Gauss: Q(μp)

In today's blog, I will review some basic properties of Q(μ_p) which is the set of complex numbers which are rational expressions in the p-th roots of unity.

The content in today's blog is taken straight from Jean-Pierre Tignol's Galois Theory of Algebraic Equations. I will later use the results in today's blog in the proof by Carl Friedrich Gauss that all roots of unity are expressible as radicals.

Definition 1: μ_p

Let μ_p denote the set of p-th roots of unity.

Example:

μ₁ = {1}
μ₂ = {1, -1}
μ₃ = {1, (1/2)[-1 + √-3], 1/2)[-1 - √-3])
μ₄ = {1, -1, i, -i}

Now, let's use this to define a field (see Definition 3, here for definition of a field if needed)

Definition 2: Q(μ_p)

Let Q(μ_p) denote the set of complex numbers that are rational expressions in these p-th roots of unity.

This gives us that:



Now, we can show that Q(μ_p) = Q(ζ).

Lemma 1:

Q(μ_p) = Q(ζ)

Proof:

(1) From Defintion 1 above:

μ_p = { ρ₁, ..., ρ_p}

(2) Let ζ = a primitive p-th root of unity (see Definition 2, here for definition of a primitive p-th root of unity if needed)

(3) Using Theorem 3, here, we have:

μ_p = { 1, ζ, ζ², ..., ζ^p-1 }

(4) Then we have:



QED

Lemma 2:

Let P,Q be polynomials with coefficients in field F.

If:

P is irreducible in F[X]. (See Definition 1, here, for definition of irreducible polynomials)

P,Q have a common root u in field K which contains F

Then:

P divides Q.

Proof:

(1) Assume that P does not divide Q.

(2) Then P,Q are relatively prime polynomials since P is irreducible. [See Definition 1, here for definition of irreducible polynomials]

(3) Then (see Corollary 3.1, here), there exists polynomials U,V in F[X] such that:

P(X)U(X) + Q(X)V(X) = 1

(4) Substituting the common root u into the polynomials, we get:

P(u)U(u) + Q(u)V(u) = 1 in K

(5) Since u is the root for P(X) and Q(X), this gives us that:

P(u) = 0, Q(u) = 0 in K

(6) But then:

o*U(u) + 0*V(u) = 1 in K which is impossible.

(7) Therefore we reject our assumption in step #1.

QED

Lemma 3:

If u ∈ field K is a root of an irreducible polynomial P ∈ F[X] of degree d

and



Then:

every element in F(u) can be uniquely written in the form:

a₀ + a₁u + a₂u² + ... + a_d-1u^d-1 with a_i ∈ field F.

Proof:

(1) Let f(u)/g(u) be an arbitrary element in F(u)

(2) Since g(u) ≠ 0, it follows that g(u) is not divisible by P since:

(a) Assume that g(u) is divisible by P.

(b) Since P(u) = 0, if follows that (X - u) divides P. [See Theorem, here]

(c) But if P divides g(u), then (X -u) divides g(u).

(d) But this is impossible since this implies g(u) = 0 [See Theorem, here] but g(u) ≠ 0.

(e) So we reject the assumption in step #2a.

(3) Then, g(u) is relatively prime to P since P is irreducible. [See Definition 1, here for definition of irreducible polynomials]

(4) Then (see Corollary 3.1, here), there exists polynomials h,U in F such that:

g(X)h(X) + P(X)U(X) = 1 in F[X]

(5) Since P(u) = 0, substituting u for X gives us:

g(u)h(u) + P(u)U(u) = g(u)h(u) + 0*U(u) = g(u)h(u) = 1 in K.

(6) Since g(u) ≠ 0, we have:

h(u) = 1/g(u) in K

which gives us that:

f(u)/g(u) = f(u)h(u) in K

(7) Using the Division Algorithm for Polynomials (see Theorem, here), there exists Q, R such that:

fh = PQ + R in F[X] with deg R ≤ d - 1.

(8) Since P(u) = 0, it follows that:

f(u)h(u) = P(u)Q(u) + R(u) = 0*Q(u) + R(u) = R(u) in K

(9) Since R(u) is a polynomial of degree at most d-1, we have converted an arbitrary expression f(u)/g(u) ∈ F(u) into a polynomial expression of the type:

a₀ + a₁u + ... + a_d-1u^d-1

with a_i ∈ F

(10) Now, I will prove uniqueness.

