Chapter 7

Question 1

strategy to solve x³+y³=z³

Answer 1

1. factorise x³+y³ in a ring ℤ[ω] = {a+bω|a,b ϵℤ}. ω = e^2πi/3
2. Find a machine and a formula via the method of decent

Question 2

Lemma. if a solution to x³+y³=z³ exists then…

Answer 2

Suppose that there exists a non-trivial integer solution to x³+y³=z³. Then there exists a non-trivial solution in which

(i) x and y are coprime in ℤ
(ii) 3|z but 3 is not a factor of x and y

Question 3

Idea to prove that if there exists a non-trivial integer solution to x³+y³=z³ then there exists a non-trivial solution in which x and y are coprime

Answer 3

1. Let d be a HCF of x,y
2. Then we get that d|z (d³|x³,y³,z³)
3. divide the eqn by d
4. x/d and y/d are coprime

Question 4

Idea to prove that if there exists a non-trivial integer solution to x³+y³=z³ then there exists a non-trivial solution in which 3|z and 3 is not a factor of x and y

Answer 4

1. calculate cubes modulo 9 for a CSR modulo 9
2. no choice of signs gives us ±1 + ±1 ≡ ±1 mod 9.
3. so at least one of x³,y³,z³≡ 0 mod 9. hence is a multiple of 3.
4. if 3|x replace (x,y,z) with (-x,y,-z) so 3| the new z. Do the same with y.

Question 5

α* where α= a+bω

Answer 5

α*=(a-b)-bω

Question 6

N(α) where α= a+bω

Answer 6

N(α) = a² + b² -ab

Question 7

Units of ℤ[ω]

Answer 7

±1, ±ω, ±ω²

Question 8

Theorem. ℤ[ω] is a..

Answer 8

The ring ℤ[ω] is a ER and hence a UFD

Question 9

Theorem. product associate to a cube of a member of ℤ[ω].

Answer 9

Let α,β,γ∈ℤ[ω], any two of which are coprime. Suppose that
αβγ is associate to a cube of a member of ℤ[ω]
Then α,β,γ are each associate to cubes of members of ℤ[ω]

Question 10

λ =

Answer 10

λ = 1 - ω

Question 11

λ² is associate to 3 proof

Answer 11

N(λ)=3.
3 is is irreducible in ℤ so λ is irreducible and hence prime in ℤ[ω] .
λ²=-3ω=3(-ω)
since -ω is a unit λ² is associate to 3

Question 12

Lemma. CSR mod λ

Answer 12

{0,1,2} is a complete set of residues mod λ in ℤ[ω]

Question 13

idea of proof for showing {0,1,2} is a CSR modulo λ in ℤ[ω]

Answer 13

1. Let α = a + bω then α ≡ a + b mod λ
2. since {0,1,2} is a CSR mod 3 in ℤ then there exists r∈{0,1,2} s.t. a+b=r+3A (a,b are integers)
3. Now λ|3 so a + b ≡ r mod λ ≡ α
   Now show no two distinct members of {0,1,2} are congruent mod λ
4. taking norms of 0,1,2 we see that no two of them are congruent modulo λ

Question 14

Lemma . integer congruent to a cube

Answer 14

Let α ∈ ℤ[ ω ]. Then there is an integer n such that

α³ ≡n mod λ³

Question 15

idea to prove there is an integer n such that

α³ ≡n mod λ³

Answer 15

1. α = x + ωy and ω = 1 − λ so α=a+bλ
2. calculate α³
3. recall that λ²|3 thus α³≡a³modλ³
4. put n=a³

Question 16

Lemma u is equal to the cube of a unit

Answer 16

Let u ∈ ℤ[ω] be a unit and let α ∈ ℤ[ω]. Suppose that

u≡α³modλ³. Then u is equal to the cube of a unit.

