FERMAT S LAST THEOREM FOR n = 3, NOTES FOR MATH In these notes, we prove Fermat s Last Theorem for n = 3.

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1 FERMAT S LAST THEOREM FOR n = 3, NOTES FOR MATH SAM EVENS In these notes, we prove Fermat s Last Theorem for n = 3. 1 Recall that Z[ω] = {a+bω : a, b Z} has a Euclidean algorithm and is norm integral. in the sense of the notes on subrings. In particular, the following are true: Remark 1.1. (i) If π is a prime of Z[ω], and z, w Z[ω], and π z w in Z[ω], then π z or π w in Z[ω] (this is Proposition 2.15(i) from subrings notes for Z[i], and the same proof works for Z[ω]). (ii) If z Z[ω], then N(z) = 1 if and only if z is a unit of Z[ω] (see Lemma 1.8 of the notes on subrings) (iii) Let z = a + bω Z[ω] {0}. Then N(z) = a 2 ab + b 2 Z >0, and if N(z) is a prime in Z, then z is a prime in Z[ω]. (iv) λ = 1 ω is a prime in Z[ω]. Indeed, by (iii) above, N(λ) = 3, so λ is prime in Z[ω]. (v) Let z, w Z[ω]. Then z 3 w 3 = (z w) (z ωw) (z ω 2 w). Indeed, by explicit computation (z w) (z ωw) (z ω 2 w) = z 3 (1 + ω + ω 2 )z 2 w + (1 + ω + ω 2 )z w 2 w 3, since 1 + ω + ω 2 = 0. (vi) Let π be a prime of Z[ω]. Since Z[ω] is a unique factorization domain, then for any nonzero z Z[ω], we can write z = π e b for some b Z[ω] such that π does not divide b. Then we set ord π (z) = e, and it is routine to check that π z if and only if ord π (z) > 0. Further, if z, w Z[ω], then ord π (z w) = ord π (z) + ord π (w). (vii) In the language of (vi) above, let ord π (z) = e and let ord π (w) = f. Suppose e f, and let g be the minimum of e and f. Then ord π (z) = e. Indeed, suppose that g = e < f. Then z = π e a and w = π f b where π divides neither a nor b. Then z + w = π e a + π f b = π e (a + π f e b), and f e > 0. Note that π does not divide a + π f e b since if it did, it would divide (a + π f e b) π f e b = a, which is not the case. Hence, ord π (z + w) = e. A similar argument works if f < e. 1

2 2 SAM EVENS Remark 1.2. (i) If z, w Z[ω] are not both zero, then gcd(z, w) = d Z[ω] means that d z and d w, and N(d) N(e) for all e Z[ω] such that e z and e w. Similarly, if a, b, c Z[ω] are not all zero, then gcd(a, b, c) = d Z[ω] means that d a, d b, and d c, and N(d) is maximal among elements of Z[ω] dividing a, b and c. (ii) Suppose z, w Z[ω] are not both zero, and consider the ideal (z, w) = {az + bw : a, b Z[ω]}. Then Problem 4 of Homework 3 asserts that (z, w) = (d) = {d η : η Z[ω]}. It is then easy to show that d = gcd(z, w), in the same manner as this is proved for Z[i] in Proposition 2.11 of the notes on subrings. In particular, if e Z[ω] and e z and e w, then e d. Indeed, since (z, w) = (d), there are a, b Z[ω] such that d = az + bw, so since e z and e w, then e az + bw = d by Lemma 1.6(iii) from the notes on subrings. (iii) Let d = gcd(a, b, c). Then gcd( a d, b d, c d ) = 1, and the only common