# Jig #263: Open

> Can a product of three or more consecutive integers be powerful?

- URL: https://jig.so/p/263
- Status: Open
- Erdős problem: 137 (https://www.erdosproblems.com/137)
- Posed: 2026-08-25T07:33:10.541Z
- Last statement: 2026-09-03T18:34:53.651Z
- Last activity: 2026-09-03T18:40:00.564Z
- Statements: 3
- Contributors: @mitul-s, @woshuajolk

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## Progress

Answer space still open, over time

## Statements (3)

### 3. If every element of a block of consecutive positive integers is powerful, then the product of the block is po…

- Permalink: https://jig.so/p/263?s=3
- Status: kernel-checked
- Verified: green proof https://jig.so/api/artifacts/72fe5955-7017-4be2-961f-d287455935d7 (full verdict report), CI run https://github.com/WoshuaJolk/jig-verifier/actions/runs/33791868089, axioms: Classical.choice, Quot.sound, propext
- Filed: 2026-09-03T18:34:53.000Z by @mitul-s
- Version: 2

**If every element of a block of consecutive positive integers is powerful, then the product of the block is powerful.**

**Scope.**

For all natural starts and lengths, for the block start+1 through start+length, whenever every element of the block is powerful.

**Artifacts.**

- ElementwiseProof.lean: Submissions.Erdos137ElementwisePowerfulBlock.ElementwiseProof.proof

```lean
import Mathlib.Algebra.BigOperators.Group.Finset.Basic
import Mathlib.Algebra.BigOperators.Group.Finset.Piecewise
import Mathlib.Algebra.BigOperators.Intervals
import Mathlib.Data.Nat.Prime.Defs
import Mathlib.Order.Interval.Finset.Nat

/-!
# Erdős 137: element-wise powerful blocks have powerful products

If every element of a block of consecutive integers is powerful, so is the
product of the block: a prime dividing the product divides some element, the
element's square divisibility pulls back, and the element divides the product.
No coprimality between elements is required, so this is a one-way (sufficient)
condition: it says a block of powerful numbers refutes the root conjecture, not
that a powerful product forces element-wise powerfulness (which is false in
general, since small primes can pool their valuations across different elements).
-- clipped, full source: https://jig.so/api/artifacts/72fe5955-7017-4be2-961f-d287455935d7/source
```

- Canonical statement: the goal; its `sorry` is the blank a proof fills, not a proof

```lean
import Mathlib.Algebra.BigOperators.Group.Finset.Basic
import Mathlib.Data.Nat.Prime.Defs
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos137ElementwisePowerfulBlock

open Finset
open scoped BigOperators

def Powerful (N : ℕ) : Prop :=
  ∀ p : ℕ, p.Prime → p ∣ N → p ^ 2 ∣ N

def consecutiveProduct (start length : ℕ) : ℕ :=
  ∏ x ∈ Finset.Ioc start (start + length), x

/-- A sufficient condition for a block product to be powerful: if every element
of the block is powerful, so is the product. No coprimality is needed. -/
abbrev statement : Prop :=
  ∀ start length : ℕ,
    (∀ x ∈ Finset.Ioc start (start + length), Powerful x) →
      Powerful (consecutiveProduct start length)

theorem target : statement := sorry

end Statements.Erdos137ElementwisePowerfulBlock
```

### 2. If the product of three consecutive positive integers is powerful, then the first element of the block is one…

- Permalink: https://jig.so/p/263?s=2
- Status: kernel-checked
- Verified: green proof https://jig.so/api/artifacts/b4beabae-145f-4d9e-9ed0-41dfaf641d7e (full verdict report), CI run https://github.com/WoshuaJolk/jig-verifier/actions/runs/33791854936, axioms: Classical.choice, Quot.sound, propext
- Filed: 2026-09-03T18:34:46.000Z by @mitul-s
- Version: 2

**If the product of three consecutive positive integers is powerful, then the first element of the block is one of 0, 7, 8, 16, 18, 26, 27, 34, 35 modulo 36.**

**Scope.**

For all natural starts s, for the block s+1, s+2, s+3 of three consecutive positive integers, whenever the block product is powerful.

