Sunday, August 20, 2006

Kummer's Proof for Regular Primes: Case I

In today's blog, I continue the proof of Fermat's Last Theorem for regular primes. Case I makes the assumption that x,y,z,λ are relatively prime and that each of the factors are relatively prime. For context and definitions, please start here at the beginning of this proof.

The details of today's content is taken from Harold M. Edwards Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Lemma 1: If n is odd, then xn + yn = zn implies that (x)n + (y)n + (-z)n = 0.

Proof:

This works because n is odd. n is odd implies that (-z)n = -(z)n

QED

Lemma 2: Every cyclotomic integer is congruent mod (α - 1) to a rational integer

Proof:

(1) Let g(α) be any cyclotomic integer.

(2) g(α) = a0 + a1α + ... + aλ-1αλ - 1 [See Lemma 1, here]

(3) α ≡ 1 (mod α - 1)

(4) g(α) ≡ g(1) ≡ a0 + a1 + ... + aλ-1 = rational integer

QED

Lemma 3: For any given cyclotomic integer g(α) and any given integer k greater than 0, there exists integers a0, a1, ..., ak-1 such that:

g(α) ≡ a0 + a1(α - 1) + ... + ak-1(α - 1)k-1 (mod (α - 1)k)

Proof:

(1) Let a0 ≡ g(α) (mod α - 1) [See Lemma 2 above]

(2) Let h(α) = [g(α) - a0]/(α - 1)

(3) Let a1 ≡ h(α) (mod α -1 ) [See Lemma 2 above]

(4) We can see that g(α) ≡ a0 + a1*(α - 1) mod (α - 1)2

(5) Using step #4, let a2 ≡ [g(α) - a0 - a1(α - 1)]/(α - 1)2 (mod α -1)

(6) Then g(α) ≡ a0 + a1(α - 1) + a2(α - 1)2 (mod (α - 1)3).

(7) We can continue this process up to any integer k so we are done.

QED

Lemma 4: Coefficients of corresponding powers of (α - 1) must be congruent mod λ provided all powers are less than the (λ - 1)st.

if:

a0 + a1(α - 1) + a2(α-1)2 + ... + aλ-2(α-1)λ-2 ≡ b0 + b1(α - 1) + b2(α-1)2 + ... + bλ-2(α-1)λ-2 (mod λ)

then:

a0 ≡ b0 (mod λ)
a1 ≡ b1 (mod λ)
...
aλ-2 ≡ bλ-2 (mod λ)

Proof:

(1) If a0 + a1(α - 1) + ... + aλ-2(α - 1)λ-2 ≡ b0 + b1(α -1) + ... + bλ-2(α-1)λ-2 (mod (α - 1)λ-1), then:

c0 + c1(α-1) + ... ≡ 0 (mod (α - 1)λ-1) where ci = ai - bi

(2) Then c0 ≡ 0 (mod α - 1) which also gives us that c0 ≡ 0 (mod λ)

(3) Then c1(α-1) ≡ 0 (mod (α-1)2) so that (α-1) divides c1 which means that c1 ≡ 0 (mod λ).

(4) We can use the same logic to show that all ci ≡ 0 (mod λ)

(5) So, ai - bi = 0, this implies that ai = bi

QED

Theorem: x,y,z,λ are coprime and all factors (x + αjy) are relatively prime, then xλ + yλ = zλ has no integer solutions except for where xyz=0.

Proof:

(1) zλ = xλ + yλ = (x + y)(x + αy)(x + α2y)*...*(x + αλ-1y). [See Lemma 1 here for details on this refactoring]

(2) From the given, we know that all the factors (x + αjy) are relatively prime, so we can conclude that the divisor of each (x + αjy) is λth power. [See Lemma 1, here]

(3) Further, we can conclude that each factor (x + αjy) is a unit*λth power. [This follows from Corollary, here]

(4) So, there exists a unit e and a cyclotomic integer t such that:

x + αy = etλ

(5) If permute α to α-1 on each side, we get:

x + α-1y = e*tλ where e is the inverse of e and t is the inverse of t.

(6) Now, there exists an r such that e/e = αr. [See Lemma 4, here]

(7) tλ ≡ C ≡ Ctλ mod λ because mod λ of every λth power is a rational integer (see Theorem 6, here) and rational integers are invariant under α → α-1.

(8) Since e = e*α-r, we have:

x + α-1y = α-retλ

(9) From tλtλ (mod λ) [See step #7], we have:

α-retλ ≡ α-retλ (mod λ)

(10) Applying x + αy = etλ [see step #4] give us:

α-retλ ≡ α-r(x + αy) (mod λ)

(11) This then gives us:

αr(x + α-1y) ≡ αr-retλ) ≡ etλ (mod λ) [from step #8]

Further, from tλtλ (mod λ) [See step #7], we have:

etλ ≡ etλ (mod λ)

Combining this result with x + αy = etλ in step #4 gives us:

αr(x + α-1y) ≡ x + αy (mod λ)

(12) We know that r not ≡ 0 (mod λ) since:

(a) Assume r ≡ 0 (mod λ)

(b) Then from step #11, we have:

(x + α-1y) ≡ x + αy (mod λ)

(c) Subtracting (x + α-1y) to both sides gives us:

0 ≡ (x + αy) - (x + α-1y) ≡ αy - α-1y (mod λ)

(d) Multiplying both sides by (α) gives us:

0 ≡ α2y - y ≡ (α2 - 1)y (mod λ)

(e) Since λ = (α-1)λ-1*unit (see Corollary 3.2, here), we have:

0 ≡ (α2 - 1)y (mod (α - 1)λ - 1)

(f) Since (α-1)λ - 1 divides 2 - 1)y, it follows that (α - 1) must divide y which contradicts our assumption that y and λ are coprime.

