to construct a seventeen-sided polygon using only compass and ruler. Gauss later generalized these results on two levels:

(1) He proved that all roots of unity are expressible as radicals.
(2) A regular polygon is constructible by ruler and compass if and only if its number of sides is a product of distinct Fermat primes Fn (where Fn = 22n + 1) and a power of 2.

In today's blog, I will show Gauss's solution for the seventeenth root of unity. In the next blog, I will prove that it is possible to construct a seventeen sided polygon known as the heptadecagon using only compass and ruler.

Gauss was so fascinated by the construction of the heptadecagon that he decided that he would dedicate his life to mathematics instead of philology.

Theorem 1: The seventeen root of unity is expressible as radicals

Proof:

(1) Let ζ be a primitive seventeen-root of unity.

(2) Then, the sixteen roots of unity not equal to 1 are:

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

(3) We can divide the above roots into the following sums:

x1 = ζ1 + ζ9 + ζ13 + ζ15 + ζ16 + ζ8 + ζ4 + ζ2

x2 = ζ3 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6

(4) Now, since (see Lemma 1, here):

1 + ζ + ζ2 + ... + ζ16 = 0

We have:

x1 + x2 + 1 = 0

So that:

x1 + x2 = -1

(5) x1 * x2 = ( ζ1 + ζ9 + ζ13 + ζ15 + ζ16 + ζ8 + ζ4 + ζ2)(ζ3 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) =

= ζ13 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ93 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ133 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ153 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ163 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ83 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ43 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ23 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) =

= (ζ4 + ζ11 + ζ6 + ζ12 + ζ15 + ζ8 + ζ13 + ζ7) + (ζ12 + ζ2 + ζ14 + ζ3 + ζ6 + ζ16 + ζ4 + ζ15) +
16 + ζ6 + ζ + ζ7 + ζ10 + ζ3 + ζ8 + ζ2) + (ζ + ζ8 + ζ3 + ζ9 + ζ12 + ζ5 + ζ10 + ζ4) +
2 + ζ9 + ζ4 + ζ10 + ζ13 + ζ6 + ζ11 + ζ5) + (ζ11 + ζ + ζ13 + ζ2 + ζ5 + ζ15 + ζ3 + ζ14) +
7 + ζ14 + ζ9 + ζ15 + ζ + ζ11 + ζ16 + ζ10) + (ζ5 + ζ12 + ζ7 + ζ13 + ζ16 + ζ9 + ζ14 + ζ8) =

= 4[ζ1 + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16] = 4[x1 + x2] = -4

(6) We can now use the two equations to form a quadratic equation:

x2 = -1 -x1

x1*x2 = -4

x1(-1 -x1) = -4

which gives us:

-x12 - x1 + 4 = 0

or if we multiply each side by -1

x12 + x1 - 4 = 0

(7) Now, it should also be clear that since each equation is symmetric, we could easily have said that:

x1 = -1 - x2

so it is clear that we also have:

x22 + x2 - 4 = 0

(8) This means that x1, x2 are found using the quadratic equation (see Theorem, here) to get:

x = (-1 ±√1 - (-16))/2 =

= (-1 ±√17)/2

So, we that we have:

x1 = (-1 +√17)/2

x2 = (-1 -√17)/2

(9) Now, let's define the following values:

y1 = ζ1 + ζ13 + ζ16 + ζ4

y2 = ζ9 + ζ15 + ζ8 + ζ2

y3 = ζ3 + ζ5 + ζ14 + ζ12

y4 = ζ10 + ζ11 + ζ7 + ζ6

(10) Now, it is clear that:

y1 + y2 = x1

and

y3 + y4 = x2

(11) Further, we have:

y1y2 = (ζ1 + ζ13 + ζ16 + ζ4)(ζ9 + ζ15 + ζ8 + ζ2) =

= ζ19 + ζ15 + ζ8 + ζ2) + ζ139 + ζ15 + ζ8 + ζ2) +
ζ16 9 + ζ15 + ζ8 + ζ2) + ζ49 + ζ15 + ζ8 + ζ2) =

= (ζ10 + ζ16 + ζ9 + ζ3) + (ζ5 + ζ11 + ζ4 + ζ15) +
8 + ζ14 + ζ7 + ζ) + (ζ13 + ζ2 + ζ12 + ζ6) =

= ζ + ζ2 + ζ3 + ζ4 + ζ56 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 = x1 + x2 = -1

y3y4 = (ζ3 + ζ5 + ζ14 + ζ12)(ζ10 + ζ11 + ζ7 + ζ6) =

= ζ310 + ζ11 + ζ7 + ζ6) + ζ510 + ζ11 + ζ7 + ζ6) +
ζ1410 + ζ11 + ζ7 + ζ6) + ζ1210 + ζ11 + ζ7 + ζ6) =

