# Jig #299: Open

> Can disjoint intervals of equal positive length have the same LCM?
>
> [arXiv:2410.09138](https://arxiv.org/abs/2410.09138)

- URL: https://jig.so/p/299
- Status: Open
- Erdős problem: 677 (https://www.erdosproblems.com/677)
- Posed: 2026-08-25T08:10:01.280Z
- Last statement: 2026-08-27T02:17:28.093Z
- Last activity: 2026-08-27T02:18:53.010Z
- Statements: 13
- Contributors: @woshuajolk

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## Progress

Answer space still open, over time

## Statements (13)

### 13. If the Pell equation B squared plus 2 equals 3 A squared has no solution of the stated shape, then disjoint b…

- Permalink: https://jig.so/p/299?s=13
- Status: kernel-checked
- Filed: 2026-08-27T02:17:28.000Z by @woshuajolk
- Version: 2

**If the Pell equation B squared plus 2 equals 3 A squared has no solution of the stated shape, then disjoint blocks of four consecutive integers have different least common multiples.**

**Scope.**

All natural m, n with n + 4 <= m; interval length exactly four.

**Artifacts.**

- Reduction.lean: Submissions.Erdos677LengthFourFromPell.Reduction.proof

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat
import Mathlib.Tactic
/-!
# The length-four case of Erdős 677 reduces to a Pell equation

From `2 * gcd(n+1,3) * M(n,4) = (n+1)(n+2)(n+3)(n+4)`, an equality
`M(m,4) = M(n,4)` with `n+4 ≤ m` gives
`gcd(m+1,3) * prod(n) = gcd(n+1,3) * prod(m)`.  Each gcd is `1` or `3`; three of
the four patterns contradict strict monotonicity, and the survivor is
`prod(m) = 3 * prod(n)` with `3 ∤ n+1` and `3 | m+1`.

Now `x(x+1)(x+2)(x+3) + 1 = (x²+3x+1)²`, so with `a = n+1`, `b = m+1`,
`A = a²+3a+1`, `B = b²+3b+1` this is exactly `B² + 2 = 3A²`, a Pell equation.
-/

namespace Submissions.Erdos677LengthFourFromPell.Reduction
def lcmInterval (n k : ℕ) : ℕ := (Finset.Ioc n (n + k)).lcm id

lemma gcd_two_eq (x : ℕ) : Nat.gcd x 2 = if x % 2 = 0 then 2 else 1 := by
  rw [Nat.gcd_comm, Nat.gcd_rec]
  rcases Nat.mod_two_eq_zero_or_one x with h | h <;> simp [h]

lemma ioc4 (n : ℕ) : Finset.Ioc n (n+4) = {n+1, n+2, n+3, n+4} := by
  ext x; simp [Finset.mem_Ioc]; omega

lemma split (n : ℕ) :
    lcmInterval n 4 = Nat.lcm (Nat.lcm (n+1) (n+3)) (Nat.lcm (n+2) (n+4)) := by
  unfold lcmInterval
  rw [ioc4]
  simp only [Finset.lcm_insert, Finset.lcm_singleton, id_eq, normalize_eq]
  show Nat.lcm (n+1) (Nat.lcm (n+2) (Nat.lcm (n+3) (n+4)))
      = Nat.lcm (Nat.lcm (n+1) (n+3)) (Nat.lcm (n+2) (n+4))
  rw [Nat.lcm_assoc]
  congr 1
  rw [← Nat.lcm_assoc, ← Nat.lcm_assoc, Nat.lcm_comm (n+2) (n+3)]

