Lemma 1: (a + b)

^{2}= a

^{2}+ 2ab + b

^{2}

(a + b)

^{2}= (a + b)(a + b) = a(a + b) + b(a + b) = a

^{2}+ ab + ab + b

^{2}= a

^{2}+ 2ab + b

^{2}.

QED

Lemma 2: (a + b)(a - b) = a

^{2}- b

^{2}

(a + b)(a - b) = a(a - b) + b(a - b) = a

^{2}- ab + ab - b

^{2}= a

^{2}- b

^{2}

Corollary 2.1: a

^{4}- b

^{4}= (a - b)(a + b)(a

^{2}+ b

^{2})

By Lemma 2, a

^{4}- b

^{4}= (a

^{2}- b

^{2})(a

^{2}+ b

^{2})

Applying Lemma 2 again, gives us:

a

^{4}- b

^{4}= (a - b)(a + b)(a

^{2}+ b

^{2})

Lemma 3: a

^{3}+ b

^{3}= (a + b)(a

^{2}- ab + b

^{2})

(a + b)(a

^{2}- ab + b

^{2}) = a(a

^{2}- ab + b

^{2}) + b(a

^{2}- ab + b

^{2}) =

a

^{3}- a

^{2}b + ab

^{2}+ a

^{2}b -ab

^{2}+ b

^{3}= a

^{3}+ b

^{3}

Corollary 3.1: a

^{3}- b

^{3}= (a - b)(a

^{2}+ ab + b

^{2})

Let b' = -b so that a

^{3}- b

^{3}= a

^{3}+ (b')

^{3}

By Lemma 3: a

^{3}+ (b')

^{3}= (a + b')(a

^{2}- ab' + (b')

^{2}) = (a - b)(a

^{2}+ ab + b

^{2})

QED

Lemma 4: a

^{5}+ b

^{5}= (a + b)(a

^{4}- a

^{3}b +a

^{2}b

^{2}- ab

^{3}+ b

^{4})

(a + b)(a

^{4}- a

^{3}b +a

^{2}b

^{2}- ab

^{3}+ b

^{4}) =

=a(a

^{4}- a

^{3}b +a

^{2}b

^{2}- ab

^{3}+ b

^{4}) + b(a

^{4}- a

^{3}b +a

^{2}b

^{2}- ab

^{3}+ b

^{4}) =

=a

^{5}- a

^{4}b + a

^{3}b

^{2}- a

^{2}b

^{3}+ ab

^{4}+ a

^{4}b - a

^{3}b

^{2}+ a

^{2}b

^{3}- ab

^{4}+ b

^{5}=

=a

^{5}+ b

^{5}

QED

Lemma 4: n ≥ 5 → a

^{n}+ b

^{n}= (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)}b

^{2}... -ab

^{(n-2)}+ b

^{(n-1)}) +

b(b

^{(n-1)}+a

^{(n-1)}- a

^{(n-2)}b + ...+a

^{2}b

^{(n-3)}-ab

^{(n-2)}) =

a

^{n}- a

^{(n-1)}b + a

^{(n-2)}b

^{2}+ ... -a

^{2}b

^{(n-2)}+ ab

^{(n-1)}+

b

^{n}+ a

^{(n-1)}b - a

^{(n-2)}b

^{2}+ ... + a

^{2}b

^{(n-2)}- ab

^{(n-1)}=

a

^{n}+ b

^{n}

QED

Lemma 5: a - b divides a

^{n}- b

^{n}

Proof:

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

(2) We know that n is at least 2 since a

^{2}- b

^{2}= (a - b)(a + b)

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

^{n+1}- b

^{n+1}

(4) We know that (a-b)(a

^{n}) = a

^{n+1}- a

^{n}b and we know that (a-b)(b

^{n}) = ab

^{n}- b

^{n+1}

(5) So that a

^{n+1}- b

^{n+1}- (a - b)(a

^{n}) - (a-b)(b

^{n}) = a

^{n}b - ab

^{n}= ab(a

^{n-1}- b

^{n-1})

(6) So that:

a

^{n+1}- b

^{b+1}= (a-b)(a

^{n}+ b

^{n}) - ab(a

^{n-1}- b

^{n-1})

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

^{n-1}- b

^{n-1}.

QED