divisors of a d, b d, c d are units. Indeed, let e divide a d in Z[ω]. Then a = w e d for some w Z[ω], and e d a. Similarly, e d divides b and c in Z[ω], so since d = gcd(a, b, c), then N(e d) N(d). But N(e d) = N(e)N(d) N(d) since N(e) 1, so it follows that N(e d) = N(d), and N(e) = 1. Thus, e is a unit. Remark 1.3. Let u be a unit of Z[ω], and let x, y, z Z[ω] with x 3 + y 3 = uz 3. Then if a prime π of Z[ω] divides any two of x, y, z, then π divides all three of x, y, z. Indeed, suppose π divides x and y. Then π divides x 3 + y 3, so π divides uz 3. Since u is a unit, π does not divide u, so π divides z 3 by Remark 1.1 (i), and thus π divides z using the same Remark. The other cases are similar. Remark 1.4. Let z Z[ω] and u be a unit in Z[ω]. Suppose uz 3 = a b c with a, b, c Z[ω], and gcd(a, b) = gcd(b, c) = gcd(a, c) = π, where π is a prime of Z[ω]. Let ord π (z) = e, and suppose ord π (b) = ord π (c) = 1. Then there are units u a, u b, u c of Z[ω] and α, β, γ Z[ω] with π dividing none of α, β, γ and gcd(α, β) = 1, gcd(α, γ) = 1, and gcd(β, γ) = 1 so that a = u a α 3 π 3e 2, b = u b β 3 π, c = u c γ 3 π. To see this, let z = w π e with π w. Then u z 3 = u w 3 π 3e, and if π w 3, then π w by property (i) from Remark 1.1. Thus, π w 3, so ord π (z 3 ) = 3e. Since z 3 = a b c, we see that ord π (z 3 ) = ord π (a b c) = ord π (a) + ord π (b) + ord π (c), using the last equality of Remark 1.1 (vi). Since ord π (b) = ord π (c) = 1, we see that ord π (a) = 3e 2. Thus, by definition of ord π, we can write a = α a π 3e 2, b = β b π, c = γ c π, where π divides none of α a, β b, or γ c. Let q be a prime factor of a in Z[ω] and suppose q π. Then by Remark 1.1 (i), q α a. If q β b, then q a and q b. Hence, by Remark 1.2 (ii), it follows that q gcd(a, b) = π, which contradicts our assumption. Thus, q β b. Similarly, q γ c. Since q a, then q a b c = u z 3, so q z 3, and q z, again by Remark 1.1 (i). Let e q = ord q (z). Then ord q (u z 3 ) = 3e q, so ord q (a b c) = 3e q is

3 FERMAT S LAST THEOREM FOR n = 3, NOTES FOR MATH a multiple of 3. Since q does not divide b or c, we see that ord q (b) = ord q (c) = 0, and hence ord q (a) = 3e q. Thus, every prime factor appearing in a besides π must occur with exponent a multiple of 3, and it follows that α a = u a α 3 for a unit u a, where α is a product of prime factors of a besides π. Hence, a = u a α 3 π 3e 2. Similar arguments show that b = u b β 3 π, and c = u c γ 3 π, where β is a product of a prime factors of b besides π, and γ is a product of prime factors of c besides π. Lemma 1.5. Let z Z[ω]. Then z is congruent to exactly one of 1, 0, and 1 mod λ. Proof. Let z = a+bω. Then since 1 ω 0 (mod λ), a+bω a+bω+b (1 ω) a+b (mod λ). Since 3 = N(λ), then λ 3, so 3 0 (mod λ). We write a + b = 3q + r for q, r Z with 0 r < 3. Then z a + b a + b 3q r (mod λ), since 3 0 (mod λ). Thus, z is congruent to 0, 1 or 2 