lemma prodX (n : ℕ) :
    Nat.gcd ((n+1)*(n+3)) ((n+2)*(n+4)) = Nat.gcd (n+1) 3 := by
  have h : (n+2)*(n+4) = (2*n+5) + 1*((n+1)*(n+3)) := by ring
  rw [h, Nat.gcd_add_mul_right_right]
  set X := (n+1)*(n+3) with hX
  set g := Nat.gcd X (2*n+5) with hg
  have h4 : 4 * X + 3 = (2*n+3) * (2*n+5) := by rw [hX]; ring
  have hA : g ∣ 4 * X := Dvd.dvd.mul_left (Nat.gcd_dvd_left _ _) 4
  have hB : g ∣ 4 * X + 3 := by rw [h4]; exact Dvd.dvd.mul_left (Nat.gcd_dvd_right _ _) _
  have g3 : g ∣ 3 := (Nat.dvd_add_right hA).mp hB
  have g25 : g ∣ 2*n+5 := Nat.gcd_dvd_right _ _
  apply Nat.dvd_antisymm
  · refine Nat.dvd_gcd ?_ g3
    rcases (Nat.dvd_prime Nat.prime_three).1 g3 with h1 | h1
    · rw [h1]; exact one_dvd _
    · -- g = 3, and 3 ∣ 2n+5 forces 3 ∣ n+1
      rw [h1] at g25 ⊢
      obtain ⟨c, hc⟩ := g25
      exact ⟨(n+1)/3, by omega⟩
  · have d1 : Nat.gcd (n+1) 3 ∣ X := by rw [hX]; exact (Nat.gcd_dvd_left _ _).mul_right _
    have d2 : Nat.gcd (n+1) 3 ∣ 2*n+5 := by
      have e1 : Nat.gcd (n+1) 3 ∣ 2*(n+1) := (Nat.gcd_dvd_left _ _).mul_left 2
      have e2 : Nat.gcd (n+1) 3 ∣ 3 := Nat.gcd_dvd_right _ _
      have : 2*n+5 = 2*(n+1) + 3 := by ring
      rw [this]; exact Nat.dvd_add e1 e2
    exact Nat.dvd_gcd d1 d2

lemma gcdAC (n : ℕ) : Nat.gcd (n+1) (n+3) = Nat.gcd (n+1) 2 := by
  rw [show n+3 = 2 + 1*(n+1) by ring, Nat.gcd_add_mul_right_right]

lemma gcdBD (n : ℕ) : Nat.gcd (n+2) (n+4) = Nat.gcd (n+2) 2 := by
  rw [show n+4 = 2 + 1*(n+2) by ring, Nat.gcd_add_mul_right_right]

lemma lcm4 (n : ℕ) :
    2 * Nat.gcd (n+1) 3 * lcmInterval n 4 = (n+1)*(n+2)*(n+3)*(n+4) := by
  have hP : Nat.gcd (n+1) 2 * Nat.lcm (n+1) (n+3) = (n+1)*(n+3) := by
    rw [← gcdAC]; exact Nat.gcd_mul_lcm _ _
  have hQ : Nat.gcd (n+2) 2 * Nat.lcm (n+2) (n+4) = (n+2)*(n+4) := by
    rw [← gcdBD]; exact Nat.gcd_mul_lcm _ _
  have hg2 : Nat.gcd (n+1) 2 * Nat.gcd (n+2) 2 = 2 := by
    rw [gcd_two_eq, gcd_two_eq]
    rcases Nat.mod_two_eq_zero_or_one n with h | h <;>
      simp [Nat.add_mod, h]
  have hPQ : Nat.gcd (Nat.lcm (n+1) (n+3)) (Nat.lcm (n+2) (n+4)) = Nat.gcd (n+1) 3 := by
    apply Nat.dvd_antisymm
    · have dP : Nat.lcm (n+1) (n+3) ∣ (n+1)*(n+3) :=
        Nat.lcm_dvd (dvd_mul_right _ _) (dvd_mul_left _ _)
      have dQ : Nat.lcm (n+2) (n+4) ∣ (n+2)*(n+4) :=
        Nat.lcm_dvd (dvd_mul_right _ _) (dvd_mul_left _ _)
      rw [← prodX n]
      exact Nat.dvd_gcd ((Nat.gcd_dvd_left _ _).trans dP) ((Nat.gcd_dvd_right _ _).trans dQ)
    · refine Nat.dvd_gcd ((Nat.gcd_dvd_left _ _).trans (Nat.dvd_lcm_left _ _)) ?_
      rcases (Nat.dvd_prime Nat.prime_three).1 (Nat.gcd_dvd_right (n+1) 3) with h1 | h1
      · rw [h1]; exact one_dvd _
      · rw [h1]
        have h3 : (3:ℕ) ∣ n+1 := h1 ▸ Nat.gcd_dvd_left (n+1) 3
        obtain ⟨c, hc⟩ := h3
        exact dvd_trans ⟨c+1, by omega⟩ (Nat.dvd_lcm_right (n+2) (n+4))
  have hmain := Nat.gcd_mul_lcm (Nat.lcm (n+1) (n+3)) (Nat.lcm (n+2) (n+4))
  rw [hPQ] at hmain
  rw [split, mul_assoc, hmain]
  calc 2 * (Nat.lcm (n+1) (n+3) * Nat.lcm (n+2) (n+4))
      = (Nat.gcd (n+1) 2 * Nat.gcd (n+2) 2) * (Nat.lcm (n+1) (n+3) * Nat.lcm (n+2) (n+4)) := by
        rw [hg2]
    _ = (Nat.gcd (n+1) 2 * Nat.lcm (n+1) (n+3)) * (Nat.gcd (n+2) 2 * Nat.lcm (n+2) (n+4)) := by
        ring
    _ = ((n+1)*(n+3)) * ((n+2)*(n+4)) := by rw [hP, hQ]
    _ = (n+1)*(n+2)*(n+3)*(n+4) := by ring

