FormalConjectures/ErdosProblems/974.lean

GitHub source file

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:

Imports (1)
  1. def Erdos974.IsTuranConfiguration Documented declaration: IsTuranConfiguration line 38 source

    Source documentation

    The configuration described by Tijdeman: if n is odd then the z i are exactly the nth roots of unity, and if n is 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 of 1 together with the (n / 2)th roots of some other point c on the unit circle.

    standard axioms: Classical.choice, Quot.sound, propext what this means

    Statement uses: Nat, Fin, Complex

    Proof uses: And, Odd, Eq, Set, Set.range, Set.ofPred, HPow.hPow, OfNat.ofNat, Even, Exists, Real, Norm.norm, Ne, Union.union, HDiv.hDiv

    15 instances, coercions and auxiliary constants left out.

    def IsTuranConfiguration {n : ℕ} (z : Fin n → ℂ) : Prop :=
    …
  2. theorem Erdos974.erdos_974 Let $z_1,\ldots,z_n\in \mathbb{C}$ be a sequence such that $z_1=1$ sorry line 60 source

    Source 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.IsTuranConfiguration

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    40 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

  3. theorem Erdos974.erdos_974.variants.two_tuples Erdős 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 source

    Source 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.IsTuranConfiguration

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    40 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