Gauss: Constructibility of Heptadecagon

In the previous blog, I showed an algorithm for constructing the heptadecagon (a regular 17-sided figure). In today's blog, I will show a proof that based on the Gauss's solution for the seventeenth roots of unity in terms of radicals, the algorithm in the previous blog really works.

Lemma 1:

if α = 2π/17

and

ζ^k = cos kα + isin k α

then

ζ^k + ζ^{17 - k} = 2 cos kα

Proof:

(1) Using De Moivre's Formula for roots of unity (see Corollary 1.1, here), we note that:

if α = 2π/17, then:

ζ = cos α + isinα is a seventeenth root of unity.

(2) Using De Moivre's Formula (see Theorem 1, here), we note that:

ζ^k = (cos α + isin α)^k = cos kα + isin k α

ζ^17-k = (cos α + isin α)^17-k = cos ([17 - k]α) + isin([17 -k]α) = cos ([17 - k]2π/17) + isin([17 - k]2π/17) = cos(2π - kα) + isin(2π - kα) = cos(-kα) + isin(-kα) = cos(kα) - isin(kα)

(3) So that we have:

ζ^k + ζ^17-k = cos kα + isin kα + cos kα - isink α = 2cos kα

QED

Lemma 2:

If C is the least positive acute angle such that tan 4C = 4

with

α = 2π/17

ζ^k = cos kα + isin k α

and:

x₁ = ζ + ζ⁹ + ζ¹³ + ζ¹⁵ + ζ¹⁶ + ζ⁸ + ζ⁴ + ζ² x₂ = ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶

then:

x₁ = 2 tan 2C x₂ = -2cot 2C

Proof:

(1) Using step #1 thru step #7 in Theorem 1, here, we have:

x₁ and x₂ are the solutions to:

x² + x - 4 = 0

(2) Since tan 4C = 4, it follows that cot 4C = 1/(tan 4C) = 1/4

(3) So 4xcot 4C = 4x(1/4) = x
(4) And we have:

x² + 4xcot 4C - 4 = 0
(5) Using the quadratic equation (see Theorem, here), we have:

x = (1/2)[-4cot4c ± √(4cot 4c)² + 16]

(6) Using Lemma 1, here, we have:

x = (1/2)[-4(1/2)(cot 2c - tan2c) ± √16(1/4)(cot 2c - tan 2c)² + 16) ]= = (tan 2c - cot2c) ± (1/2)√4[cot²(2c) - 2cot(2c)tan(2c) + tan²(2c)] + 16 = = (tan 2c - cot 2c) ± √cot²(2c) - 2 + tan²(2c) + 4 = = (tan 2c - cot 2c) ± √cot²(2c) + tan²(2c) + 2 = = (tan 2c - cot 2c) ± √[cot(2c) + tan(2c)]² = = (tan 2c - cot 2c) ± [cot(2c) + tan(2c)]

(7) Since x₁ is greater than x₂, we have:

x₁ = 2 tan 2c x₂ = -2 cot 2c

QED

Lemma 4:

If C is the least positive acute angle such that tan 4C = 4

with:

α = 2π/17

ζ^k = cos kα + isin k α

and:

y₁ = ζ¹ + ζ¹³ + ζ¹⁶ + ζ⁴

y₂ = ζ⁹ + ζ¹⁵ + ζ⁸ + ζ²

y₃ = ζ³ + ζ⁵ + ζ¹⁴ + ζ¹²

y₄ = ζ¹⁰ + ζ¹¹ + ζ⁷ + ζ⁶

then:

y₁ = tan(C + π/4) y₂ = tan(C - π/4) y₃ = tan C y₄ = -cot C

Proof:

(1) From steps #9 thru #13 in Theorem 1, here, we have:

y₁, y₂ are solutions to:

y² - x₁y - 1 = 0

and

y₃, y₄ are solutions to:

y² - x₂y - 1 = 0

(2) Let's solve first for y₁, y₂

(3) Using the quadratic equation (see Theorem 1, here), we get:
y = (1/2)(x₁ ± √x₁² + 4)