(11) Assume that:

a₀ + a₁u + ... + a_d-1u^d-1 = b₀ + b₁u + ... + b_d-1u^d-1

(12) Let us define V(X) such that:

V(X) = (a₀ - b₀) + (a₁ - b₁)X + ... + (a_d-1 - b_d-1)X^d-1 ∈ F[X]

(13) It is clear that V(u) = 0 from step #11.

(14) Using Lemma 2 above, it is clear that P divides V.

(15) But since deg V ≤ d-1 and since P is of degree d, this is impossible unless V=0.

(16) Therefore, it follows that:

a₀ - b₀ = a₁ - b₁ = ... = a_d-1 - b_d-1 = 0

QED

Theorem 4: Every element in Q(μ_p) can be expressed in one and only one way as a linear combination with rational coefficients of the p-th roots of unity other than 1:

a₁ζ + a₂ζ² + ... + a_p-1ζ^p-1

with a_i ∈ Q.

Proof:

(1) Let ζ be a root of Φ_p [See Definition 1, here]

(2) Q(μ_p) = Q(ζ) [From Lemma 1 above]

(3) Since p is a prime, Φ_p is irreducible over Q . [See Corollary 1.1., here]

(4) Using Lemma 3 above, since the degree of ζ is p-1, it follows that every element a ∈ Q(μ_p) can be uniquely expressed in the form:

a = a₀ + a₁ζ + a₂ζ² + ... + a_p-2ζ^p-2

for some a_i ∈ Q.

(5) Using the cyclotomic equation (see Lemma 1, here), we have:

Φ_p(ζ) = 1 + ζ + ζ² + ... + ζ^p-1 = 0

which means that:

ζ + ζ² + ... + ζ^p-1 = -1

(6) This gives us that:

a₀ = -a₀(ζ + ζ² + ... + ζ^p-1)

(7) Combining step #6 with step #4 gives us:

a = (a₁ - a₀)ζ + (a₂ - a₀)ζ² + ... + (a_p-2 - a₀)ζ^p-2 + (-a₀)ζ^p-1

(8) To prove uniqueness, let's assume that:

a₁ζ + ... + a_p-1ζ^p-1 = b₁ζ + ... + b_p-1ζ^p-1

(9) From step #5, we also have:

ζ^p-1 = -1 - ζ - ζ² + ... -ζ^p-2

(10) Putting this into step #8 gives us:

a₁ζ + ... + a_p-1(-1 - ζ - ζ² + ... -ζ^p-2) = b₁ζ + ... + b_p-1(-1 - ζ - ζ² + ... -ζ^p-2)

which reduces to:

-a_p-1 + (a₁ - a_p-1)ζ + (a₂ - a_p-1)ζ² + ... + (a_p-2 - a_p-1)ζ^p-2 = -b_p-1 + (b₁ - b_p-1)ζ + (b₂ - b_p-1)ζ² + ... + (b_p-2 - b_p-1)ζ^p-2

(11) From Lemma 3 above, we can conclude that the coefficients on both sides are equal which gives us:

a_p-1 = b_p-1

which then gives us:

a₁ = b₁
...
a_p-2 = b_p-2

QED

References

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

Eisenstein: Criteria for Irreducibility of Polynomials

In today's blog, I will go over a classical result from Ferdinand Eisenstein. The content in today's proof is taken directly from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

I will later use this result in Carl Friedrich Gauss's proof that all roots of unity are expressible as radicals.

For those who are not familiar, a monic polynomial is a polynomial whose leading coefficient is 1.

Theorem 1: Eisenstein's Criteria for Irreducibility of Polynomials

Let P be a monic polynomial with integral coefficients such that:

P = X^t + c_t-1X^t-1 + ... + c₁X + c₀ ∈ Z[X]

If there is a prime number p which divides c_i for i = 0, ..., t-1 but p² does not divide c₀

then:

P is irreducible over Q

Proof

(1) Assume that P is not irreducible over Q.

(2) Then there exist two non-constant polynomials f,g ∈ Q[X] such that:

P = fg

(3) Assume that f has degree n with n ≥ 1 and g has degree m with m ≥ 1 so that:

f = Xⁿ + a_n-1X^n-1 + ... + a₁X + a₀ ∈ Z[X]

g = X_m + b_m-1X^m-1 + ... + b₁X + b₀ ∈ Z[X]

(4) Since P is monic, we can assume that f,g are monic [if f,g are not monic, then (f/a_n is monic and a_n*g is monic since P = (f/a_n)*(g*a_n)]

(5) From step #2, it is also clear that c₀ = a₀*b₀

(6) Using Euclid's Lemma (see Lemma 2, here), it is clear that for the prime p, p divides a₀ or b₀.