= (ζ13 + ζ14 + ζ10 + ζ9) + (ζ15 + ζ16 + ζ12 + ζ11) +
7 + ζ8 + ζ4 + ζ3) + (ζ5 + ζ6 + ζ2 + ζ) =

= ζ +ζ23 + ζ4 + ζ56 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 = -1

(12) So, this gives us:

y2 = x1 - y1
y4 = x2 - y3

y1(x1 - y1) = -1
y3(x2 - y3) =-1

So that:

x1y1 - y12 + 1 = 0
x2y3 - y32 + 1 = 0

(13) So that we can find the value for y1, y2 by solving:

y2 - (x1)y - 1 = 0

So that y = [(x1) ± √(x12 + 4)]/2

So we assign:

y1 = [(x1) + √(x12 + 4)]/2

y2 = [(x1) - √(x12 + 4)]/2

(14) We can find the value for y3, y4 by solving:

Y2 - (x2)Y - 1 = 0

So that Y = [(x2) ± √(x22 + 4)]/2

So we assign:

y3 = [(x2) + √(x22 + 4)]/2

y4 = [(x2) - √(x22 + 4)]/2

(15) Now, let's define the following values:

z1 = ζ + ζ16

z2 = ζ13 + ζ4

z3 = ζ8 + ζ9

z4 = ζ2 + ζ15

z5 = ζ3 + ζ14

z6 = ζ5 + ζ12

z7 = ζ7 + ζ10

z8 = ζ6 + ζ11

(16) We note that:

z1 + z2 = y1

z3 + z4 = y2

z5 + z6 = y3

z7 + z8 = y4

(17) If we multiply combinations together, we get:

z1*z2 = (ζ + ζ16)(ζ13 + ζ4) = ζ14 + ζ5 + ζ12 + ζ3 = y3

z3*z4 = (ζ8 + ζ9)( ζ2 + ζ15) = ζ10 + ζ6 + ζ11 + ζ7 = y4

z5*z6 = (ζ3 + ζ14)(ζ5 + ζ12) = ζ8 + ζ15 + ζ2 + ζ9 = y2

z7*z8 = (ζ7 + ζ10)(ζ6 + ζ11) = ζ13 + ζ + ζ16 + ζ4 = y1

(18) So, this gives us:

z2 = y1 - z1

z4 = y2 - z3

z6 = y3 - z5

z8 = y4 - z7

which means:

z1(y1 - z1) = y3

z3(y2 - z3) = y4

z5(y3 - z5) = y2

z7( y4 - z7 ) = y1

(19) So, we find z1, z2 by solving for:

z12 - z1y1 + y3 = 0

so that:

z1 = [y1 + √(y1)2 - 4y3)]/2

z2 = [y1 - √(y1)2 - 4y3)]/2

(20) Finally, we can now solve for the seventeen root of unity since:

ζ16 = z1 - ζ [From step #15 above]

ζ(z1 - ζ) = 1 [This follows since ζ17 = 1]

z1ζ - ζ2 - 1 = 0

ζ2 - z1ζ + 1 = 0 [Multiplying both sides by -1]

ζ = (z1 + √(z1)2 - 4)/2 [Using the solution to the quadratic equation, see here]

Working all this out gives us:

ζ =

References

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Bob Dillon said...

I have a question for Larry Freeman regarding this proof. Larry, would you be willing to send your e-mail address to me at bdillon@aurora.edu ? My question is a bit long to post as a comment.

Thanks,

Bob Dillon
Aurora, IL
(I wasn't able to find any contact info on this blog site for you)

Bob Dillon said...

Actually, the final result is only the real part of zeta (which is all that's needed to construct the heptadecagon), and the final result cannot be found from step 20. The final result is half of z_1, and that is where the back substitution should start.

Unknown said...

Why is the real part of primitive seventeenth root of unity half of Z-1? Is there any explanation or proof for this?

Bob Dillon said...

I will use "w" for the primitive 17th root of unity (I don't know how to use Greek letters on this blog).

The primitive 17th root of unity is cos(2*pi/17) + i sin(2*pi/17) [See an article on roots of unity for an explanation]. By DeMoivre's Theorem, w^n = cos(2n*pi/17)+i sin(2n*pi/17) So
w^16 = cos(32*pi/17) + i sin(32*pi/17) = cos(-2*pi/17) + i sin(-2*pi/17) since 32*pi/17 = -2*pi/17 [32*pi/17 measures the angle in the counter-clockwise direction, -2*pi/17 measures it in the clockwise direction] But recall that cos(-A) = cos(A) and sin(-A) = - sin(A), so
w + w^16 = 2*cos(2*pi/17)