FormalConjectures/ErdosProblems/939.lean
Explanation legend Lean4 Domain Explanation Language (L4DEL)
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- 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
9 declarations, 6 contain sorry, 0 declared as axioms
Erdős Problem 939
References:
- erdosproblems.com/939
- [Ni95] Nitaj, A., On a conjecture of Erdős on 3-powerful numbers. Bull. London Math. Soc. (1995), 317-318.
- [Co98] Cohn, J. H. E., A conjecture of Erdős on 3-powerful numbers. Math. Comp. (1998), 439-440.
- [Wa24] Walsh, P., A question of Erdős on 3-powerful numbers and an elliptic curve analogue of the Ankeny-Artin-Chowla conjecture. arXiv:2404.03970 (2024).
- [LaPa67] Lander, L. J. and Parkin, T. R., A counterexample to Euler's sum of powers conjecture. Math. Comp. (1967), 101-103.
Imports (1)
- def
Erdos939.Erdos939SumsA setSbelongs toErdos939Sums rif it meets the following criteria: - The elements are positive line 53 sourceSource documentation
A set
Sbelongs toErdos939Sums rif it meets the following criteria:- The elements are positive.
0has no prime factors, so it is vacuouslyr-powerful, and the source means positive integers. - The size of the set is
$|S| = r - 2$. - The elements of the set are coprime (their greatest common divisor is 1).
- Every element in
Sis an$r$-powerfulnumber. - The sum of the elements in
S, i.e.,$\sum_{s \in S} s$, is also an$r$-powerfulnumber.
The summands are taken to be distinct (
Sis aFinset). The source does not say whether repeated summands are allowed; all known examples and constructions use distinct summands.standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Nat,Set,FinsetProof uses:
Set.ofPred,And,Eq,Finset.card,HSub.hSub,OfNat.ofNat, Finset.Coprime, Nat.Full,Finset.sum,Membership.mem,LT.lt7 instances, coercions and auxiliary constants left out.
def Erdos939Sums (r : ℕ) := …
- The elements are positive.
- theorem
Erdos939.erdos_939If $r≥4$ then can the sum of $r-2$ coprime $r$-powerful numbers ever be itself $r$-powerful? sorry line 61 sourceSource documentation
If $r≥4$ then can the sum of $r-2$ coprime $r$-powerful numbers ever be itself $r$-powerful?
depends on sorry what this means
Statement uses:
Iff,True,Nat,GE.ge,OfNat.ofNat,Set.Nonempty,Finset, Erdos939.Erdos939SumsProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous4 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_939 : answer(sorry) ↔ ∀ r ≥ 4, (Erdos939Sums r).Nonempty := by …
Mentions: Erdos939.Erdos939Sums
- theorem
Erdos939.erdos_939.variants.finiteIf $r≥4$, are there at most finitely many sums of $r-2$ coprime $r$-powerful numbers that are themselves $r$-powerful? sorry line 77 sourceSource documentation
If $r≥4$, are there at most finitely many sums of $r-2$ coprime $r$-powerful numbers that are themselves $r$-powerful?
The answer is no: for every $r \ge 6$ there are infinitely many such sums, see
erdos_939.variants.infinite_of_six_le. (For $r = 4$ and $r = 5$ the question is open; for $r = 4$ no example is known at all, seeerdos_939.) A construction in the site's comments, from GPT-5.5 Pro prompted by Price, gives infinitely many for every $r \ge 6$. This statement quantifies over every $r \ge 4$, so it stays open at $r = 4$ and $r = 5$. The category is unchanged because the construction is recorded in the comments and not in the literature.depends on sorry what this means
Statement uses:
Iff,False,Nat,GE.ge,OfNat.ofNat,Set.Finite,Finset, Erdos939.Erdos939SumsProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous4 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_939.variants.finite : answer(False) ↔ ∀ r ≥ 4, (Erdos939Sums r).Finite := by …
Mentions: Erdos939.Erdos939Sums
- theorem
Erdos939.erdos_939.variants.infinite_of_six_leFor every $r \ge 6$ there are infinitely many sums of $r - 2$ coprime $r$-powerful numbers that are themselves $r$-powerful sorry line 92 sourceSource documentation
For every $r \ge 6$ there are infinitely many sums of $r - 2$ coprime $r$-powerful numbers that are themselves $r$-powerful.
A construction, found by GPT-5.5 Pro prompted by Liam Price and recorded in the comments on erdosproblems.com/939, expands $(X+Y)^r = (X-Y)^r + \sum_{j \text{ odd}} 2\binom{r}{j} X^{r-j} Y^j$, splits the $j = 3$ term into $\lfloor r/2 \rfloor - 2$ distinct pieces to obtain exactly $r - 2$ summands, and takes $X = q^r$, $Y = B^r$ with $B$ divisible by all primes in the coefficients and $q > B$ a prime not dividing $B$; varying $q$ gives infinitely many solutions.
L4DEL explanation
For every $r \ge 6$ there are infinitely many sums of $r - 2$ coprime $r$-powerful numbers that are themselves $r$-powerful. A construction, found by GPT-5.5 Pro prompted by Liam Price and recorded in the comments on erdosproblems.com/939, expands $(X+Y)^r = (X-Y)^r + \sum_{j \text{ odd}} 2\binom{r}{j} X^{r-j} Y^j$, splits the $j = 3$ term into $\lfloor r/2 \rfloor - 2$ distinct pieces to obtain exactly $r - 2$ summands, and takes $X = q^r$, $Y = B^r$ with $B$ divisible by all primes in the coefficients and $q > B$ a prime not dividing $B$; varying $q$ gives infinitely many solutions.