Question 17

idea to prove u is equal to the cube of a unit

Answer 17

1. by lemma u≡n mod λ³
2. so λ³|u-n. Now λ² is associate to 3 so 3|u-n i.e. u≡nmod3
3. take cases for each of the units to check if u-n/3 ∈ ℤ[ω]. This shows the only units which hold are ±1. so just set v=u=u³=v³

Question 18

formula for fermats last theorem

Answer 18

ordλ(Z)

Question 19

Theorem C

Answer 19

Suppose we have the following
(i) x³+y³=uz³
(ii) x,y,z ∈ ℤ[ω]{0} and u is a unit in ℤ[ω]
(iii) x,y are coprime in ℤ[ω]
(iv) λ|z
consider the numbers
A={x+y,x+ωy,x+ω²y}
Then,
(a) Each member of A is a multiple of λ
(b) λ is a HCF of any two distinct members of A
(c) Exactly one member of A is a multiple of λ²

Question 20

Factorisation of x³+y³

Answer 20

(x+y)(x+ωy)(x+ω²y)

Question 21

idea of proof for (a) Each member of A is a multiple of λ (Theorem C)

Answer 21

1. calculate the differences of elements of A. The differences are all multiples of λ so either every element of A is a multiple of λ or no element of A is a multiple of λ.
2. show λ|(x+y)(x+ωy)(x+ω²y) since (x+y)(x+ωy)(x+ω²y) = x³+y³=uz³ λ|z
3. since λ is irreducible it is prime so λ is a factor of at least one element of A => λ is a factor of all the elements of A

Question 22

idea of proof for (b) λ is a HCF of any two distinct members of A (Theorem C)

Answer 22

1. Let d be a HCF of two elements of A
2. by (a) we have λ|d
3. also from the differences and since ω, ω² are units then d|λy so e|y and e is a common factor of two members of A
4. But then e|x so as x and y are coprime we deduce that e is a unit. Thus d is associate to λ (i.e. λ is a HCF)

Question 23

idea of proof for (c) Exactly one member of A is a multiple of λ² (Theorem C)

Answer 23

1. Let B = {x+y /λ, x+ωy /λ , x+ω²y/λ }
2. find the differences of elements in B (y,ωy, ω²y)
3. λ|z and x,y are coprime so λ∤ y. Also λ∤ ωy and λ∤ ω²y. Thus no two members of B are congruent.
4. we know {0,1,2} is a CSR. Since B has 3 members we deduce that exactly one member of B is congruent to 0,1,2 respectively. In particular, exactly one member of B is a multiple of λ. Hence exactly, one member of A is a multiple of λ²

Question 24

Theorem B (the machine for FLT)

Answer 24

Consider the equations
(i) x³+y³=uz³
(ii) x,y,z ∈ ℤ[ω]{0} and u is a unit in ℤ[ω]
(iii) x,y are coprime in ℤ[ω]
(iv) λ|z, λ∤y, λ∤x
If (x,y,z,u) is a solution to the above then there exists another solution (X,Y,Z,U) in which ordλ(Z) = ordλ(z)-1

Question 25

Idea of proof of theorem B

Answer 25

1. recall from theorem C that λ is a HCF of any pair of the factors (x+y)(x+ωy)(x+ω²y) = x³+y³=uz³ and exactly one of the factors is a multiple of λ²
2. suppose that λ²|x+ω²y
3. divide (x+y)(x+ωy)(x+ω²y) = uz³ by λ³
4. by theorem each of these new factors is associate to a cube
5. Note ω²+ω+1=0 take a comb of the new factors to remove x,ys to get 0 = ωαX³+ ω²βY³ = ωγZ³
6. divide through by the unit ωα to get X³ + δY³ = εZ³
7. Now argue δ is the cube of a unit. Find a multiplicative inverse (Y and λ³ are coprime). So there exists a unit δ₀ s.t. δ=δ₀³ so X³ + (δ₀Y)³ = εZ³. We claim that (X,δ₀Y,Z,ε) is a solution.
8. check all the hypotheses hold for this solution.
9. Prove ordλ(Z) = ordλ(z)-1 by multiplying the factors and taking ords.

Question 26

How to show Y and λ³ are coprime (for proof of theorem B when finding multiplicative inverse)

Answer 26

Since λ²|x+ω²y then from theorem C we know that λ²∤x+ωy so λ∤x+ωy/λ so λ∤Y. since λ is irreducible it follows that λ³ and Y are coprime.