Lemma 1:

if:

x

^{10}+ x

^{9}+ x

^{8}+ x

^{7}+ x

^{6}+ x

^{5}+ x

^{4}+ x

^{3}+ x

^{2}+ x + 1 = 0

then there exists y such that:

y

^{5}- y

^{4}-4y

^{3}+ 3y

^{2}+ 3y - 1 = 0

Proof:

(1) Let y = -(x + x

^{-1}) = -x + -x

^{-1}

(2) Then:

y

^{2}= x

^{2}+ 2(-x)*(-x)

^{-1}+ x

^{-2}= x

^{2}+ 2 + x

^{-2}

y

^{3}= y*y

^{2}= -x

^{3}+ -2x + -x

^{-1}+ -x + -2x

^{-1}+ -x

^{-3}= -x

^{3}+ -3x + -3x

^{-1}+ -x

^{-3}

y

^{4}= y*y

^{3}= x

^{4}+ 3x

^{2}+ 3 + x

^{-2}+ x

^{2}+ 3 + 3x

^{-2}+ x

^{-4}= x

^{4}+ 4x

^{2}+ 6 + 4x

^{-2}+ x

^{-4}

y

^{5}= y*y

^{4}= -x

^{5}+ -4x

^{3}-6x - 4x

^{-1}+ -x

^{-3}+ -x

^{3}+ -4x + -6x

^{-1}+ -4x

^{-3}+ -x

^{-5}=

= -x

^{5}- 5x

^{3}- 10x -10x

^{-1}-5x

^{-3}- x

^{-5}

(3) So:

y

^{5}- y

^{4}-4y

^{3}+ 3y

^{2}+ 3y - 1 =

= [-x

^{5}- 5x

^{3}- 10x -10x

^{-1}-5x

^{-3}- x

^{-5}] - [x

^{4}+ 4x

^{2}+ 6 + 4x

^{-2}+ x

^{-4}] - 4[-x

^{3}+ -3x + -3x

^{-1}+ -x

^{-3}] + 3[x

^{2}+ 2 + x

^{-2}] + 3[-x + -x

^{-1}] - 1 =

= -x

^{5}- x

^{4}+ [-5x

^{3}+4x

^{3}] + [-4x

^{2}+ 3x

^{2}] + [-10x +12x -3x] + [-6 + 6 - 1] + [-10x

^{-1}+ 12x

^{-1}-3x

^{-1}] + [-4x

^{-2}+ 3x

^{-2}] + [-5x

^{-3}+ 4x

^{-3}] -x

^{-4}-x

^{-5}=

= -x

^{5}-x

^{4}-x

^{3}-x

^{2}-x - 1 -x

^{-1}- x

^{-2}-x

^{-3}-x

^{-4}-x

^{-5}

(4) We assume that:

x

^{10}+ x

^{9}+ x

^{8}+ x

^{7}+ x

^{6}+ x

^{5}+ x

^{4}+ x

^{3}+ x

^{2}+ x + 1 = 0

(5) We divide each side of the equation by -x

^{5}to get:

-x

^{5}-x

^{4}-x

^{3}-x

^{2}-x - 1 -x

^{-1}- x

^{-2}-x

^{-3}-x

^{-4}-x

^{-5 }= 0

(6) Now, we apply step #3 to get:

y

^{5}- y

^{4}-4y

^{3}+ 3y

^{2}+ 3y - 1 = 0

QED

Lemma 2: The eleventh roots of unity using trigonometric methods

Using trigonometric methods, the eleventh roots of unity are:

cos 0 + isin 0

cos 2π/11 + isin 2π/11

cos 4π/11 + isin 4π/11

cos 6π/11 + isin 6π/11

cos 8π/11 + isin 8π/11

cos 10π/11 + isin 10π/11

cos 12π/11 + isin 12π/11

cos 14π/11 + isin 14π/11

cos 16π/11 + isin 16π/11

cos 18π/11 + isin 18π/11

cos 20π/11 + isin 20 π/11

Proof:

This follows directly from a previous result [See Corollary 1.1, here].