Sources and attribution: Pinned Lean source (1646ca16afd6)
depends on sorry what this means
Statement uses:
Nat,GE.ge,OfNat.ofNat,Set.Infinite,Finset, Erdos939.Erdos939SumsProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous4 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_939.variants.infinite_of_six_le : ∀ r ≥ 6, (Erdos939Sums r).Infinite := by …
Mentions: Erdos939.Erdos939Sums
- theorem
Erdos939.erdos_939.variants.triplesAre there infinitely many triples of coprime $3$-powerful numbers $a, b, c$ such that $a + b = c$? sorry line 104 sourceSource documentation
Are there infinitely many triples of coprime $3$-powerful numbers $a, b, c$ such that $a + b = c$?
The answer is yes. Nitaj [Ni95] proved it, with $2^3\cdot 3^5\cdot 73^3 + 271^3 = 919^3$ as an example. In Nitaj's construction at least two of $a, b, c$ are perfect cubes. Cohn [Co98] constructed infinitely many triples of which none is a perfect cube, and Walsh [Wa24] gave a further construction.
depends on sorry what this means
Statement uses:
Iff,True,Set.Infinite,Prod,Nat,Set.ofPred,Erdos939.erdos_939.variants.triples.match_1,And, Finset.Coprime,Insert.insert,Finset,Singleton.singleton,LT.lt,OfNat.ofNat, Nat.Full,Eq,HAdd.hAddProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous14 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_939.variants.triples : answer(True) ↔ {(a,b,c) | ({a, b, c} : Finset ℕ).Coprime ∧ 0 < a ∧ 0 < b ∧ (3).Full a ∧ (3).Full b ∧ (3).Full c ∧ a + b = c}.Infinite := by …Mentions: c
- theorem
Erdos939.erdos_939.variants.examplesCambie has found several examples of the sum of $r - 2$ coprime $r$-powerful numbers being itself $r$-powerful line 117 sourceSource documentation
Cambie has found several examples of the sum of $r - 2$ coprime $r$-powerful numbers being itself $r$-powerful. For example when $r=5$ we have $$3^7\cdot 61^5 = 2^8\cdot3^{10}\cdot 5^7 + 2^{12}\cdot 23^6 + 11^5\cdot 13^5$$.
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Exists,Nat,And,GE.ge,OfNat.ofNat,Set.Nonempty,Finset, Erdos939.Erdos939SumsProof uses:
Exists.intro,Eq.mpr,id,Eq,Eq.trans,True,congrFun',congrArg,LE.le,eq_true_of_decide,Eq.refl,Bool,Bool.true,true_and,Set.ofPred,Finset.card,HSub.hSub, Finset.Coprime, Nat.Full,Finset.sumand 73 more62 instances, coercions and auxiliary constants left out.
Tactic trace: 12 steps
Open to load the checked before-and-after goal states.
theorem erdos_939.variants.examples : (∃ r ≥ 4, (Erdos939Sums r).Nonempty) := by …
Mentions: Erdos939.Erdos939Sums
- theorem
Erdos939.erdos_939.variants.sevenCambie has also found solutions when $r=7$ sorry line 132 sourceSource documentation
Cambie has also found solutions when $r=7$.
depends on sorry what this means
Statement uses:
Set.Nonempty,Finset,Nat, Erdos939.Erdos939Sums,OfNat.ofNatProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous2 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_939.variants.seven : (Erdos939Sums 7).Nonempty := by …
Mentions: Erdos939.Erdos939Sums
- theorem
Erdos939.erdos_939.variants.eightCambie has also found solutions when $r=8$ sorry line 143 sourceSource documentation
Cambie has also found solutions when $r=8$.
The source adds that the $r=8$ solution works "even with the sum of $5$ $8$-powerful numbers". That is a stronger result than this statement, which asks for the $r - 2 = 6$ summands of
Erdos939Sums.depends on sorry what this means
Statement uses:
Set.Nonempty,Finset,Nat, Erdos939.Erdos939Sums,OfNat.ofNatProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous2 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_939.variants.eight : (Erdos939Sums 8).Nonempty := by …
Mentions: Erdos939.Erdos939Sums
- theorem
Erdos939.erdos_939.variants.eulerDocumented declaration: euler line 156 sourceSource documentation
Euler had conjectured that the sum of $k - 1$ many $k$-th powers is never a $k$-th power, but this is false for $k=5$, as Lander and Parkin [LaPa67] found $$27^5+84^5+110^5+133^5=144^5$$.
The summands must be positive. Without that condition a set containing
0would count, so the negation would be satisfied by a sum of fewer than $k-1$ powers and would claim less than the refutation of Euler's conjecture that this theorem records.standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Not,Nat,GE.ge,OfNat.ofNat,Finset,Eq,Finset.card,HSub.hSub,Membership.mem,LT.lt,Exists,Finset.sum,HPow.hPowProof uses:
Eq.mpr,And,id,Eq.trans,Mathlib.Tactic.Push.not_forall_eq,congrArg,funext,Exists.intro,True,congr,eq_true,Mathlib.Meta.NormNum.isNat_le_true,Mathlib.Meta.NormNum.isNat_ofNat,Eq.refl,Bool,Bool.true,congrFun',Mathlib.Meta.NormNum.IsNat.to_eq,Mathlib.Meta.NormNum.isNat_natSub,true_andand 27 more51 instances, coercions and auxiliary constants left out.
Tactic trace: 8 steps
Open to load the checked before-and-after goal states.
theorem erdos_939.variants.euler : ¬ (∀ k ≥ 4, ∀ S : Finset ℕ, S.card = k - 1 → (∀ s ∈ S, 0 < s) → ¬ (∃ q, ∑ s ∈ S, s ^ k = q ^k)) := by …