328.lean`

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Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

1 declarations, 1 contain sorry, 0 declared as axioms

Erdős Problem 328

References:

erdosproblems.com/328
[Er80] Erdős, Paul, A survey of problems in combinatorial number theory. Ann. Discrete Math. (1980), 89-115.
[ErGr80] Erdős, P. and Graham, R., Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathématique (1980).
[Er80e] Erdős, P., Some applications of Ramsey's theorem to additive number theory. European J. Combin. (1980), 43-46.
[NeRo85] J. Nešetřil and V. Rödl, Two proofs in combinatorial number theory. Proc. Amer. Math. Soc. (1985), 185-188.

Imports (1)

FormalConjecturesUtil

theorem Erdos328.erdos_328 Suppose $A\subseteq\mathbb{N}$ and $C>0$ is such that $1_A\ast 1_A(n)\leq C$ for all $n\in\mathbb{N}$ sorry line 62 source
Source documentation
Suppose $A\subseteq\mathbb{N}$ and $C>0$ is such that $1_A\ast 1_A(n)\leq C$ for all $n\in\mathbb{N}$. Can $A$ be partitioned into $t$ many subsets $A_1,\ldots,A_t$ (where $t=t(C)$ depends only on $C$) such that $1_{A_i}\ast 1_{A_i}(n)<C$ for all $1\leq i\leq t$ and $n\in \mathbb{N}$?
The answer is no. Asked by Erdős and Newman. Nešetřil and Rödl [NeRo85] have shown the answer is no for all $C$ (even if $t$ is also allowed to depend on $A$).
Erdős [Er80e] had previously shown the answer is no for $C=3,4$ and infinitely many other values of $C$.
See also [774].
The linked proof writes the representation function as Set.ncard {p : ℕ × ℕ | p.1 ∈ A ∧ p.2 ∈ A ∧ p.1 + p.2 = n}, which counts the same ordered pairs as sumRep, and states the partition condition as a named definition with the same two conjuncts used below. Its ∃ t additionally carries 1 ≤ t, which costs nothing: t = 0 forces A = ∅, and A = {1} has all representation counts at most C for C ≥ 1.

depends on sorry what this means
Statement uses: Iff, False, Nat, LT.lt, OfNat.ofNat, Exists, Set, LE.le, AdditiveCombinatorics.sumRep, Fin, And, Eq, Set.iUnion, Set.PairwiseDisjoint, Set.univ
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
18 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_328 : answer(False) ↔
    ∀ C : ℕ, 0 < C →
      ∃ t : ℕ, ∀ A : Set ℕ, (∀ n, sumRep A n ≤ C) →
        ∃ P : Fin t → Set ℕ, (⋃ i, P i) = A ∧
          Set.univ.PairwiseDisjoint P ∧
          ∀ i, ∀ n, sumRep (P i) n < C := by
…
```