QED

Corollary 2.1:

The 10 solutions for:

x

^{10}+ x

^{9}+ x

^{8}+ x

^{7}+ x

^{6}+ x

^{5}+ x

^{4}+ x

^{3}+ x

^{2}+ x + 1 = 0

are:

cos 2π/11 + isin 2π/11

cos 4π/11 + isin 4π/11

cos 6π/11 + isin 6π/11

cos 8π/11 + isin 8π/11

cos 10π/11 + isin 10π/11

cos 12π/11 + isin 12π/11

cos 14π/11 + isin 14π/11

cos 16π/11 + isin 16π/11

cos 18π/11 + isin 18π/11

cos 20π/11 + isin 20 π/11

Proof:

(1) From Lemma 2 above, we can see that the solutions for:

x

^{11}- 1 = 0

are:

cos 0 + isin 0

cos 2π/11 + isin 2π/11

cos 4π/11 + isin 4π/11

cos 6π/11 + isin 6π/11

cos 8π/11 + isin 8π/11

cos 10π/11 + isin 10π/11

cos 12π/11 + isin 12π/11

cos 14π/11 + isin 14π/11

cos 16π/11 + isin 16π/11

cos 18π/11 + isin 18π/11

cos 20π/11 + isin 20 π/11

(2) Using the Fundamental Theorem of Algebra (see Theorem, here), we can see that:

x

^{11}- 1 = (x - cos0 - isin 0)(x - cos 2π/11 - isin 2π/11)*...*(x - cos 20π/11 - isin 20π/11)

(3) Since, cos 0 - isin 0 = 1 - i*0 = 1, dividing both sides by (x - 1) gives us:

(x

^{11}-1)/(x - 1) = (x - cos 2π/11 - isin 2π/11)*...*(x - cos 20π/11 - isin 20π/11)

(4) From a previous result (see Lemma 1, here):

(x

^{11}-1)/(x-1) = x

^{10}+ x

^{9}+ x

^{8}+ x

^{7}+ x

^{6}+ x

^{5}+ x

^{4}+ x

^{3}+ x

^{2}+ x + 1

(5) So, combining step #3 and step #4 gives us:

(x

^{11}-1)/(x-1) = x

^{10}+ x

^{9}+ x

^{8}+ x

^{7}+ x

^{6}+ x

^{5}+ x

^{4}+ x

^{3}+ x

^{2}+ x + 1 = (x - cos 2π/11 - isin 2π/11)*...*(x - cos 20π/11 - isin 20π/11)

QED

Lemma 3:

-2cos 12π/11 = -2cos(10π/11)

-2 cos 14π/11 = -2cos(8π/11)

-2 cos 16π/11 = -2cos(6π/11)

-2 cos 18π/11 = -2cos(4π/11)

-2cos 20π/11 = -2cos(2π/11)

Proof:

Using Property 5 here and the fact that 2π = 360 degrees [see Definition, here], we have:

-2cos 12π/11 = -2 cos (12π/11 - 2π) = -2cos([12π - 22π]/11) = -2cos(-10π/11) = -2cos(10π/11)

-2 cos 14π/11 = -2 cos (14π/11 - 2π) = -2cos([14π - 22π]/11) = -2cos(-8π/11) = -2cos(8π/11)

-2 cos 16π/11 = -2cos (16π/11 - 2π) = -2cos([16π - 22π]/11) = -2cos(-6π/11) = -2cos(6π/11)

-2 cos 18π/11 = -2cos (18π/11 - 2π) = -2cos([18π - 22π]/11) = -2cos(-4π/11) = -2cos(4π/11)

-2cos 20π/11 = -2cos(20π/11 - 2π) = -2cos([20π - 22π]/11) = -2cos(-2π/11) = -2cos(2π/11)

QED

Lemma 4

if:

x = cos u + isin u

then:

1/x = cos u - isin u

Proof:

(1) Let x = cos u + isin u

(2) 1/x = 1/[cos u + isin u] = [cos u - isinu]/[cos

^{2}u - i

^{2}sin

^{2}u] = [cos u - isin u]/[cos

^{2}u + sin

^{2}u] = cos u - isin u

QED

Corollary 4.1

if:

x = cos u + isin u

then:

-x + -x

^{-1}= -2cos u

Proof:

(1) Let x = cos u + isin u

(2) Then x

^{-1}= cosu - isin u

(3) x + x

^{-1}= 2cos u

(4) -x + x

^{-1}= -2cos u

QED

Corollary 4.2:

if:

x

^{10}+ x

^{9}+ x

^{8}+ x

^{7}+ x

^{6}+ x

^{5}+ x

^{4}+ x

^{3}+ x

^{2}+ x + 1 = 0

and

y = -x + -x

^{-1}

then:

the possible values of y are:

-2cos 2π/11

-2cos 4π/11

-2cos 6π/11

-2cos 8π/11

-2cos 10π/11

Proof:

(1) Case 1: x = cos 2π/11 + isin 2π/11

-x + -x

^{-1}= -2cos 2π/11

(2) Case 2: x = cos 4π/11 + isin 4π/11

-x + -x

^{-1}= -2 cos 4π/11

(3) Case 3: x = cos 6π/11 + isin 6π/11

-x + -x

^{-1}= -2 cos 6π/11

(4) Case 4: x = cos 8π/11 + isin 8π/11

-x + -x

^{-1}= -2 cos 8π/11

(5) Case 5: x = cos 10π/11 + isin 10π/11

-x + -x

^{-1}= -2cos 10π/11

(6) Case 6: x = cos 12π/11 + isin 12π/11

-x + -x

^{-1}= -2cos 12π/11 = -2cos(10π/11) [See Lemma 3 above]

(7) Case 7: x = cos 14π/11 + isin 14π/11

-x + -x

^{-1}= -2 cos 14π/11 = -2cos(8π/11) [See Lemma 3 above]

(8) Case 8: x = cos 16π/11 + isin 16π/11

-x + -x

^{-1}= -2 cos 16π/11 = -2cos(6π/11) [See Lemma 3 above]

(9) Case 9: x = cos 18π/11 + isin 18π/11

-x + -x

^{-1}= -2 cos 18π/11 = -2cos(4π/11) [See Lemma 3 above]

(10) Case 10: x = cos 20π/11 + isin 20π/11

-x + -x

^{-1}= -2cos 20π/11 = -2cos(2π/11) [See Lemma 3 above]

QED

Lemma 5: 2cosαcosβ = cos(α + β) + cos(α - β)

Proof:

(1) cos(A+B) = cosA*cosB - sinA*sinB [See Theorem 2, here]

(2) Likewise:

cos (A - B) = cos (A + [-B]) = cosA*cos(-B) - sin(A)sin(-B) = cosAcosB + sinAsinB

(3) cos(A + B) + cos(A-B) = cosA*cosB - sinA*sinB + cosA*cosB + sinA*sinB = 2cosAcosB

QED

Corollary 5.1: 2cos

^{2}(α) = cos(2*α) + 1

Proof:

cos(2a) + 1 = cos(2a) + cos(0) = cos(a + a) + cos(a - a) = 2cos(α)cos(α) = 2cos

^{2}(α)

QED

Corollary 5.2:

if:

x

^{10}+ x

^{9}+ x

^{8}+ x

^{7}+ x

^{6}+ x

^{5}+ x

^{4}+ x

^{3}+ x

^{2}+ x + 1 = 0

then:

y

^{5}- y

^{4}-4y

^{3}+ 3y

^{2}+ 3y - 1 = 0 with roots a,b,c,d,e

and

a

^{2}= -b + 2

b

^{2}= -d + 2

c

^{2}= -e + 2

d

^{2}= -c + 2

e

^{2}= -a + 2

Proof:

(1) Using Corollary 4.2 above, we have:

a = -2cos 2π/11

b = -2cos 4π/11

c = -2 cos 6π/11

d = -2 cos 8π/11

e = -2 cos 10π/11

(2) Using Corollary 5.1 above, we have:

2cos

^{2}(2π/11) = cos(4π/11) + 1

2cos

^{2}(4π/11) = cos(8π/11) + 1

2cos

^{2}(6π/11) = cos(12π/11) + 1

2cos

^{2}(8π/11) = cos(16π/11) + 1

2cos

^{2}(10π/11) = cos(20π/11) + 1

(3) Multiplying both sides by 2 gives us:

4cos

^{2}(2π/11) = 2cos(4π/11) + 2

4cos

^{2}(4π/11) = 2cos(8π/11) + 2

4cos

^{2}(6π/11) = 2cos(12π/11) + 2

4cos

^{2}(8π/11) = 2cos(16π/11) + 2

4cos

^{2}(10π/11) = 2cos(20π/11) + 2

(4) So that we have:

a

^{2}= 4cos

^{2}(2π/11) = 2cos(4π/11) + 2 = -b + 2

b

^{2}= 4cos

^{2}(4π/11) = 2cos(8π/11) + 2 = -d + 2

c

^{2}= 4cos

^{2}(6π/11) = 2cos(12π/11) + 2 = 2cos(10π/11) + 2 = -e + 2

d

^{2}= 4cos

^{2}(8π/11) = 2cos(16π/11) + 2 = 2cos(6π/11) + 2 = -c + 2

e

^{2}= 4cos

^{2}(10π/11) = 2cos(20π/11) + 2 = 2cos(2π/11) + 2 = -a + 2

QED

Corollary 5.3:

if:

x

^{10}+ x

^{9}+ x

^{8}+ x

^{7}+ x

^{6}+ x

^{5}+ x

^{4}+ x

^{3}+ x

^{2}+ x + 1 = 0

then:

y

^{5}- y

^{4}-4y

^{3}+ 3y

^{2}+ 3y - 1 = 0 with roots a,b,c,d,e

and

a

^{2}= -b + 2

where at all times we can exchange each letter by one arrow:

a --> b --> d --> c --> e --> a

[That is, each letter can be changed to the next letter so that: a

^{2}= -b + 2 becomes b

^{2}= -d + 2 { a --> b and b --> d } which becomes d

^{2}= -c + 2 { b --> d and d -->c } which becomes c

^{2}= -e + 2 { d --> c and c --> e } which becomes e

^{2}= -a + 2 { c --> e and e --> a }]

Proof:

This follows directly from Corollary 6.2 above.

QED

Lemma 6:

if:

x

^{10}+ x

^{9}+ x

^{8}+ x

^{7}+ x

^{6}+ x

^{5}+ x

^{4}+ x

^{3}+ x

^{2}+ x + 1 = 0

then:

y

^{5}- y

^{4}-4y

^{3}+ 3y

^{2}+ 3y - 1 = 0 with roots a,b,c,d,e

and

ab = -a -c

bc = -a -e

cd = -a -d

de = -a-b

ac = -b -d

bd = -b -e

ce = -b -c

ad = -c - e

be = -c -d

ae = -d -e

Proof:

(1) Using Corollary 4.2 above, we have:

a = -2cos 2π/11

b = -2cos 4π/11

c = -2 cos 6π/11

d = -2 cos 8π/11

e = -2 cos 10π/11

(2) So, we have:

-a -b = (2cos 2π/11) + (2cos 4π/11)

-a -c = (2cos 2π/11) + (2cos 6π/11)

-a -d = (2cos 2π/11) + (2cos 8π/11)

-a -e = (2cos 2π/11) + (2cos 10π/11)

-b -c = (2cos 4π/11) + (2cos 6π/11)

-b -d = (2cos 4π/11) + (2cos 8π/11)

-b -e = (2cos 4π/11) + (2cos 10π/11)

-c -d = (2cos 6π/11) + (2cos 8π/11)

-c -e = (2cos 6π/11) + (2cos 10π/11)

-d -e = (2cos 8π/11) + (2cos 10π/11)

(3) Which gives us:

ab = (-2cos 2π/11)(-2cos 4π/11) = 4cos(2π/11)*cos(4π/11) = 2[2cos(2π/11)*cos(4π/11)] = 2[cos(4π/11 + 2π/11) + cos(4π/11 - 2π/11) = 2[cos(6π/11) + cos(2π/11)] = 2cos(6π/11) + 2cos(2π/11) = -a -c

ac = (-2cos 2π/11)(-2 cos 6π/11) = 4cos(2π/11)*cos(6π/11) = 2[2cos(2π/11)*cos(6π/11)] = 2[cos(6π/11 + 2π/11) + cos(6π/11 - 2π/11) = 2[cos(8π/11) + cos(4π/11)] = 2cos(8π/11) + 2cos(4π/11) = -b -d

