Chapter 6

Question 1

Wallis’ equation method (where you can factorise x,y and n has only a few factors)

Answer 1

1. factorise the x,y expression via difference of two squares
2. take cases for each of the factors of n
3. solve simultaneous equations for each set of factors with the factorisation (since the powers of x,y are even it suffices to find the non -ve soltuions then the others are found by multiplying through by -1)

Question 2

Lemma. no integer solutions to equation.

Answer 2

Let f(X₁,…,Xᵣ) be a polynomial with integer coefficients. Suppose that there is a n∈ℕ s.t. f(X₁,…,Xᵣ)≡0modn has no solutions. Then there are no integer solutions to f(X₁,…,Xᵣ)=0

Question 3

How to prove there are no integer solutions to any equation

Answer 3

1. define f(X₁,…,Xᵣ) and p ( is coefficient of X)
2. suppose there is a solution for a contradication. then Xᵏ≡cmodn
3. since k is even the legendre symbol (Xᵏ/p)-1 => (c/p)=1
4. by GLQR find another value for (c/p)
5. not equal => contradiction

Question 4

Lemma. in a UFD irreducible => prime.

Answer 4

Let a,b and p be non-zero elements of R with p irreducible. Suppose that p|ab then p|a and p|b. i.e. p is prime.

Question 5

idea for proving that in a UFD irreducible => prime

Answer 5

1. suppose p|ab then ab=pc for some c
2. find a factorisation into irreducibles by multiplying p by a factorisationof c
3. a second factorisation of ab is obtained by multiplying a factorisation of a by a factorisation of b
4. since R is a UFD and p appears in the 2st factorisation then an associate of p appears in the 2nd factorisation so p|a or p|b

Question 6

coprime

Answer 6

Two elements of R are coprime if their only common factors are units.

Question 7

Theorem. a and b associate to kth powers.

Answer 7

Let a,b,c∈R{0} and k∈ℕ. Suppose that ab is associate to cᵏ and that a and b are coprime. then a and b are associate to kth powers of elements of R.

Question 8

idea for proving that a and b associate to kth powers.

Answer 8

1. factorise ab into irreducibles by a factorisation of a x factorisation of b
2. factorise ab into irreducibles by a factorisation of c k times (times by a unit)
3. since UFD 1. and 2. are essentially the same.
4. let p be an irreducible that appears in 1. then an associate appears in 2. k times (multiple of k). so this p must appear a multiple of k times in 1.
5. now a,b are coprime so p can only appear in a factorisation of a (not b). So a is associate to a kth power of p.

Question 9

How to find the integer solutions to an equation of the form yˡ =xᵏ+n when l is odd and k is even

Answer 9

1. factorise xᵏ+n to determine what ring to work in
2. determine if the ring is a UFD, what its units are and what the norm is in this ring
3. show y is odd. work mod4/mod8 for each yˡ and xᵏ+n to show that y≡1mod4
4. show that the factors are coprime. let d be a common factor then d is a factor of the difference, take norms. also d|yˡ, take norms. note y is odd.
5. apply theorem so that factor = uαˡ
6. find a way to absorb the units or replace uαˡ with βˡ
7. chose integers a,b to define α set this equal to the factor.
8. expand using binomial method.
9. equate real and imaginary parts. factorise fully.
10. use the fact that a,b are integers to take cases for each possible value of b.

Question 10

lemma. make two elements coprime.

Answer 10

Let α,β be non-zero members of a ring R and suppose that δ is a HCF of α,β. Then,
α/δ,β/δ are coprime.

Question 11

idea of proof for making two elements of a ring coprime.

Let α,β be non-zero members of a ring R and suppose that δ is a HCF of α,β. Then,
α/δ,β/δ are coprime.

Answer 11

1. let γ be a common factor of α,β. We must show gamma is a unit.
2. γ|α/δ and γ|β/δ so α/δ=aγ β/δ=bγ
3. rearranging we get that γδ is a common factor of alpha and beta
4. since delta is a HCF this forces γδ|δ
5. so δ = γδε => 1 = γε so gamma is a unit

Question 12

ordₚ(a)

Answer 12

Let a∈R{0} and let p∈R be irreducible. Define, ordₚ(a) to be the number of associates of p that occur in a factorisation of a into irreducibles.

Question 13

if a’ is associate to a and p’ is associate to p then ords…

Answer 13

ordₚ’(a’) = ordₚ(a)

Question 14

ordₚ(ab) =

Answer 14

ordₚ(a)+ordₚ(b)

Question 15

if δ|α then ordₚ(α/δ) =

Answer 15

ordₚ(α) - ordₚ(δ)

Question 16

Lemma. factors and ords.

Answer 16

Let α,β be non-zero elements of a ring, Then α|β iff ordₚα ≤ ordₚβ for every irreducible p in R

Question 17

Lemma. HCF and ord.

Answer 17

Let α,β be non-zero elements of a ring, let δ be a HCF of α and β. Then ordₚδ = min(ordₚα, ordₚβ) for every irreducible r in R

Question 18

Lemma. factors of an irreducible to a power.

Answer 18

Suppose that π∈R is irreducible and that n is a natural number. Then the factors of πⁿ are the associates of πᵏ for 0≤k≤n

Question 19

How to solve a Diophantine equations when the factors are not coprime y³=x²+n

Answer 19

1. factorise the lower power and constant part. Work in the ring implied by this factorisation.
2. let δ be the highest common factor of these factors
3. divide the equation through by δ. now the factors are coprime.
4. δ| difference of factors (find explicit value)
5. Let π be an irreducible. Where πᵉˣᵖˡᶦᶜᶦᵗ ᵛᵃˡᵘᵉ is associate to the explicit value.
6. Consequently, δ is associate to πᵗ for some t, 0≤t≤4. Since associates of HCFs are HCFs we may replace δ with πᵗ
7. Now get more information about t using the ord function to show 3|t
8. Now we have the LHS is a cube but still need to show the value being cubed is an element of the ring.
   Apply the theorem to get the factor/ delta = uα³
9. absorb the unit
10. chose α=a+bi and equate real and imaginary parts.
11. take cases for each value

Question 20

Method of descent

Answer 20

To show f has no non-trivial integer solutions we need
1. a formula for measuring the size of a solution
2. a machine which when given a solution produces another smaller solution
The machine produces infinitely many solutions which is a contradiction since only a finite number of non-negative integers less than a.

Question 21

How to prove an equation has no non-trivial integer solutions by the method of descent (hard example)

(since all exponents are even only consider positive solutions)

Answer 21

1. Divide through by a highest common factor to get that x and y are coprime.
2. Suppose (x,y,z) is a solution. Swap x,y if neccessary suppose x is odd.
3. Then find integers a and b that represent x²,y²,z by theorem
4. apply theorem again to equation in correct form to get integers u,v to replace a,b.
5. Manipulate to get a solution smaller than the previous solution (machine).
6. find a formula

Question 22

How to prove an equation has no non-trivial integer solutions by the method of descent (easier example)

Answer 22

1. suppose (x,y,z) is a solution
2. check the coefficiants. Show that there is a common factor of all of them.
3. divide through by the common factor to get a smaller solution. (machine)
4. define the formula.

Question 23

Mod in any ring

Answer 23

Let R be a ring and n̸=0, α,β∈R define α≡βmodn to mean n|(α-β) in R.

Question 24

if n|α <=>

Answer 24

α≡0modn

Question 25

CSR

Answer 25

Let R be a ring and 0 ̸= n ∈ R. A complete set of residues mod n in R is a set S of elements of R such that each member of R is congruent modulo n to exactly one member of S