FormalConjectures.ErdosProblems.«328»

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import FormalConjecturesUtil

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.


open AdditiveCombinatorics


namespace Erdos328

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].

@[category research solved, AMS 11]
theorem declaration uses 'sorry'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⊢ False ↔
  ∀ (C : ℕ),
    0 < C →
      ∃ t,
        ∀ (A : Set ℕ),
          (∀ (n : ℕ), sumRep A n ≤ C) →
            ∃ P, ⋃ i, P i = A ∧ Set.univ.PairwiseDisjoint P ∧ ∀ (i : Fin t) (n : ℕ), sumRep (P i) n < C
  sorryAll goals completed! 🐙


end Erdos328