(4) From Lemma 2 above, we know that x₁ = 2 tan 2c

(5) So that we have:

y = (1/2)(2 tan 2c ± √4(tan 2c)² + 4 = = tan 2c ± √(tan 2c)² + 1

(6) Using Lemma 4, here, we have:

tan 2c ± √(tan 2c)² + 1 = (2 tan c)/(1 - tan²c) ± √(4 tan² c)/(1 - 2tan²c + tan⁴ c) + 1 = = (2 tan c)/(1 - tan²c) ± √(4 tan² c + 1 - 2tan²c + tan⁴ c )/(1 - 2tan²c + tan⁴ c) = = (2 tan c)/(1 - tan²c) ± √(2tan² c + 1 + tan⁴ c )/(1 - 2tan²c + tan⁴ c) = = (2 tan c)/(1 - tan²c) ± √(1 + tan² c )²/(1 - tan²c)² = = (2 tan c)/(1 - tan²c) ± (1 + tan²c)/(1 - tan²c) = = (2 tan c ± [1 + tan²c])/(1 - tan²c)
(7) This then gives us:

(2 tan c + 1 + tan²c)/(1 - tan²c) = = (tan c + 1)(tan c + 1)/(1 + tan c)(1 - tan c) = (tan c + 1)/(1 - tan c)

and

(2 tan c - 1 - tan²c)/(1 - tan²c) = (tan²c - 2tan c + 1)/(tan²c - 1) = = (tan c - 1)(tan c - 1)/(tan c - 1)(tan c + 1) = (tan c - 1)/(tan c + 1)
(8) Since tan(π/4) = 1 (see Lemma 1, here), we can restate this as:

(1 + tan c)/(1 - tan c) = (tan π/4 + tan c)/(1 - (tan c)(tan π/4)) (tan c - 1)/(tan c + 1) = (tan c - tan π/4)/(1 + (tan c)(tan π/4))

(9) Using Lemma 3, here, we have:

(1 + tan c)/(1 - tan c) = (tan π/4 + tan c)/(1 - (tan c)(tan π/4)) = tan(c + π/4)

(10) Using Corollary 3.1, here, we have:

(tan c - 1)/(tan c + 1) = (tan c - tan π/4)/(1 + (tan c)(tan π/4)) = tan(c - π/4)

(11) Since y₁ is greater than y₂,

y₁ = tan(c + π/4) y₂ = tan(c - π/4)

(11) Now, let's solve for y₃, y₄

(12) Using the quadratic equation (see Theorem 1, here), we get:

y = (1/2)(x₂ ± √x₂² + 4)

(13) From Lemma 2 above, we know that x₂ = -2 cot 2c

(14) So that we have:

y = (1/2)(-2 cot 2c ± √4(cot 2c)² + 4 = = -cot 2c ± √(cot 2c)² + 1

(15) Using Lemma 1, here, we have:

-cot 2c ± √(cot 2c)² + 1 = -(1/2)(cot c - tan c) ± √(1/4)(cot c - tan c)² + 1 = = (1/2)(tan c - cot c) ± √(1/4)(cot²c + tan²c - 2(cot c)(tan c)) + (4/4) = = (1/2)(tan c - cot c) ± √(1/4)(cot²c + tan²c - 2 + 4) = = (1/2)(tan c - cot c) ± √(1/4)(cotc + tanc)² = = (1/2)(tan c - cot c) ± (1/2)(tan c + cot c)
(16) Since y₃ is greater than y₄, this gives us:

y₃ = tan c y₄ = -cot c

QED
Corollary 4.1:

If C is the least positive acute angle such that tan 4C = 4

with:

α = 2π/17 tan c = 2 cos 3α + 2 cos 5α tan (c - π/4) = 2 cos 2α + 2 cos8α

Proof:

(1) Let us define the following:

y₃ = ζ³ + ζ⁵ + ζ¹⁴ + ζ¹²

and

y₂ = ζ⁹ + ζ¹⁵ + ζ⁸ + ζ²

where

ζ^k = cos kα + isin k α

(2) Using Lemma 1 above:

y₃ =(ζ³ + ζ¹⁴) + (ζ⁵ + ζ¹²) = 2cos 3α + 2cos 5α

y₂ = (ζ⁸ + ζ⁹) + (ζ² + ζ¹⁵) = 2 cos 8α + 2 cos 2 α

(3) Using Lemma 5, here, we have:

2 cos(3α) * 2cos(5α) = 2*[2 cos(3α)*cos(5α)] = 2*[cos(5 + 3)α + cos(5 - 3)α] = 2[cos(8α) + cos(2α)] = 2cos(8α) + 2cos(2α)