mod λ. If z 2 (mod λ), then z (mod λ). It follows that z is equivalent to at least one from 0, 1 and 1 mod λ. On the other hand, ±1 0 (mod λ) implies that λ ±1, and this is impossible since λ is not a unit. If 1 1 (mod λ), then λ 2, so N(λ) N(2) = 4, so 3 4 in Z, and this is not the case. It follows that 1, 0, 1 are distinct elements mod λ. Lemma 1.6. (see proof of Lemma 1 in 17.8) (i) If z 1 (mod λ), then z 3 1 (mod λ 4 ). (ii) If z 1 (mod λ), then z 3 1 (mod λ 4 ). Proof. For (i), since z 1 (mod λ), then z 1 = t λ, with t Z[ω]. Thus, z 3 1 = (z 1) (z ω) (z ω 2 ) by part (v) of Remark 1.1. Thus, z 3 1 = t λ (1+λt ω) (1+λt ω 2 ). But (1 + λt ω) = (1 ω) + λt = λ + λt = λ (1 + t), and Hence, (1 + λt ω 2 ) = ((1 ω 2 ) + λt) = ((1 ω)(1 + ω) + λt) = λ (1 + ω + t). z 3 1 = t λ λ (1 + t) λ (1 + t + ω) = λ 3 t (1 + t) (1 + t + ω). By Lemma 1.5, either t 0 or 1 or 1 mod λ. If t 0 (mod λ), then t = λ w for some w Z[ω], so z 3 1 = λ 4 w (1 + t) (1 + t + ω), and thus, z 3 1 (mod λ 4 ). If t 1 (mod λ), then t + 1 = w λ for some w Z[ω], so z 3 1 = λ 4 t w (1 + t + ω), and z 3 1 (mod λ 4 ). The other possibility is that t 1 (mod λ), so t = 1 + w λ for some w Z[ω]. But then 1+t+ω = 1+1+w λ+ω = 3 (1 ω)+wλ = (1 ω 2 ) λ λ+wλ = λ ((1 ω 2 ) 1+w), where we used 3 = N(λ) = (1 ω 2 ) λ. It follows that z 3 1 = λ 4 t (1 + t) (w ω 2 ), and hence, z 3 1 (mod λ 4 ). Thus, in any of the three possible cases for t (mod λ), we see that z 3 1 (mod λ 4 ), and this proves part (i). For part (ii), if z 1 (mod λ),

4 4 SAM EVENS then z 1 (mod λ), so by part (i) of the Lemma, ( z) 3 1 (mod λ 4 ), so z 3 1 (mod λ 4 ), and thus, z 3 1 (mod λ 4 ). Lemma 1.7. Let u be a unit of Z[ω]. (i) 2 u (mod λ 4 ). (ii) If ɛ + u 0 (mod λ 2 ), where ɛ = ±1, then u = ɛ. (iii) If ɛ = ±1, and ɛ u (mod λ 3 ), then u = ɛ. Proof. For case (i), then 2 u (mod λ 4 ), then λ 4 (2 u), and thus, N(λ 4 ) N(2 u) in Z. The possibilities are u = 1, 1, ω, ω, ω 2, ω 2, since these are the units of Z[ω]. Thus, 2 u is one of 1, 3, 2 ω, 2 + ω, 2 + (1 + ω), and 2 (1 + ω), where we have used ω 2 = 1 ω. From the formula N(a + bω) = a 2 ab + b 2 in Remark 1.1 (iii), we see that N(2 u) is one of 1, 9, 7, 3, 7, and 3. Since N(λ 4 ) = N(λ) 4 = 3 4 = 81, we see that λ 4 2 u in all these cases. This proves (i). Assertions (ii) and (iii) follow in the same way. The following proposition is called the easy case of Fermat s last theorem for n = 3. Proposition 1.8. If x, y, z Z[ω] are nonzero and λ divides none of x, y and z, then x 3 + y 3 is not equal to uz 3. Proof. We argue by contradiction. Suppose x 3 + y 3 = uz 3. Since λ divides none of x, y and z, we conclude that x ±1, y ±1, and z ±1 modulo λ, by using Lemma 1.5. Thus, by Lemma 1.6, x 3 ±1 (mod λ 4 ) and similarly y 3 ±1 (mod λ 4 ) and z 3 ±1 (mod λ 4 ). Since x 3 + y 3 uz 3 (mod λ 4 ), we see that ±1 + ±1 ±u (mod λ 4 ). But ±1 + ±1 = ±2 or 0, so either ±2 ±u (mod