Today's blog continues the proof for Fermat's Last Theorem: n = 5. If you are interested in the history behind the proof, start here. If you are just interested in the mathematical details behind the proof, start here.
Today's blog continues the discusssion from here. The argument offered today comes from Harold M. Stark's An Introduction to Number Theory.
To establish the ideas behind the Key Lemmas, we will need to factor a2 - 5b2 into (a + b√5)(a - b√5). To do this, we will need unique factorization. For those who need a review of quadratic integers, start here.
Unfortunately, it turns out that Z[√5] is not characterized by unique factorization. (Note: Z[√5] notations describes the set of algebraic integers formed by a + b√5 where a,b are rational integers). To show this, I will introduce two lemmas:
Lemma 1: If a set Z[√d] is characterized by unique factorization, then 2 is not a prime.
(1) Assume that Z[√d] has unique factorization and 2 is a prime.
(2) d is either odd or even so either d or d-1 is divisible by 2.
(3) So, 2 divides (d)(d-1) = d2 - d = (d + √d)(d - √d) [Note, we can factor this up since we are assuming that Z[√d] has unique factorization]
(4) In Z[√d], 2 can't divide (d + √d) because 2 doesn't divide √d. Likewise, 2 can't divide (d - √d)
(5) But this is a contradiction, since by Euclid's Lemma (one of the properties of unique factorization), 2 must divide either (d + √d) or (d - √d) if 2 divides (d + √d)(d - √d).
(6) So we can reject our assumption.
QED
Lemma 2: if d ≡ 1 (mod 4), then Z[√d] never has unique factorization.
(1) Suppose that 2 is not a prime in Z[√d]
(2) Then, there exists values: α, β such that:
2 = αβ
(3) Now, we note that Norm(2) = (2 - 0√d)(2 + √d) = 4 [See here for review of norms and conjugates]
(4) Since 2 = αβ, we also note that 4 = Norm(α)*Norm(β)
(5) Since 2 is not a prime and since it is not a unit, we can suppose that absolute(Norm(α)) is greater than 1 and absolute(Norm(β)) is greater than 1.
(6) This means that Norm(α) and Norm(β) must be equal to ±2.
(7) Since α is in Z[√d], there must exist integers a,b such that α = a + b√d.
(8) So, Norm(α) = (a + b√d)(a - b√d) = a2 + b2d = ±2.
(9) Since d ≡ 1 (mod 4), we note that:
a2 + b2d ≡ a2 + b2(1) ≡ ±2 ≡2 (mod 4). [See here for a review of modular arithmetic]
(10) But this is impossible:
(odd) 2 ≡ (2u+1)2 ≡ 4u2 + 4u + 1 ≡ 1 (mod 4).
(even)2 ≡ (2u)2 ≡ 4u2 ≡ 0 (mod 4).
Case I: a is odd, b is odd: a2 - b2 ≡ 1 - 1 ≡ 0 (mod 4).
Case II: a is odd, b is even: a2 - b2 ≡ 1 - 0 ≡ 1 (mod 4).
Case III: a is even, b is odd: a2 - b2 ≡ 0 - 1 ≡ 3 (mod 4).
Case IV: a is even, b is even: a2 - b2 ≡ 0 - 0 ≡ 0 (mod 4).
QED
Luckily, there is a way out. From a previous result, we know that Z[(1 + √5)/2] has unique factorization.
So in my next blog, I will talk more about Z[(1 + √5)/2].
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