(7) It is also clear that since p² does not divide c₀ that p cannot divide both a₀ and b₀.

(8) We can assume that p divides a₀ but not b₀ (if it did not, we could switch f,g since they are interchangeable with respect to p and make the same assumption)

(9) Let i be the largest index of a such that p divides a_i. It is clear that 0 ≤ i ≤ n-1. So, we can conclude that for any i, p divides a₀, a₁, ..., a_i.

(10) Let k = i+1, it is clear that 1 ≤ k ≤ n and p does not divide a_k.

(11) So:

c_i+1 = a_i+1b₀ + a_ib₁ + a_i-1b₂ + ... + a₀b_i+1.

where b_j = 0 if j is greater than m.

(12) Now i+1 ≤ n which is less than t since n+1 ≤ m+n = t [since m ≥ 1 and n ≥ 1]

(13) So it follows that i+1 ≤ t-1 and p divides c_i+1. [From the given in the theorem that p divides c₀, ..., c_t-1]

(14) So, from step #13, it follows that since p divides c_i+1 and p divides a₀, a₁, ..., a_i [step #9] that p must also divide a_i+1b₀.

(15) But this is impossible since p doesn't divide b₀ (step #8) and p doesn't divide a_i+1 [step #10]

(16) So we have a contradiction and we must reject our assumption in step #1.

QED

Corollary 1.1: For every prime p, the cyclotomic polynomial

Φ_p(x) = x^p-1 + x^p-2 + ... + x + 1

is irreducible over the field of rational numbers

Proof:

(1) From the definition of the cyclotomic polynomial (see Definition 1, here):

Φ_p(x) = (x^p - 1)/(x - 1)

(2) Setting x = y + 1, we get:

Φ_p(y+1) = ([y + 1]^p - 1)/([y + 1] - 1) =

= ([y + 1]^p - 1)/y

(3) Using the Binomial Theorem (see Theorem, here):



(4) So, combining this with step #2 gives us:












(5) Now, it is clear that for (y+1)^p all the resulting coefficients are integral (since integers are closed on additional and multiplication, see Lemma 1 and Lemma 2, here).

(6) Further, we know that for p!/[(m!)(p-m)!] where 1 ≤ m ≤ p-2, p is not divisible by (m!) or by (p-1)! [Since p is a prime and only divisible by p and m! and (p-m)! does not include p]

(7) So, we can conclude that for m=1, ..., p-2, we have p divides p!/[(m!)(p-m)!]

(8) Now, we can use Eisenstein's Criterion above to conclude that the cyclotomic polynomial is irreducible in the field Q since:

the cyclotomic polynomial is monic, p divides all coefficients except for the leading coefficient, and p² does not divide the last coefficient (p² does not divide p.)

QED

References

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

Gauss: Gauss's Lemma for Polynomials

Today's proof is taken from Carl Friedrich Gauss' Disquisitiones Arithmeticae (Article 42). This is Gauss' classic work on number theory which he wrote when he was 21. It is a masterpiece that includes the first use of modular arithmetic, the first complete proof of the fundamental theorem of algebra, a proof for quadratic reciprocity, Gauss' analysis of cyclotomic equations, as well as the first systematic presentation of number theory.

Lemma 1:

if a prime p doesn't divide an integer a, then (a +bp)/cp is not an integer

Proof:

(1) There exists an integer d such that a ≡ d (mod p) where 1 ≤ d ≤ p-1.

(2) So, p divides a - d. (see here for review of modular arithmetic)

(3) And p divides (a - d) so p divides (a -d) + bp = (a + bp) -d

(4) So therefore a + bp ≡ d (mod p)

(5) So it follows that since p doesn't divide a+bp, then (a +bp)/cp is not an integer.

QED

A monic polynomial is a polynomial whose leading coefficient is 1 (see Definition 5, here)

Theorem 2: Gauss's Lemma for Polynomials

Let P,Q be two monic polynomials such that:

P = x^m + a₁x^m-1 + a₂x^m-2 + a₃x^m-3 + ... + a_m

Q = xⁿ + b₁x^n-1 + b₂x^n-2 + b₃x^n-3 + ... + b_n

If the coefficients of P are rational but not all integers and if the product is also monic such that:

PQ = x^m+n + c₁x^m+n-1 + c₂x^m+n-2 + c₃x^m+n-3 + ... + c_m+n

Then, the coefficients c₁, c₂, ..., c_m+n cannot all be integers.