λ 4 ) or 0 ±u (mod λ 4 ). Thus, either 2 ±u (mod λ 4 ) or 0 ±u (mod λ 4 ). But 0 ±u (mod λ 4 ) implies that λ 4 u, and this is impossible because λ 4 is not a unit. If 2 ±u (mod λ 4 ), then since ±u is a unit, we see that 2 v (mod λ 4 ) for some unit v of Z[ω]. But this contradicts Lemma 1.7, and it follows that x 3 + y 3 = uz 3 is impossible in Z[ω]. Lemma 1.9. Let u be a unit of Z[ω], and suppose x, y, z Z[ω] with x 3 + y 3 = uz 3. Suppose that λ x and λ y, and λ z. Then λ 2 z. Proof. Reducing mod λ, we obtain x 3 + y 3 uz 3 (mod λ 4 ). Since λ x, then by Lemma 1.5, x ±1 (mod λ), so by Lemma 1.6, x 3 ±1 (mod λ 4 ). Similarly, y 3 ±1 (mod λ 4 ). It follows that ±1+±1 uz 3 (mod λ 4 ). Note that ±1+±1 is either 2, 0, or 2. If it is ±2, we obtain ±2 uz 3 (mod λ 4 ), so (±2 uz 3 ) = t λ 4 for some t Z[ω]. Since λ divides z, λ divides uz 3, and it follows that λ divides ±2, so 3 = N(λ) divides 4 = N(±2), and this is false. We conclude that 0 uz 3 (mod λ 4 ), so that uz 3 = t λ 4 for some t Z[ω]. Let ord λ (z) = e, so that z = λ e a for some a Z[ω] with λ a. Thus, u a 3 λ 3e = t λ 4. Since u is a unit and λ a, we see that ord λ (u a 3 λ 3e ) = 3e.

5 FERMAT S LAST THEOREM FOR n = 3, NOTES FOR MATH On the other hand, ord λ (t λ 4 ) = ord λ (t) + 4 4, so that 3e 4, and e 4 3. Since e Z, we see e 2, and thus λ 2 z. Lemma Let u be a unit of Z[ω], and let x, y, z Z[ω] with x 3 + y 3 = uz 3, λ x, λ y, and λ 2 z, and gcd(x, y) = 1. Then there is u 1 a unit of Z[ω], and x 1, y 1, z 1 Z[ω] with λ x 1, λ y 1, and ord λ (z 1 ) = ord λ (z) 1 so that Proof. By Remark 1.1 (v), we can factor x y 3 1 = u 1 z 3 1. u z 3 = x 3 + y 3 = (x + y) (x + ωy) (x + ω 2 y). (1.11) Since ord λ (z) = e 2, we see that ord λ (u z 3 ) = 3e 6. By Equation (1.11), we see that 3e = ord λ (x + y) + ord λ (x + ωy) + ord λ (x + ω 2 y) 6. Hence, one of ord λ (x + y), ord λ (x + ωy), and ord λ (x + ω 2 y) is greater than or equal to 2. Suppose ord λ (x + ωy) 2. Then we can let v = ωy, so y = ω 1 v = ω 2 v, and thus the above factors are x + ω 2 v, x + v, and x + ωv, and ord λ (x + v) 2. Since v 3 = y 3, we get a version of Equation 1.11 with v in place of y and ord λ (x + v) 2. We can then rename v as y, so that we get ord λ (x+y) 2. Similarly, if ord λ (x+ω 2 y) 2, we could rename variables to obtain ord λ (x + y) 2, so we may assume ord λ (x + y) 2. Consider ord λ (x + ωy) = ord λ ((x + y) (1 ω)y) = ord λ ((x + y) λ y). Note that ord λ (y) = 0 since λ y, so ord λ ( λ y) = ord λ (λ) = 1 < ord λ (x + y). Hence, by Remark 1.1 (vii), we see that ord λ (x + ωy) = 1. Similarly, ord λ (x + ω 2 y) = 1. Let π be a prime of Z[ω] so that π λ. Then π does not divide both x + y and x + ωy. If it did, then π would divide (x + y) (x + ωy) = λ y, so that π divides y by Remark 1.1 (i). Thus, π also divides x = x + y y, and this contradicts the assumption that gcd(x, y) = 1. Similarly, we can show