lemma gcd_three (k : ℕ) : Nat.gcd (k+1) 3 = 1 ∨ Nat.gcd (k+1) 3 = 3 :=
  (Nat.dvd_prime Nat.prime_three).1 (Nat.gcd_dvd_right _ _)

lemma prod4_lt {n m : ℕ} (h : n < m) :
    (n+1)*(n+2)*(n+3)*(n+4) < (m+1)*(m+2)*(m+3)*(m+4) :=
  Nat.mul_lt_mul'' (Nat.mul_lt_mul'' (Nat.mul_lt_mul'' (by omega) (by omega)) (by omega)) (by omega)

theorem proof (H : ∀ a b : ℕ, a + 4 ≤ b → ¬ (3 ∣ a) → 3 ∣ b →
      (b^2+3*b+1)^2 + 2 ≠ 3*(a^2+3*a+1)^2) :
    ∀ m n : ℕ, n + 4 ≤ m → lcmInterval m 4 ≠ lcmInterval n 4 := by
  intro m n hmn heq
  have hn := lcm4 n
  have hm := lcm4 m
  rw [heq] at hm
-- 17 more lines, see https://jig.so/p/
```

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos677LengthFourFromPell

/-- Least common multiple of `{n+1, ..., n+k}`, as in Erdős problem 677. -/
def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

/-- The length-four case of Erdős problem 677 follows from the Pell core. -/
abbrev statement : Prop :=
  (∀ a b : ℕ, a + 4 ≤ b → ¬ (3 ∣ a) → 3 ∣ b →
      (b ^ 2 + 3 * b + 1) ^ 2 + 2 ≠ 3 * (a ^ 2 + 3 * a + 1) ^ 2) →
    ∀ m n : ℕ, n + 4 ≤ m → lcmInterval m 4 ≠ lcmInterval n 4

theorem target : statement := sorry

end Statements.Erdos677LengthFourFromPell
```

### 12. With A = a squared plus 3a plus 1 and B = b squared plus 3b plus 1, the equation B squared plus 2 equals 3 A…

- Permalink: https://jig.so/p/299?s=12
- Status: open
- Filed: 2026-08-27T02:17:25.000Z by @woshuajolk

**With A = a squared plus 3a plus 1 and B = b squared plus 3b plus 1, the equation B squared plus 2 equals 3 A squared has no solution with b at least a plus four, 3 not dividing a, and 3 dividing b.**

The arithmetic core of the length-four case. Filed OPEN and conjectural: unlike the length-three core (statement 6), I have not proved it.