If (α - 1) divides y, then λ must divide y since α - 1 divides λ and y is a rational integer.

(g) So we reject our assumption at #12a.

(13) From step #12, we can assume that r is greater than 0 and less than λ.

NOTE: It is less than λ because it is a power of α and αλ = 1 = α0.

(14) Since r is greater than 0, we can rearrange step #11 to give us:

αr-1(αx + y) ≡ x + αy (mod λ)

(15) Since α = [1 + (α - 1)], we have:

αr-1(αx + y) ≡ [1 + (α - 1)]r-1[x(1 + [α - 1]) + y] ≡ [1 + (α - 1)]r-1[x + y + x(α - 1) ] (mod λ)

x + αy ≡ x + (1 + [α - 1])y ≡ x + y + y(α - 1) (mod λ)

(16) Putting the result of step #15 together with the fact that λ = (α - 1)λ - 1*unit (see Corollary 3.2, here) gives us:

[1 + (α - 1)]r-1[x + y + x(α - 1) ] ≡ x + y + y(α - 1) (mod (α - 1)λ - 1)

(17) If we carry out this equation using the Binomial Theorem (see here), we get:

[1 + (α - 1)]r-1 = 1 + (r-1)(α - 1) + [(r-1)(r-2)/(2!)](α - 1)2 + ... + (α - 1)r-1

Multiplying this with [x + y + x(α - 1)] gives us:

[x + y + x(α-1)] + (r-1)[x(α - 1) + y(α - 1) + x(α-1)2] + ... + [x(α-1)r-1 + y(α-1)r-1 + x(α-1)r]

(18) Coefficients of corresponding powers of (α - 1) must be congruent mod λ provided all powers are less than the (λ - 1)st. [See Lemma 4 above]

(19) The congruence in step #16 is impossible when 2 ≤ r ≤ λ - 2 because then the highest order term on the left is x(α - 1)r and step #18 gives us 0 ≡ x (mod λ) which contradicts x,λ being relatively prime.

(20) r = λ - 1 is also impossible since:

The last two terms are:

(λ-2)[x(α-1)λ-3 + y(α-1)λ-3 + x(α-1)λ-2] + [x(α-1)λ-2 + y(α-1)λ-2 + x(α-1)λ-1]

We only need to consider (α-1)λ-2 which gives us:

(λ-2+1)x(α-1)λ-2

Using step #18, we have:

(λ-1)x(α-1)λ-2 ≡ 0 (mod λ)

This implies that:

x ≡ 0 (mod λ)

Which is impossible.

(21) This leaves r = 1 which gives us:

[1 + (α-1)]0[x + y + x(α -1)] ≡ x + y + y(α - 1) (mod λ)

Since [1 + (α-1)]0 = 1, we have:

[x + y + x(α-1)] ≡ x + y + y(α-1) (mod λ)

Subtracting x+y from both sides gives us:

x(α-1) ≡ y(α-1) (mod λ)

Appying step #18 gives us:

x ≡ y (mod λ)

(22) We can put xλ + yλ = zλ into a symmetric form (see Lemma 1 above) where:

xλ + yλ + (z')λ = 0.

(23) By the fact that x,y,z,λ are relatively prime (from the given above), we know that:

x not ≡ 0 (mod λ)
y not ≡ 0 (mod λ)
z' not ≡ 0 (mod λ)

(24) We further know that x ≡ y (mod λ) (from step #21) but because the equation is symmetric we can further conclude that x ≡ z' (mod λ) and y ≡ z' (mod λ)

(25) Using Fermat's Little Theorem (see here), we can conclude that:

xλ ≡ x (mod λ)
yλ ≡ y (mod λ)
z' λ ≡ z' (mod λ)

since:

(a) Fermat's Little Theorem gives us:

gcd(a,p)=1 → ap-1 ≡ 1 (mod p)

(b) So we have:

xλ-1 ≡ 1 (mod λ)

(c) Multiplying x to both sides gives us:

xλ ≡ x (mod λ)

(d) We can make the same argument for y and z.

(26) Putting this all together gives us:

0 = xλ + yλ + z'λ ≡ x + y + z' ≡ 3x (mod λ)

(27) Since x not ≡ 0 (mod λ), this implies that λ = 3. For all other values we have a contradiction and case I is proved.

(28) Using Sophie's Theorem (see Theorem, here), we know that if 2λ+1 is a prime, then λ divides xyz.

(29) But 2*3+1 = 7 which is a prime so this means that λ must divide xyz.

(30) But it does not since x,y,z,λ are relatively prime.

(31) Therefore we have a contradiction and Case I is proved.

QED

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