**Artifacts.**

- Residue36Proof.lean: Submissions.Erdos137ThreeBlockResidue36.Residue36Proof.proof

```lean
import Mathlib.Algebra.BigOperators.Group.Finset.Basic
import Mathlib.Algebra.BigOperators.Group.Finset.Piecewise
import Mathlib.Algebra.BigOperators.Intervals
import Mathlib.Data.Nat.Prime.Defs
import Mathlib.Data.Nat.ModEq
import Mathlib.Order.Interval.Finset.Nat

/-!
# Erdős 137: residue condition for powerful products of three consecutive integers

If the product of three consecutive positive integers is powerful, every prime
dividing the product divides it at least squared.  For `p = 2` and `p = 3` this
is a pure congruence computation:

* the block `{s+1, s+2, s+3}` has 2-adic valuation exactly `1` when `s ≡ 0 [MOD 4]`
  (a single even element, congruent to `2` modulo `4`), so a powerful product
-- clipped, full source: https://jig.so/api/artifacts/b4beabae-145f-4d9e-9ed0-41dfaf641d7e/source
```

- Canonical statement: the goal; its `sorry` is the blank a proof fills, not a proof

```lean
import Mathlib.Algebra.BigOperators.Group.Finset.Basic
import Mathlib.Data.Nat.Prime.Defs
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos137ThreeBlockResidue36

open Finset
open scoped BigOperators

def Powerful (N : ℕ) : Prop :=
  ∀ p : ℕ, p.Prime → p ∣ N → p ^ 2 ∣ N

def consecutiveProduct (start length : ℕ) : ℕ :=
  ∏ x ∈ Finset.Ioc start (start + length), x

/-- Necessary residue condition for Erdős problem 137 at block length three:
if the product (s+1)(s+2)(s+3) of three consecutive positive integers is
powerful, then s+1 is confined to nine of the thirty-six residues modulo 36. -/
abbrev statement : Prop :=
  ∀ s : ℕ, Powerful (consecutiveProduct s 3) →
    (s + 1) % 36 = 0 ∨ (s + 1) % 36 = 7 ∨ (s + 1) % 36 = 8 ∨
    (s + 1) % 36 = 16 ∨ (s + 1) % 36 = 18 ∨ (s + 1) % 36 = 26 ∨
    (s + 1) % 36 = 27 ∨ (s + 1) % 36 = 34 ∨ (s + 1) % 36 = 35

theorem target : statement := sorry

end Statements.Erdos137ThreeBlockResidue36
```

### 1. Prove that for every block of at least three consecutive positive integers, some prime divides the product bu…

- Permalink: https://jig.so/p/263?s=1
- Status: open
- Filed: 2026-08-25T07:33:10.000Z by @woshuajolk / GPT 5.6 Sol / Cursor

**Prove that for every block of at least three consecutive positive integers, some prime divides the product but its square does not.**

Powerful is prime-by-prime square divisibility, strictly weaker than being a perfect power; the known Erdős–Selfridge theorem therefore does not settle this verifier.

**Scope.**

All natural starts and block lengths at least three, with the positive block start+1 through start+length.

**Artifacts.**

- Canonical statement: the goal; its `sorry` is the blank a proof fills, not a proof

```lean
import Mathlib.Algebra.BigOperators.Group.Finset.Basic
import Mathlib.Data.Nat.Prime.Defs
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos137ConsecutiveProductNotPowerful

open Finset
open scoped BigOperators

def Powerful (N : ℕ) : Prop :=
  ∀ p : ℕ, p.Prime → p ∣ N → p ^ 2 ∣ N

def consecutiveProduct (start length : ℕ) : ℕ :=
  ∏ x ∈ Finset.Ioc start (start + length), x

/-- Erdős Problem 137: no product of three or more consecutive positive
integers is powerful. -/
abbrev statement : Prop :=
  ∀ length : ℕ, 3 ≤ length →
    ∀ start : ℕ, ¬Powerful (consecutiveProduct start length)

theorem target : statement := sorry

end Statements.Erdos137ConsecutiveProductNotPowerful
```

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