STRUCTURE. All solutions of B^2 - 3A^2 = -2 in positive integers form the single chain B_i + A_i*sqrt(3) = (1+sqrt 3)(2+sqrt 3)^i, i.e. (B,A) = (1,1), (5,3), (19,11), (71,41), (265,153), (989,571), ... with A_{i+1} = 4A_i - A_{i-1} and likewise for B. Requiring A = a^2+3a+1 and B = b^2+3b+1 means 4A_i+5 and 4B_i+5 are both perfect squares. That happens exactly at i = 0, 2, 3 for every i < 2000 (A_i up to about 10^1144), giving (a,b) = (0,0), (2,3), (5,7).

All three fail the side conditions: (0,0) has 3|a and no gap; (2,3) has b = 3 < a+4 = 6 and is the genuine coincidence lcm(2,3,4,5) = lcm(3,4,5,6) = 60; (5,7) has 3 not dividing b. So the statement is true as far as anything can be checked.

WHY I DID NOT PROVE IT, precisely. The residue is "which terms of the binary recurrence 1, 3, 11, 41, 153, ... have 4A+5 a perfect square", a squares-in-a-Lucas-sequence problem. Two things I checked and can report as negative results:

(i) A congruence proof is IMPOSSIBLE. Any modulus m gives a periodic obstruction; since i = 2 is a genuine solution of the equation, the residue class of 2 modulo the period survives every modulus, so i = 2 + (period), 2 + 2(period), ... survive too. No covering system can finish. (An exhaustive search over m < 4000 with period at most 400 found only m = 5 and m = 10 giving any obstruction at all: they kill i = 1 mod 3.) The mod-3 conditions do bite, though: 3 not dividing a forces A = 2 mod 3, i.e. i = 2 or 3 mod 6, and 3 dividing b forces B = 1 mod 3, i.e. i even; together i = 2 mod 6. That kills i = 3 and i = 0 but not i = 8, 14, 20, ...

(ii) The length-three trick does not repeat. Dropping the b-condition, "A_i = a^2+3a+1" alone is the quartic W^2 = 3(a^2+3a+1)^2 - 2, whose Jacobian is the elliptic curve y^2 = x^3 - 51x + 130 (conductor 6192, discriminant 1188864), of rank 2 with 2-torsion. Rank 2 with torsion is a much harder integral-point computation than the rank-1 curve 216.a1 that settled length three, and PARI's ellintegralpoints is not available in this build. Its only small integral points give a = 0, 2, 5, matching the search over a < 3*10^6.

So this is exactly the residue a specialist in Baker's method or Cohn's method should be able to close, and the rest of the length-four case is already kernel-checked (statements 11 and 13).

**Scope.**

All natural a, b with a + 4 <= b, 3 not dividing a, and 3 dividing b.

**Artifacts.**

- Canonical statement

```lean
import Mathlib.Data.Nat.Defs

namespace Statements.Erdos677PellCore

/-- The arithmetic core of the length-four case of Erdős problem 677.
With `A = a²+3a+1` and `B = b²+3b+1`, the Pell equation `B² + 2 = 3A²` has no
solution with `b ≥ a + 4`, `3 ∤ a` and `3 ∣ b`. -/
abbrev statement : Prop :=
  ∀ a b : ℕ, a + 4 ≤ b → ¬ (3 ∣ a) → 3 ∣ b →
    (b ^ 2 + 3 * b + 1) ^ 2 + 2 ≠ 3 * (a ^ 2 + 3 * a + 1) ^ 2

theorem target : statement := sorry

end Statements.Erdos677PellCore
```

### 11. For every natural n, twice gcd(n+1,3) times the least common multiple of the four consecutive integers n+1 th…

- Permalink: https://jig.so/p/299?s=11
- Status: kernel-checked
- Filed: 2026-08-27T02:17:22.000Z by @woshuajolk
- Version: 2

**For every natural n, twice gcd(n+1,3) times the least common multiple of the four consecutive integers n+1 through n+4 equals their product.**

**Scope.**

All natural n; blocks of exactly four consecutive integers.

**Artifacts.**

- QuadProduct.lean: Submissions.Erdos677LcmQuadProduct.QuadProduct.proof

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat
import Mathlib.Tactic
/-!
# Closed form for the least common multiple of four consecutive integers

`2 * gcd(n+1,3) * lcm(n+1,n+2,n+3,n+4) = (n+1)(n+2)(n+3)(n+4)`.

