FormalConjectures/ErdosProblems/974.lean
Explanation legend Lean4 Domain Explanation Language (L4DEL)
L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.
- L4DEL
- Lean4 Domain Explanation Language
Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.
Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1
3 declarations, 2 contain sorry, 0 declared as axioms
Erdős Problem 974
References:
- erdosproblems.com/974
- [Ti66] Tijdeman, R., On a conjecture of Turán and Erdős. Indag. Math. (1966), 374-383.
Imports (1)
- def
Erdos974.IsTuranConfigurationDocumented declaration: IsTuranConfiguration line 38 sourceSource documentation
The configuration described by Tijdeman: if
nis odd then thez iare exactly thenth roots of unity, and ifnis even then they are the vertices of two regular(n / 2)-gons with the same circumscribed circle centred at the origin, that is, the(n / 2)th roots of1together with the(n / 2)th roots of some other pointcon the unit circle.standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Nat,Fin,ComplexProof uses:
And,Odd,Eq,Set,Set.range,Set.ofPred,HPow.hPow,OfNat.ofNat,Even,Exists,Real,Norm.norm,Ne,Union.union,HDiv.hDiv15 instances, coercions and auxiliary constants left out.
def IsTuranConfiguration {n : ℕ} (z : Fin n → ℂ) : Prop := … - theorem
Erdos974.erdos_974Let $z_1,\ldots,z_n\in \mathbb{C}$ be a sequence such that $z_1=1$ sorry line 60 sourceSource documentation
Let $z_1,\ldots,z_n\in \mathbb{C}$ be a sequence such that $z_1=1$. Suppose that the sequence of $$s_k=\sum_{1\leq i\leq n}z_i^k$$ contains infinitely many $(n-1)$-tuples of consecutive values of $s_k$ which are all $0$. Then (essentially) $$z_j=e(j/n),$$ where $e(x)=e^{2\pi ix}$.
A conjecture of Turán.
This is true (in the stronger form with only two such tuples) - in fact if $n$ is odd then the $z_i$ must be exactly the $n$th roots of unity, and if $n$ is even they must be the vertices of two regular $(n/2)$-gons with the same circumscribed circle centred at the origin. This was first proved by Tijdeman [Ti66]. An independent proof of this was given in the comments section by Hu, Tang, and Zhang.
depends on sorry what this means
Statement uses:
Nat,NeZero,Fin,Complex,Eq,OfNat.ofNat,Set.Infinite,Set.ofPred,LT.lt,HSub.hSub,Finset.sum,Finset.univ,HPow.hPow,HAdd.hAdd, Erdos974.IsTuranConfigurationProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous40 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_974 {n : ℕ} [NeZero n] (z : Fin n → ℂ) (hz : z 0 = 1) (hs : Set.Infinite {k : ℕ | ∀ j < n - 1, ∑ i, z i ^ (k + j) = 0}) : IsTuranConfiguration z := by …Mentions: Erdos974.IsTuranConfiguration
- theorem
Erdos974.erdos_974.variants.two_tuplesErdős speculates that this may be true if there are two distinct $(n-1)$-tuples of consecutive values of $s_k$ which are $0$ sorry line 76 sourceSource documentation
Erdős speculates that this may be true if there are two distinct $(n-1)$-tuples of consecutive values of $s_k$ which are $0$. He does not elaborate on what the 'essentially' may mean precisely.
This is true (in the stronger form with only two such tuples) - in fact if $n$ is odd then the $z_i$ must be exactly the $n$th roots of unity, and if $n$ is even they must be the vertices of two regular $(n/2)$-gons with the same circumscribed circle centred at the origin. This was first proved by Tijdeman [Ti66]. An independent proof of this was given in the comments section by Hu, Tang, and Zhang.
depends on sorry what this means
Statement uses:
Nat,NeZero,Fin,Complex,Eq,OfNat.ofNat,Ne,LT.lt,HSub.hSub,Finset.sum,Finset.univ,HPow.hPow,HAdd.hAdd, Erdos974.IsTuranConfigurationProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous40 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_974.variants.two_tuples {n : ℕ} [NeZero n] (z : Fin n → ℂ) (hz : z 0 = 1) {a b : ℕ} (hab : a ≠ b) (ha : ∀ j < n - 1, ∑ i, z i ^ (a + j) = 0) (hb : ∀ j < n - 1, ∑ i, z i ^ (b + j) = 0) : IsTuranConfiguration z := by …Mentions: Erdos974.IsTuranConfiguration