Cyclotomic Equation

Lemma 1:

Let ζ = the n-th root of unity where ζ ≠ 1

Then:

1 + ζ + ζ² + ... + ζ^n-1 = 0

Proof:

(1) Since ζⁿ = 1, we have:

1 + ζ + ζ² + ... + ζ^n-1 = ζ(1 + ζ + ζ² + ... + ζ^n-1)

(2) Now ζ ≠ 0 [since 0ⁿ ≠ 1] and ζ ≠ 1

(3) Let x = 1 + ζ + ζ² + ... + ζ^n-1

(4) Assume that x ≠ 0

(5) So, x = ζx [from step #1]

(6) Since we assume x ≠ 0, we can divide both sides by x to get 1 = ζ

(7) But this contradicts step #2 so we reject our assumption in step #3 and conclude that:

1 + ζ + ζ² + ... + ζ^n-1 =0

QED

Corollary 1.1:

if ζ is a primitive n-th root of unity, then:

a = (-a) ζ + (-a)ζ² + ... + (-a)ζ^n-1

Proof:

(1) From Lemma 1 above:

-1 = ζ + ζ² + ... + ζ^n-1

(2) So, a = (-a)(ζ + ζ² + ... + ζ^n-1 ) = (-a) ζ + (-a)ζ² + ... + (-a)ζ^n-1

QED

Lemma 2:

if i ≡ j (mod q)

and

ζ is a primitive qth root of unity

Then:

ζⁱ = ζ^j

Proof:

(1) i ≡ j (mod q) implies there exists integers r,s such that:

i +r*q = j + s*q

(2) So, it follows that:

ζⁱ*(ζ^q)^r = ζ^j*(ζ^q)^s

(3) Since ζ^q = 1, we have:

ζⁱ*1^r = ζ^j*1^s

which means that:

ζⁱ = ζ^j

QED