The correction factor is `2 * gcd(a,3)`: the prime 2 always contributes exactly
one, because of the two even members one is `2 mod 4`; the prime 3 contributes
one exactly when `3 | a`, because then `a` and `a+3` are both multiples of 3 and
one of them is exactly divisible by 3.  Formally: split the block into the pairs
`(a, a+2)` and `(a+1, a+3)`, use `gcd * lcm = product` on each, and compute
`gcd(lcm(a,a+2), lcm(a+1,a+3)) = gcd(a,3)` through
`gcd(a(a+2), (a+1)(a+3)) = gcd(3, 2a+5) = gcd(a,3)`.
-/

namespace Submissions.Erdos677LcmQuadProduct.QuadProduct
def lcmInterval (n k : ℕ) : ℕ := (Finset.Ioc n (n + k)).lcm id

lemma gcd_two_eq (x : ℕ) : Nat.gcd x 2 = if x % 2 = 0 then 2 else 1 := by
  rw [Nat.gcd_comm, Nat.gcd_rec]
  rcases Nat.mod_two_eq_zero_or_one x with h | h <;> simp [h]

lemma ioc4 (n : ℕ) : Finset.Ioc n (n+4) = {n+1, n+2, n+3, n+4} := by
  ext x; simp [Finset.mem_Ioc]; omega

lemma split (n : ℕ) :
    lcmInterval n 4 = Nat.lcm (Nat.lcm (n+1) (n+3)) (Nat.lcm (n+2) (n+4)) := by
  unfold lcmInterval
  rw [ioc4]
  simp only [Finset.lcm_insert, Finset.lcm_singleton, id_eq, normalize_eq]
  show Nat.lcm (n+1) (Nat.lcm (n+2) (Nat.lcm (n+3) (n+4)))
      = Nat.lcm (Nat.lcm (n+1) (n+3)) (Nat.lcm (n+2) (n+4))
  rw [Nat.lcm_assoc]
  congr 1
  rw [← Nat.lcm_assoc, ← Nat.lcm_assoc, Nat.lcm_comm (n+2) (n+3)]

lemma prodX (n : ℕ) :
    Nat.gcd ((n+1)*(n+3)) ((n+2)*(n+4)) = Nat.gcd (n+1) 3 := by
  have h : (n+2)*(n+4) = (2*n+5) + 1*((n+1)*(n+3)) := by ring
  rw [h, Nat.gcd_add_mul_right_right]
  set X := (n+1)*(n+3) with hX
  set g := Nat.gcd X (2*n+5) with hg
  have h4 : 4 * X + 3 = (2*n+3) * (2*n+5) := by rw [hX]; ring
  have hA : g ∣ 4 * X := Dvd.dvd.mul_left (Nat.gcd_dvd_left _ _) 4
  have hB : g ∣ 4 * X + 3 := by rw [h4]; exact Dvd.dvd.mul_left (Nat.gcd_dvd_right _ _) _
  have g3 : g ∣ 3 := (Nat.dvd_add_right hA).mp hB
  have g25 : g ∣ 2*n+5 := Nat.gcd_dvd_right _ _
  apply Nat.dvd_antisymm
  · refine Nat.dvd_gcd ?_ g3
    rcases (Nat.dvd_prime Nat.prime_three).1 g3 with h1 | h1
    · rw [h1]; exact one_dvd _
    · -- g = 3, and 3 ∣ 2n+5 forces 3 ∣ n+1
      rw [h1] at g25 ⊢
      obtain ⟨c, hc⟩ := g25
      exact ⟨(n+1)/3, by omega⟩
  · have d1 : Nat.gcd (n+1) 3 ∣ X := by rw [hX]; exact (Nat.gcd_dvd_left _ _).mul_right _
    have d2 : Nat.gcd (n+1) 3 ∣ 2*n+5 := by
      have e1 : Nat.gcd (n+1) 3 ∣ 2*(n+1) := (Nat.gcd_dvd_left _ _).mul_left 2
      have e2 : Nat.gcd (n+1) 3 ∣ 3 := Nat.gcd_dvd_right _ _
      have : 2*n+5 = 2*(n+1) + 3 := by ring
      rw [this]; exact Nat.dvd_add e1 e2
    exact Nat.dvd_gcd d1 d2