(4) From Lemma 4 above, we then have:

tan c = 2 cos 3α + 2 cos 5α tan (c - π/4) = 2 cos 2α + 2 cos8α = 2cos(3α)*2cos(5α)
QED

Theorem 5: Constructibility of Heptadecagon using compass and ruler




































Proof:

(1) Let CD, CP0 two radii of a circle C1 that are perpendicular to each other.
(2) Let CG be 1/4 the length of CD (bisector of a bisector).
(3) Let ∠ CGL be 1/4 ∠ CGP0 with L on CP0 (bisector of a bisector).
(4) Let ∠ LGP = π/4 (bisector a right angle since right angle = π/2, see here for review of radians) where P is on CP0.
(5) Let Q be the midpoint of PP0 and draw a circle C2 with center Q and radius QP0.
(6) Let R be the point where the circle C1 intersects with CD.
(7) Draw a circle C3 with center L and radius equal to RL.
(8) Let S,Q be the points where the circle C3 intersects with CP0.
(9) Let SP5, QP3 be lines parallel to CP0 that intersect with circle C1.
(10) Let C = measurement ∠ CGL
(11) Then 4C = measurement ∠ CGP0
(12) cos P0CP3 = CQ/CP3 = CQ/CP0 [See here for definition of cosine]
(13) cos P0CP5 = cos (π/2 + π/2 - SCP5) = cos (π - SCP5)
(14) Using cos(a + b) = (cos a)(cos b) - (sin a)(sin b) (see Theorem 2, here) and cos(-x) = cos(x) (see Property 9, here), we have:

cos(π - SCP5) = (cos π)(cos -SCP5) - (sin π)(sin -SCP5) = (-1)cos(SCP5) - (0)(-SCP5) = -cos(SCP5)

(15) So, cos P0CP5 = -cos(SCP5) = -SC/CP5 = -SC/CP0

(16) This gives us that:

2cos P0CP3 + 2cos P0CP5 = 2(CQ - SC)/CP0

(17) Since CQ = CL + LQ and SC = SL - CL and SL=LQ:

SC = SL - CL = LQ - CL

CQ - SC = (CL + LQ) - (LQ - CL) = CL + LQ - LQ + CL = 2CL

(18) This gives us that:

2(CQ - SC)/CP0 = 2(2CL)/CP0 = 4CL/CP0

(19) Since CG is 1/4 of CD and CD=CP0, we have:

4CL/CP0 = 4CL/CD = 4CL/(4CG) = CL/CG

(20) Since GCL is a right angle and C = measurement ∠ CGL, we have:

tan CGL = tan c = sin CGL/cos CGL = (LC/GL)/(GC/GL) = LC/GC = 2cos P0CP3 + 2cos P0CP5

(21) Since cos P0CP3 =CQ/CP0 and cos P0CP5 = -SC/CP0, we have:

2 cos(P0CP3)*2cos(P0CP5) = -4*(SC)(CQ)/(CP0)(CP0)

(22) Using the Pythagorean Theorem (see Theorem, here):

CR² = CL² + RL² = (CL - RL)(CL + RL)

(23) Since LQ=LR and LS=RL:

CL + RL = CL + LQ = CQ

CL - RL = CL - LS = SC

(24) So we have:

CR² = (CL - LR)(CL + LR) = (CQ)(SC)

(25) Using the Pythagorean Theorem:

(RC)² = RQ² - CQ² = (RQ - CQ)(RQ + CQ)

(26) Now RQ = QP and RQ=QP0 so that:

(RQ - CQ) = (QP - CQ) = PC

(RQ + CQ) = (QP1 + CQ) = CP1

(27) This gives us that (RC)² = (PC)(CP0) so that:

-4(RC)²/(CP0)² = -4(PC)(CP0)/(CP0)(CP0) = -4(PC)/(CP0)

(28) Since CG = (1/4)CD = (1/4)CP0, we have:

-4(PC)/(CP0) = -4(PC)/(4CG) = -PC/CG

(29) Also:

tan ∠ CGP = sin ∠ CGP/cos ∠ CGP = (PC/GP)/(CG/GP) = PC/GC

(30) Since ∠ CGP = ∠ LGP - ∠ CGL, we have:

tan ∠ CGP = tan (∠ LGP - ∠ CGL) = tan(π/4 - C)

and this means:

-PC/GC = -tan(π/4 - C) = tan(C - π/4)

(31) Using Corollary 4.1 above, we have:

2cos ∠ P0CP3 + 2cos ∠ P0CP5 = 2 cos 3α + 2 cos 5α = tan c where α = 2π/17

2 cos(∠P0CP3)*2cos(∠P0CP5) = 2cos(3α)*2cos(5α) = tan (c - π/4)

(32) Now, it is clear that ∠ P0CP3 and ∠ P0CP5 map onto 3α and 5α.

