Erdős Problem 351

Reference: erdosproblems.com/351

def Erdos351.imageSet {α : Type u_1} [Semifield α] (P : Polynomial α) : Set α

The set of rational numbers of the form P(n) + 1 / n where n is a natural number and P is a polynomial with rational coefficients.

Note: We include P 0 in there (since 1 / 0 = 0), but this doesn't change the validity of the conjecture

Equations

• Erdos351.imageSet P = Set.range fun (n : ℕ) => Polynomial.eval (↑n) P + 1 / ↑n

def Erdos351.IsStronglyComplete {α : Type u_1} [Semiring α] (A : Set α) : Prop

The predicate that a set A is strongly complete, i.e. that for every finite set B, every sufficiently large integer is a sum of elements of the set A \ B.

Equations

• Erdos351.IsStronglyComplete A = ∀ (B : Finset α), ∀ᶠ (m : ℕ) in Filter.atTop, ↑m ∈ {x : α | ∃ (X : Finset α) (_ : ↑X ⊆ A \ ↑B), ∑ n ∈ X, n = x}

def Erdos351.HasCompleteImage (P : Polynomial ℚ) : Prop

The predicate that the rational polynomial P has a complete image.

Equations

• Erdos351.HasCompleteImage P = Erdos351.IsStronglyComplete (Erdos351.imageSet P)

theorem Erdos351.erdos_351 : sorry ↔ ∀ (P : Polynomial ℚ), 0 < P.natDegree → 0 < P.leadingCoeff → HasCompleteImage P

Let $p(x) \in \mathbb{Q}[x]$ be a non-constant rational polynomial with positive leading coefficient. Is it true that [A={ p(n)+1/n : n \in \mathbb{N}}] is strongly complete, in the sense that, for any finite set $B$, [\left{\sum_{a \in X} a : X \subseteq A \setminus B, X \textrm{ is finite}\right}] contains all sufficiently large integers?

theorem Erdos351.erdos_351.variants.X : HasCompleteImage Polynomial.X

Let $p(x) = x \in \mathbb{Q}[x]$. It has been shown that [A={ p(n)+1/n : n \in \mathbb{N}}] is strongly complete, in the sense that, for any finite set $B$, [\left{\sum_{a \in X} a : X \subseteq A \setminus B, X \textrm{ is finite}\right}] contains all sufficiently large integers.

theorem Erdos351.erdos_351.variants.X_sq : HasCompleteImage (Polynomial.X ^ 2)

Let $p(x) = x ^ 2 \in \mathbb{Q}[x]$. It has been shown that [A={ p(n)+1/n : n \in \mathbb{N}}] is strongly complete, in the sense that, for any finite set $B$, [\left{\sum_{a \in X} a : X \subseteq A \setminus B, X \textrm{ is finite}\right}] contains all sufficiently large integers.