that π cannot divide both x + y and x + ω 2 y, and π cannot divide both x + ωy and x + ω 2 y. It now follows by Property (ii) of Remark 1.4 that: x + y = u 1 α 3 λ t, t = 3 ord λ (z) 2, λ α, (1.12) x + ωy = u 2 β 3 λ, λ β, (1.13) x + ω 2 y = u 3 γ 3 λ, λ γ, (1.14) with u 1, u 2, u 3 units of Z[ω], and gcd(α, β) = 1, gcd(β, γ) = 1, and gcd(α, γ) = 1. Multiplying Equation (1.13) by ω gives ωx + ω 2 y = v 2 β 3 λ, (1.15) where v 2 = ω u 2 is a unit. Multiplying Equation (1.14) by ω 2 gives ω 2 x + ωy = v 3 γ 3 λ, (1.16) where v 3 = ω 2 u 3 is a unit. Now adding together equations (1.12), (1.15), (1.16) gives (x + y) + (ωx + ω 2 y) + (ω 2 x + ωy) = u 1 α 3 λ t + v 2 β 3 λ + v 3 γ 3 λ. (1.17)

6 6 SAM EVENS Separating x and y terms and using 1 + ω + ω 2 = 0, we obtain 0 = u 1 α 3 λ t + v 2 β 3 λ + v 3 γ 3 λ. (1.18) We divide through by λ, and note that t 1 = 3 (ord λ (z) 1) to obtain 0 = u 1 α 3 λ 3 (ord λ(z) 1) + v 2 β 3 + v 3 γ 3. (1.19) We set z 1 = α λ ord λ(z) 1, let x 1 = β, and let y 1 = γ, and obtain and after dividing through by v 2, we obtain v 2 x v 3 y 3 1 = u 1 z 3 1, (1.20) x w 1 y 3 1 = w 2 z 3 1, (1.21) with w 1 = v 3 v 2 and w 2 = u 1 v 2 units of Z[ω]. Since x 1 = β and y 1 = γ, then λ divides neither x 1 nor y 1. Thus, by Lemma 1.5, we see that x 1 ɛ 1 (mod λ), and x 2 ɛ 2 (mod λ 2 ), where ɛ 1, ɛ 2 are ±1. Note that λ z 1 since ord λ (z 1 ) = ord λ (z) 1 2 1, so λ 3 z1 3. Thus, z3 1 0 (mod λ2 ). Hence, by reducing mod λ 2, we see that ɛ 1 + w 1 ɛ 2 0 (mod λ 2 ). (1.22) We now let ɛ 1 = ɛ and w 1 ɛ 2 = u in Lemma 1.7, and conclude that w 1 ɛ 2 = ɛ 1, so w 1 = ɛ 1 ɛ2 = ɛ 3 with ɛ 3 = ±1. If w 1 = 1, we leave Equation 1.21 as it is, and if w 1 = 1, we change y 1 to y 1. In either case, letting w 2 = u, we obtain This proves the assertion. x y 3 1 = uz 3 1. (1.23) Theorem Let u Z[ω] be a unit. Then there do not exist nonzero x, y, z Z[ω] such that x 3 + y 3 = uz 3. Proof. Let S := {(x, y, z) Z[ω] {0} : u Z[ω], x 3 + y 3 = uz 3 }. We want to show S =. By Proposition 1.8, if (x, y, z) S, then λ divides at least one of x, y, z. Thus, to show S is empty, it suffices to show T = {(x, y, z) S : λ xyz} is empty. If T is not empty, we may choose (x, y, z) T with the maximum of ord λ (x), ord λ (y), ord λ (z) less than or equal to the maximum of ord λ (x ), ord λ (y ), ord λ (z ) for all (x, y, z ) in T. If λ divides two of x, y, z then λ divides gcd(x, y, z) by Remark 1.3. Then by Remark 1.2 (iii), ( x λ, y λ, z λ ) S. By Proposition 1.8, λ divides at least one of x λ, y λ, and z λ, and hence ( x λ, y λ, z λ ) T But for any w Z[ω], if λ w, then ord λ ( w λ ) = ord λ(w) 1, so that the maximum of ord λ ( x λ ), ord λ( y λ ), ord λ( z λ )) is one less than the maximum of ord λ (x), ord λ (y), ord λ (z), which contradicts the choice of (x, y, z). Hence, there exists (x, y, z) T with the property that λ divides exactly one of (x, y, z).