(33) Since ∠ P0CP5 is greater than ∠P0CP3, it is clear that:

∠ P0CP3 = 3α

∠ P0CP5 = 5α

QED

References

• G. H. Hardy & E. M. Wright, An Introduction to the Theory of Numbers, Oxford Science Publications, 1979

Gauss: Construction of the Heptadecagon

Carl Friedrich Gauss not only solved the seventeenth root of unity in terms of radicals but also realized that his solution indicated that a heptadecagon (a regular, seventeen-sided polygon) could be constructed by compass and ruler.

In my previous blog, I showed Gauss's solution of the seventeenth root of unity. In today's blog, I will show how this solution provides a recipe for constructing a heptadecagon.

For those ready to cut to the chase, a YouTube video of construction of the heptadecagon can be found here.

The algorithm below is a bit more complicated than the above YouTube video. It is taken from Hardy and Wright's Introduction to the Theory of Numbers (see the reference below).

Algorithm: Construction of a heptadecagon using compass and ruler

























































































(1) Draw a line connecting point A and P1
(2) Let C be the midpoint
(3) Draw a circle with C as the center and the radius equal to CP0
(4) Let DE be a line perpendicular to CP0 which passes through C.
(5) Let F be the midpoint on CD and G be be the midpoint on CF.
(6) Draw a line connecting G and P0.
(7) Find a point H on GE where GH = GP0
(8) Draw an arc with center G and radius GP0 from P0 to H.
(9) Draw a line connecting P0 to H
(10) Let I be the midpoint on HP0.
(11) Draw a line passing from G through point I onto the arc and label the point on the arc J.
(12) Draw a line connecting J and H.
(13) Let K be the midpoint on JH.
(14) Draw a line from G to K and let L be the point where GK intersects with CP0.
(15) Draw a line from point G that is perpendicular to GL and label the point N where this line intersects with AC.
(16) Bisect the angle ∠ NGL and label P the point where this angle bisector intersects with AC.
(17) Let Q be the midpoint between P and P0.
(18) Draw a circle with center Q and radius QP0.
(19) Label R the point where this circle intersects with CD
(20) Draw a circle with center L and radius LR
(21) Label S the point where this circle intersects with AC.
(22) Draw a line from S that is perpendicular to AC and label P5 the point where this line intersects with the top of the circle.
(23) Draw a line from Q that is perpendicular to CP1 and label P3 the point where this line intersects with the top of the circle.
(24) Draw a line connecting P3 and P5 and label the midpoint W.
(25) Draw a line from C to W and label the point P4 where CW intersects with the top of the circle.
(26) Now, all the rest of the points can be completed in the following way:
(a) Draw a circle with center P3 and radius P3P4.
(b) Label P2 where this circle intersects with the larger circle.
(27) Repeat this same step for all missing remaining points.

In my next blog, I will show the proof that the above construction really works. This proof is based on the solution for the seventeenth root of unity.

References

• "The 17-gon", Jim Loy's Home Page
  
• G. H. Hardy & E. M. Wright, An Introduction to the Theory of Numbers, Oxford Science Publications, 1979.
  

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 F_n (where F_n = 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) =

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

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

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

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

= ζ + ζ² + ζ³ + ζ⁴ + ζ⁵ +ζ⁶ + ζ⁷ + ζ⁸ + ζ⁹ + ζ¹⁰ + ζ¹¹ + ζ¹² + ζ¹³ + ζ¹⁴ + ζ¹⁵ + ζ¹⁶ = x₁ + x₂ = -1

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

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

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

= ζ +ζ² +ζ³ + ζ⁴ + ζ⁵ +ζ⁶ + 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

• Tom Rike, "Fermat Numbers and the Heptadecagon", San Jose Math Circle.Org
• Heinrich Dorrie, 100 Great Problems of Elementary Mathematics
• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001