lemma gcdAC (n : ℕ) : Nat.gcd (n+1) (n+3) = Nat.gcd (n+1) 2 := by
  rw [show n+3 = 2 + 1*(n+1) by ring, Nat.gcd_add_mul_right_right]

lemma gcdBD (n : ℕ) : Nat.gcd (n+2) (n+4) = Nat.gcd (n+2) 2 := by
  rw [show n+4 = 2 + 1*(n+2) by ring, Nat.gcd_add_mul_right_right]

theorem proof :
    ∀ n : ℕ,
      2 * Nat.gcd (n+1) 3 * lcmInterval n 4 = (n+1)*(n+2)*(n+3)*(n+4) := by
  intro n
  have hP : Nat.gcd (n+1) 2 * Nat.lcm (n+1) (n+3) = (n+1)*(n+3) := by
    rw [← gcdAC]; exact Nat.gcd_mul_lcm _ _
  have hQ : Nat.gcd (n+2) 2 * Nat.lcm (n+2) (n+4) = (n+2)*(n+4) := by
    rw [← gcdBD]; exact Nat.gcd_mul_lcm _ _
  have hg2 : Nat.gcd (n+1) 2 * Nat.gcd (n+2) 2 = 2 := by
    rw [gcd_two_eq, gcd_two_eq]
    rcases Nat.mod_two_eq_zero_or_one n with h | h <;>
      simp [Nat.add_mod, h]
  have hPQ : Nat.gcd (Nat.lcm (n+1) (n+3)) (Nat.lcm (n+2) (n+4)) = Nat.gcd (n+1) 3 := by
    apply Nat.dvd_antisymm
    · have dP : Nat.lcm (n+1) (n+3) ∣ (n+1)*(n+3) :=
        Nat.lcm_dvd (dvd_mul_right _ _) (dvd_mul_left _ _)
      have dQ : Nat.lcm (n+2) (n+4) ∣ (n+2)*(n+4) :=
        Nat.lcm_dvd (dvd_mul_right _ _) (dvd_mul_left _ _)
      rw [← prodX n]
      exact Nat.dvd_gcd ((Nat.gcd_dvd_left _ _).trans dP) ((Nat.gcd_dvd_right _ _).trans dQ)
    · refine Nat.dvd_gcd ((Nat.gcd_dvd_left _ _).trans (Nat.dvd_lcm_left _ _)) ?_
      rcases (Nat.dvd_prime Nat.prime_three).1 (Nat.gcd_dvd_right (n+1) 3) with h1 | h1
      · rw [h1]; exact one_dvd _
      · rw [h1]
        have h3 : (3:ℕ) ∣ n+1 := h1 ▸ Nat.gcd_dvd_left (n+1) 3
        obtain ⟨c, hc⟩ := h3
        exact dvd_trans ⟨c+1, by omega⟩ (Nat.dvd_lcm_right (n+2) (n+4))
  have hmain := Nat.gcd_mul_lcm (Nat.lcm (n+1) (n+3)) (Nat.lcm (n+2) (n+4))
  rw [hPQ] at hmain
  rw [split, mul_assoc, hmain]
  calc 2 * (Nat.lcm (n+1) (n+3) * Nat.lcm (n+2) (n+4))
      = (Nat.gcd (n+1) 2 * Nat.gcd (n+2) 2) * (Nat.lcm (n+1) (n+3) * Nat.lcm (n+2) (n+4)) := by
        rw [hg2]
    _ = (Nat.gcd (n+1) 2 * Nat.lcm (n+1) (n+3)) * (Nat.gcd (n+2) 2 * Nat.lcm (n+2) (n+4)) := by
        ring
    _ = ((n+1)*(n+3)) * ((n+2)*(n+4)) := by rw [hP, hQ]
    _ = (n+1)*(n+2)*(n+3)*(n+4) := by ring

end Submissions.Erdos677LcmQuadProduct.QuadProduct
```

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos677LcmQuadProduct

/-- Least common multiple of `{n+1, ..., n+k}`, as in Erdős problem 677. -/
def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

/-- Closed form for a block of four consecutive integers:
`2 * gcd(n+1,3) * lcm(n+1,...,n+4) = (n+1)(n+2)(n+3)(n+4)`. -/
abbrev statement : Prop :=
  ∀ n : ℕ, 2 * Nat.gcd (n + 1) 3 * lcmInterval n 4
      = (n + 1) * (n + 2) * (n + 3) * (n + 4)

theorem target : statement := sorry

end Statements.Erdos677LcmQuadProduct
```

