Fermat Last Theorem in the case n = 3 #
The goal of this file is to prove Fermat Last theorem in the case n = 3.
Main results #
fermatLastTheoremThree_case1: the first case of Fermat Last Theorem whenn = 3: ifa b c : ℤare such that¬ 3 ∣ a * b * c, thena ^ 3 + b ^ 3 ≠ c ^ 3.
TODO #
Prove case 2.
To prove Fermat's Last Theorem for n = 3, it is enough to show that that for all a, b, c
in ℤ such that c ≠ 0, ¬ 3 ∣ a, ¬ 3 ∣ b, a and b are coprime and 3 ∣ c, we have
a ^ 3 + b ^ 3 ≠ c ^ 3.
FermatLastTheoremForThreeGen is the statement that a ^ 3 + b ^ 3 = u * c ^ 3 has no
nontrivial solutions in 𝓞 K for all u : (𝓞 K)ˣ such that ¬ λ ∣ a, ¬ λ ∣ b and λ ∣ c.
The reason to consider FermatLastTheoremForThreeGen is to make a descent argument working.
Equations
- One or more equations did not get rendered due to their size.
Instances For
To prove FermatLastTheoremFor 3, it is enough to prove FermatLastTheoremForThreeGen.
Solution' is a tuple given by a solution to a ^ 3 + b ^ 3 = u * c ^ 3,
where a, b, c and u are as in FermatLastTheoremForThreeGen.
See Solution for the actual structure on which we will do the descent.
- u : (NumberField.RingOfIntegers K)ˣ
- hc : self.c ≠ 0
- coprime : IsCoprime self.a self.b
Instances For
Solution is the same as Solution' with the additional assumption that λ ^ 2 ∣ a + b.
- u : (NumberField.RingOfIntegers K)ˣ
- hc : self.c ≠ 0
- coprime : IsCoprime self.a self.b
Instances For
For any S' : Solution', the multiplicity of λ in S'.c is finite.
Given S' : Solution', S'.multiplicity is the multiplicity of λ in S'.c, as a natural
number.
Equations
- S'.multiplicity = (multiplicity (hζ.toInteger - 1) S'.c).get ⋯
Instances For
Given S : Solution, S.multiplicity is the multiplicity of λ in S.c, as a natural
number.
Equations
- S.multiplicity = S.multiplicity
Instances For
We say that S : Solution is minimal if for all S₁ : Solution, the multiplicity of λ in
S.c is less or equal than the multiplicity in S₁.c.
Equations
- S.isMinimal = ∀ (S₁ : FermatLastTheoremForThreeGen.Solution hζ), S.multiplicity ≤ S₁.multiplicity
Instances For
If there is a solution then there is a minimal one.
Given S' : Solution', then S'.a and S'.b are both congruent to 1 modulo λ ^ 4 or are
both congruent to -1.
Given S' : Solution', we have that 2 ≤ S'.multiplicity.
Given S : Solution, we have that 2 ≤ S.multiplicity.
Given S' : Solution', the key factorization of S'.a ^ 3 + S'.b ^ 3.
Given S' : Solution', we have that λ ^ 2 divides one amongst S'.a + S'.b,
S'.a + η * S'.b and S'.a + η ^ 2 * S'.b.
Given S' : Solution', we may assume that λ ^ 2 divides S'.a + S'.b ∨ λ ^ 2 (see also the
result below).
Given S' : Solution', then there is S₁ : Solution such that
S₁.multiplicity = S'.multiplicity.