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Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Theorem: if x,y,z,n are integers and n is a regular prime, then xⁿ + yⁿ = zⁿ ? xyz = 0.

Proof:

(1) From the given, we know that n is a regular prime.

(2) We can assume that x,y,z are relatively prime.

(3) Let ? = n. [We use ? to represent n ]

(4) Let ? be a primitive root of unity such that ?ⁱ ? 1 when i is less than ? and ?^? = 1

(5) Then, using ?, we can refactor xⁿ + yⁿ to get:

z^? = x^? + y^? = (x + y)(x + ?y)(x + ?²y)*...*(x + ?^?-1y).

(6) We can assume that all elements (x + ?ⁱy) in step #4 are either relatively prime or have only ? - 1 as a common factor since:

(a) Assume that there exists some prime p that divides both x + ?ⁱy and x + ?^i+ky.

(b) Now, we note that:

(x + ?^i+ky) - (x + ?ⁱy) = ?^i+ky - ?ⁱy = ?ⁱ(?^k - 1)y

(x + ?^i+ky) - ?^k(x + ?ⁱy) = x - ?^kx = (?^k - 1)*(-1)*x

(c) So, if p divides x + ?ⁱy and x + ?^i+ky, then p divides both ?ⁱ(?^k - 1)y and (?^k - 1)*(-1)*x

This is true since if a factor a divides b and c, then a divides (b - c) and a also divides (b-dc).

(d) Now, we note that ?ⁱ is a unit since ?ⁱ * ?^?-i = 1 and since -1 is a unit.

This gives us that p divides (?^k - 1)y and (?^k - 1)*x [Since only a unit divides a unit,]
(e) Now, we know that p cannot divide both x,y (since x,y are relatively prime) so p must divide ?^k - 1 which means that p = ? - 1.

(7) If ? - 1 divides any of the element of the form (x + ?ⁱy), then it divides all the other factors of this form since:

(a) if (? - 1) divides (x + ?ⁱy), then (?-1) divides (x + ?ⁱ⁺¹y) since:

(x + ?ⁱ⁺¹y) - (x + ?ⁱy) = ?ⁱ⁺¹y - ?ⁱy = ?ⁱ(? - 1)y.

In other words, if a factor a divides c and d and b + c = d, then a must divide b.

(b) if (? - 1) divides (x + ?ⁱy), then (?-1) divides (x + ?^i-1y) since:

(x + ?ⁱy) - (x + ?^i-1y) = ?^i-1(? - 1)y.

(8) Since there are ? - 1 factors of the form (x + ?ⁱy), then (? - 1)^?-1 = ?*unit gives us that ? divides z^? which using Euclid's Generalized Lemma gives us that ? divides z.

(9) Even if ? divides x or y, we can assume that it divides z.

(a) Let's assume that ? divide x. [if ? divides y, we can use the same argument substituting y for x]

(b) Since ? is odd, we know that -(-x)^? = x^? and -(-z)^? = z^?.

(c) This gives us:

(-x)^? = y? + (-z)^?

(d) Then, if we label (-x)^? as z' and (-z)^? as x', then we have shown that if ? divides x,y,or z, we can assume an equation of the form z'^? = x'^? + y^? where ? divides z'^?

(e) Finally, if ? divides z^?, then by Euclid's Generalized Lemma, it divides z.

(10) Step #9 is useful because it means that we only have to consider two cases:

Case I: x,y,z,? are coprime and therefore each factor (a+x^jy) are relatively prime to each other.

Case II: ? divides z.

(11) We can now conclude that Fermat's Last Theorem holds for regular primes since:

(a) Fermat's Last Theorem holds for Case I.
(b) Fermat's Last Theorem holds for Case II.

Generalized Euclid's Lemma

The idea here is that if a prime divides a product of n elements, then it is necessarily divides at least one of those elements.

(1) Let's assume that we have a product of n elements.
a = a₁ * a₂ * ... * a_n(2) We can take any one of these elements and get:
a = a₁ * (a₂ * ... * a_n)
(3)So, by Euclid's Lemma, the prime either divides a₁ or one of the rest of the elements.
4) So, either it divides a₁ or a₂ etc.
(5) And we get to the last two elements, we are done by Euclid's Lemma.

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