351.lean`

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

6 declarations, 3 contain sorry, 0 declared as axioms

Erdős Problem 351

Reference: erdosproblems.com/351

Imports (1)

FormalConjecturesUtil

def Erdos351.imageSet 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 line 37 source
Source documentation
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

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Semifield, Polynomial, Set
Proof uses: Set.range, Nat, HAdd.hAdd, Polynomial.eval, HDiv.hDiv, OfNat.ofNat
18 instances, coercions and auxiliary constants left out.
```
def imageSet {α : Type*} [Semifield α] (P : α[X]) : Set α :=
…
```
def Erdos351.IsStronglyComplete 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 line 42 source
Source documentation
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.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Semiring, Set
Proof uses: Finset, Filter.Eventually, Nat, Membership.mem, Set.ofPred, Exists, LE.le, SDiff.sdiff, Eq, Finset.sum, Filter.atTop
12 instances, coercions and auxiliary constants left out.
```
def IsStronglyComplete {α : Type*} [Semiring α] (A : Set α) : Prop :=
…
```
def Erdos351.HasCompleteImage The predicate that the rational polynomial P has a complete image line 48 source
Source documentation
The predicate that the rational polynomial P has a complete image.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Polynomial, Rat
Proof uses: Erdos351.IsStronglyComplete, Erdos351.imageSet
4 instances, coercions and auxiliary constants left out.
```
def HasCompleteImage (P : ℚ[X]) : Prop := IsStronglyComplete (imageSet P)
…
```
Mentions: Erdos351.IsStronglyComplete, Erdos351.imageSet
theorem Erdos351.erdos_351 Let $p(x) \in \mathbb{Q}[x]$ be a non-constant rational polynomial with positive leading coefficient sorry line 57 source
Source documentation
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?

depends on sorry what this means
Statement uses: Iff, True, Polynomial, Rat, LT.lt, Nat, OfNat.ofNat, Polynomial.natDegree, Polynomial.leadingCoeff, Erdos351.HasCompleteImage
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
10 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_351 :
    answer(True) ↔ ∀ P : ℚ[X], 0 < P.natDegree → 0 < P.leadingCoeff → HasCompleteImage P := by
…
```
Mentions: Erdos351.HasCompleteImage
theorem Erdos351.erdos_351.variants.X Let $p(x) = x \in \mathbb{Q}[x]$ sorry line 69 source
Source documentation
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.

depends on sorry what this means
Statement uses: Erdos351.HasCompleteImage, Polynomial.X, Rat
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous, OfNat.ofNat, Nat
3 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
protected theorem erdos_351.variants.X : HasCompleteImage X := by
…
```
Mentions: Erdos351.HasCompleteImage
theorem Erdos351.erdos_351.variants.X_sq Let $p(x) = x ^ 2 \in \mathbb{Q}[x]$ sorry line 78 source
Source documentation
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.

depends on sorry what this means
Statement uses: Erdos351.HasCompleteImage, HPow.hPow, Polynomial, Rat, Nat, Polynomial.X, OfNat.ofNat
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
14 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_351.variants.X_sq : HasCompleteImage (X ^ 2) := by
…
```
Mentions: Erdos351.HasCompleteImage, Erdos351.erdos_351.variants.X