Proof:

(1) Let us assume that all rational cofficients in a₁, ..., a_m and b₁, ..., b_n are expressed in their lowest form.

(2) Let p be a prime number that divides one of the denominators of P [we can assume this since not all the coefficients of P are integers]

(3) Since Q is monic, it is clear that the at least one of the coefficients of (Q)/p will have p as a factor of its denominator (for example, bⁿ will have (1/p) as its coefficient).

(4) In P, there will always be one term, a fraction, whose denominator involves high powers of p than all the fractional coefficients that precede it and no lower powers than the denominators of all succeeding fractional coefficients.

For example, if there is only one fraction with the rest integers, than this fraction is the coefficient. One can start with the first coefficient divisible by p and then check if there are any coefficients after it that are divisible by a greater power of p.

(5) We can identify the term in step #4 as Gx^g with G divisible by 1/(p^t) with t ≥ 1.

(6) Since p divides at least one of the denominators in (Q)/p, the same term in step #4 exists in (Q)/p.

(7) We can identify the term in step #6 as Hx^h with H divisible by 1/(p^u) with u ≥ 1.

(8) Now, it follows that t + u ≥ 2.

(9) Next, I will show that the term x^g+h in the product of (P) and (Q) will have a fractional coefficient who denominator is divisible by 1/(p^t+u-1). This then will complete the proof since a coefficient that is divisible by 1/(p^t+u-1) where p is a prime is not an integer.

(10) Let the terms in P which precede Gx^g be 'Gx^g+1, ''Gx^g+2 and those which follow Gx^g be G'x^g-1, G''x^g-2

(11) Let the terms in (Q)/p which precede Hx^h be 'Hx^h+1, ''Hx^h+2 and those which follow Hx^h be H'x^h-1, H''x^h-2

(12) If we multiply P and (Q)/p, the coefficient of the term Gx^g*Hx^h is GH + 'GH' + ''GH'' + ... + 'HG' + ''HG'' + ...

(13) Because G,H have the highest power of p, there exists e₀,f₀ such that:

p does not divide e₀
p does not divide f₀

and

GH = e₀/(f₀p^t+u)

(14) Since any of the ('G, ''G, etc.) or ('H, ''H, etc.) have a power of p less than t or u and because any of the (G', G'', etc.) or (H', H'', etc) have a power of p no greater than t or u, it follows that any product of the form ('GH', ''GH'', etc.) will have the following form:

'GH' = e_i/(f_ip^t+u-d_i)

where

p does not divide e_i
p does not divide f_i
d_i ≥ 1.

(15) Now if we sum all the factors in step #14 and reduce the denominator to the lowest term, then we get:

∑ e_i/(f_ip^t+u-d_i) =

= e'/(f'p^t+u-d'i)

where

p does not divide e'
p does not divide f'
d' ≥ 1.

(16) If we add up the result of step #14 with step #15, we get:

(e₀f' + e'f₀p^d)/(f₀f'p^t+u)

where:

p does not divide e₀
p does not divide e'
p does not divide f₀
p does not divide f'
d ≥ 1
(t + u) ≥ 2

(17) Since Q = p*(Q)/p, for this same coefficient for P*Q, we have:

(e₀f' + e'f₀p^d)/(f₀f'p^t+u-1)

where:

p does not divide e₀
p does not divide e'
p does not divide f₀
p does not divide f'
d ≥ 1
(t + u - 1) ≥ 1

(18) Since p does not divide e₀ or f', it follows that p does not divide e₀f.

(19) If:

a = e₀f'
b = ef₀p^d-1
c = f₀f'p^t+u-2

Then we can use Lemma 1 above to conclude that:

(e₀f' + e'f₀p^d)/(f₀f'p^t+u-1)

is not an integer.

QED

Corollary 2.1:

If a monic polynomial in Q[X] divides a monic polynomial with integral coefficients, then its coefficients are all integral.

Proof:

(1) Let f be a monic polynomial with integral coefficients.

(2) Let P be a monic polynomial that divides f.

(3) Since P divides f, there exists a polynomial Q such that:

f = PQ

(4) Since P,f are monic, it follows that Q must also be monic. [See Lemma 1, here]

(5) Assume that the coefficient in P are not all integers.

(6) Then, using Theorem 2 above, it follows that the coefficients of f are not all integers.

(7) But this is not correct since by assumption, all the coefficients of f are integers.

(8) So we reject our assumption in step #5.

QED

References

• Carl Friedrich Gauss, Translated by Arthur A. Clarke, Disquisitiones Arithmeticae, Yale University Press, 1965.
  

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.