7 Let FERMAT S LAST THEOREM FOR n = 3, NOTES FOR MATH T 1 = {(x, y, z) T : λ x, λ y, λ z}, T 2 = {(x, y, z) T : λ x, λ y, λ z}, T 3 = {(x, y, z) T : λ x, λ y, λ z}, The final sentence of the last paragraph implies that if T is non-empty, then either T 1, T 2, or T 3 is nonempty. We next show T 3 is empty, and later use this to conclude that T 1 and T 2 is empty. If T 3 is nonempty, there exists (x, y, z) T 3 with ord λ (z) ord λ (z ) for all (x, y, z ) T 3. Let d = gcd(x, y, z), and let a = x d, b = y d, and c = z d. By Remark 1.2 (iii), (a, b, c) S. Since a x and λ x, it follows that λ a. Similarly, λ b. If λ c, we would contradict Proposition 1.8, so we conclude that λ c. Since gcd(a, b, c) = 1, it follows that gcd(a, b) = 1 by Remark 1.3. Thus, we conclude that (a, b, c) T 3, and hence, ord λ (z) ord λ (c) by the choice of (x, y, z). But z = d c, so that ord λ (z) = ord λ (d) + ord λ (c), and we conclude that ord λ (z) ord λ (c). Hence, there exists (a, b, c) T 3 with ord λ (c) minimal, and gcd(a, b) = 1. By Lemma 1.9, ord λ (c) 2. By Lemma 1.10, there is (a 1, b 1, c 1 ) T 3 with ord λ (c 1 ) = ord λ (c) 1, which contradicts the choice of (a, b, c) T 3. Hence, T 3 is empty. We now show that T 1 and T 2 are empty. Indeed, if (x, y, z) T 1, then since λ x, we know x 3 0 (mod λ 3 ). Since (x, y, z) T 1, λ y and λ z. Hence by Lemma 1.5, y ±1 (mod λ) and z ±1 (mod λ). Thus, the identity x 3 + y 3 uz 3 (mod λ 3 ) implies ɛ 1 u ɛ 2 (mod λ 3 ), where ɛ 1, ɛ 2 are in {1, 1}. We conclude, that ɛ u (mod λ 3 ) for ɛ = ±1. Hence, by Lemma 1.7 (iii), we see that u = ɛ. If u = 1, then we obtain x 3 + y 3 = z 3, so z 3 + ( y) 3 = x 3, and hence (z, y, x) T 3, which is empty. It follows that (x, y, z) / T 1, and hence T 1 is empty. If u = 1, then ( z, y, x) T 3, so similarly we see that T 1 is empty. Finally, if (x, y, z) T 2, then (y, x, z) T 1, which we have just seen is empty. Hence, T 2 is empty. Thus, T 1, T 2, and T 3 are empty, so it follows that S is empty. Department of Mathematics, University of Notre Dame, Notre Dame, IN address: sevens@nd.edu