Useful Equations

I thought it would be useful to review some basic equations. These are presented as a series of lemmas.

Lemma 1: (a + b)² = a² + 2ab + b²

(a + b)² = (a + b)(a + b) = a(a + b) + b(a + b) = a² + ab + ab + b² = a² + 2ab + b².

QED

Lemma 2: (a + b)(a - b) = a² - b²

(a + b)(a - b) = a(a - b) + b(a - b) = a² - ab + ab - b² = a² - b²

Corollary 2.1: a⁴ - b⁴ = (a - b)(a + b)(a² + b²)

By Lemma 2, a⁴ - b⁴ = (a² - b²)(a² + b²)

Applying Lemma 2 again, gives us:
a⁴ - b⁴ = (a - b)(a + b)(a² + b²)

Lemma 3: a³ + b³ = (a + b)(a² - ab + b²)

(a + b)(a² - ab + b² ) = a(a² - ab + b²) + b(a² - ab + b²) =
a³ - a²b + ab² + a²b -ab² + b³ = a³ + b³

Corollary 3.1: a³ - b³ = (a - b)(a² + ab + b²)

Let b' = -b so that a³ - b³ = a³ + (b')³

By Lemma 3: a³ + (b')³ = (a + b')(a² - ab' + (b')²) = (a - b)(a² + ab + b²)

QED

Lemma 4: a⁵ + b⁵ = (a + b)(a⁴ - a³b +a²b² - ab³ + b⁴)

(a + b)(a⁴ - a³b +a²b² - ab³ + b⁴) =
=a(a⁴ - a³b +a²b² - ab³ + b⁴) + b(a⁴ - a³b +a²b² - ab³ + b⁴) =
=a⁵ - a⁴b + a³b² - a²b³ + ab⁴ + a⁴b - a³b² + a²b³ - ab⁴ + b⁵ =
=a⁵ + b⁵

QED

Lemma 4: n ≥ 5 → aⁿ + bⁿ = (a + b)(a^(n-1) - a^(n-2)b + ... - ab^(n-2) + b^(n-1))

(a + b)(a^(n-1) - a^(n-2)b + ... -ab^(n-2) + b^(n-1)) =

a(a^(n-1) - a^(n-2)b + a^(n-3)b²... -ab^(n-2) + b^(n-1)) +
b(b^(n-1)+a^(n-1) - a^(n-2)b + ...+a²b^(n-3) -ab^(n-2)) =

aⁿ - a^(n-1)b + a^(n-2)b² + ... -a²b^(n-2) + ab^(n-1) +
bⁿ + a^(n-1)b - a^(n-2)b² + ... + a²b^(n-2) - ab^(n-1) =
aⁿ + bⁿ

QED

Lemma 5: a - b divides aⁿ - bⁿ

Proof:

(1) Let n = be the highest number where this is true.

(2) We know that n is at least 2 since a² - b² = (a - b)(a + b)

(3) To complete this proof, we need to show that a - b divides aⁿ⁺¹ - bⁿ⁺¹

(4) We know that (a-b)(aⁿ) = aⁿ⁺¹ - aⁿb and we know that (a-b)(bⁿ) = abⁿ - bⁿ⁺¹

(5) So that aⁿ⁺¹ - bⁿ⁺¹ - (a - b)(aⁿ) - (a-b)(bⁿ) = aⁿb - abⁿ = ab(a^n-1 - b^n-1)

(6) So that:

aⁿ⁺¹ - b^b+1 = (a-b)(aⁿ + bⁿ) - ab(a^n-1 - b^n-1)

(7) Since n ≥ 2, we know that a-b divides a^n-1 - b^n-1.

QED