ad = (-2cos 2π/11)(-2 cos 8π/11) = 4cos(2π/11)*cos(8π/11) = 2[2cos(2π/11)*cos(8π/11)] = 2[cos(8π/11 + 2π/11) + cos(8π/11 - 2π/11) = 2[cos(10π/11) + cos(6π/11)] = 2cos(10π/11) + 2cos(6π/11) = -c -e

ae = (-2cos 2π/11)(-2 cos 10π/11) = 4cos(2π/11)*cos(10π/11) = 2[2cos(2π/11)*cos(10π/11)] = 2[cos(10π/11 + 2π/11) + cos(10π/11 - 2π/11) = 2[cos(12π/11) + cos(8π/11)] = 2cos(12π/11) + 2cos(8π/11) = 2cos(10π/11) + 2cos(8π/11) = -d -e

bc = (-2cos 4π/11)( -2 cos 6π/11) = 4cos(4π/11)*cos(6π/11) = 2[2cos(4π/11)*cos(6π/11)] = 2[cos(6π/11 + 4π/11) + cos(6π/11 - 4π/11) = 2[cos(10π/11) + cos(2π/11)] = 2cos(10π/11) + 2cos(2π/11) = -a -e

bd = (-2cos 4π/11)(-2 cos 8π/11) = 4cos(4π/11)*cos(8π/11) = 2[2cos(4π/11)*cos(8π/11)] = 2[cos(8π/11 + 4π/11) + cos(8π/11 - 4π/11) = 2[cos(12π/11) + cos(4π/11)] = 2cos(12π/11) + 2cos(4π/11) = 2cos(10π/11) + 2cos(4π/11) = -b -e

be = (-2cos 4π/11)(-2 cos 10π/11) = 4cos(4π/11)*cos(10π/11) = 2[2cos(4π/11)*cos(10π/11)] = 2[cos(10π/11 + 4π/11) + cos(10π/11 - 4π/11) = 2[cos(14π/11) + cos(6π/11)] = 2cos(14π/11) + 2cos(6π/11) = 2cos(8π/11) + 2cos(6π/11) = -c -d

cd = ( -2 cos 6π/11)(-2 cos 8π/11) = 4cos(6π/11)*cos(8π/11) = 2[2cos(6π/11)*cos(8π/11)] = 2[cos(8π/11 + 6π/11) + cos(8π/11 - 6π/11) = 2[cos(14π/11) + cos(2π/11)] = 2cos(14π/11) + 2cos(2π/11) = 2cos(8π/11) + 2cos(2π/11) = -a -d

ce = (-2 cos 6π/11)(-2 cos 10π/11) = 4cos(6π/11)*cos(10π/11) = 2[2cos(6π/11)*cos(10π/11)] = 2[cos(10π/11 + 6π/11) + cos(10π/11 - 6π/11) = 2[cos(16π/11) + cos(4π/11)] = 2cos(16π/11) + 2cos(4π/11) = 2cos(6π/11) + 2cos(4π/11) = -b -c

de = (-2 cos 8π/11)(-2 cos 10π/11) = 4cos(8π/11)*cos(10π/11) = 2[2cos(8π/11)*cos(10π/11)] = 2[cos(10π/11 + 8π/11) + cos(10π/11 - 8π/11) = 2[cos(18π/11) + cos(2π/11)] = 2cos(18π/11) + 2cos(2π/11) = 2cos(4π/11) + 2cos(2π/11) = -a -b

QED

Corollary 6.1:

if:

x

^{10}+ x

^{9}+ x

^{8}+ x

^{7}+ x

^{6}+ x

^{5}+ x

^{4}+ x

^{3}+ x

^{2}+ x + 1 = 0

then:

y

^{5}- y

^{4}-4y

^{3}+ 3y

^{2}+ 3y - 1 = 0 with roots a,b,c,d,e

and

ab = -a -c

ac = -b -d

where at all times we can exchange each letter by one arrow:

a --> b --> d --> c --> e --> a

[That is, each letter can be changed to the next letter so that: ab = -a - c becomes bd = -b - e (a --> b and b--> d and c --> e) which becomes cd = -d -a ( b --> d and d --> c and e --> a) which becomes ce = -c -b ( c --> e and d --> c and a --> b) which becomes ae = -e -d etc.]

Proof:

This follows directly from Lemma 7 above.

QED

References

- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001