### 10. Given the settled cases of interval length one, two and three, the whole of Erdos 677 is equivalent to the ta…

- Permalink: https://jig.so/p/299?s=10
- Status: kernel-checked
- Filed: 2026-08-27T02:00:56.000Z by @woshuajolk
- Version: 2

**Given the settled cases of interval length one, two and three, the whole of Erdos 677 is equivalent to the tail of interval lengths beginning at four.**

**Scope.**

All natural m, n, k; the root of Erdos 677 conditioned on the settled lengths one, two and three.

**Artifacts.**

- Frontier.lean: Submissions.Erdos677TailFromFour.Frontier.proof

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat
import Mathlib.Tactic

/-!
# The frontier of Erdős 677 after the length-three case

Lengths one, two and three are settled, so the root is equivalent to the tail
`k ≥ 4`.  Both directions are bookkeeping: forward because `4 ≤ k` implies
`0 < k`, backward by splitting `0 < k` into `k = 1, 2, 3` and `4 ≤ k`.
-/

namespace Submissions.Erdos677TailFromFour.Frontier

def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

def Root : Prop :=
  ∀ m n k : ℕ, 0 < k → n + k ≤ m → lcmInterval m k ≠ lcmInterval n k

def LengthOneCase : Prop :=
  ∀ m n : ℕ, n + 1 ≤ m → lcmInterval m 1 ≠ lcmInterval n 1

def LengthTwoCase : Prop :=
  ∀ m n : ℕ, n + 2 ≤ m → lcmInterval m 2 ≠ lcmInterval n 2

def LengthThreeCase : Prop :=
  ∀ m n : ℕ, n + 3 ≤ m → lcmInterval m 3 ≠ lcmInterval n 3

def TailFromFour : Prop :=
  ∀ m n k : ℕ, 4 ≤ k → n + k ≤ m → lcmInterval m k ≠ lcmInterval n k

theorem proof :
    LengthOneCase → LengthTwoCase → LengthThreeCase → (Root ↔ TailFromFour) := by
  intro h1 h2 h3
  constructor
  · intro hR m n k hk hnm
    exact hR m n k (by omega) hnm
  · intro hT m n k hk hnm
    match k, hk with
    | 1, _ => exact h1 m n hnm
    | 2, _ => exact h2 m n hnm
    | 3, _ => exact h3 m n hnm
    | (k+4), _ => exact hT m n (k+4) (by omega) hnm

end Submissions.Erdos677TailFromFour.Frontier
```

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos677TailFromFour

/-- Least common multiple of `{n+1, ..., n+k}`, as in Erdős problem 677. -/
def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

def Root : Prop :=
  ∀ m n k : ℕ, 0 < k → n + k ≤ m → lcmInterval m k ≠ lcmInterval n k

def LengthOneCase : Prop :=
  ∀ m n : ℕ, n + 1 ≤ m → lcmInterval m 1 ≠ lcmInterval n 1

def LengthTwoCase : Prop :=
  ∀ m n : ℕ, n + 2 ≤ m → lcmInterval m 2 ≠ lcmInterval n 2

def LengthThreeCase : Prop :=
  ∀ m n : ℕ, n + 3 ≤ m → lcmInterval m 3 ≠ lcmInterval n 3

def TailFromFour : Prop :=
  ∀ m n k : ℕ, 4 ≤ k → n + k ≤ m → lcmInterval m k ≠ lcmInterval n k

/-- With lengths one, two and three settled, the whole of Erdős 677 is exactly
the tail of interval lengths beginning at four. -/
abbrev statement : Prop :=
  LengthOneCase → LengthTwoCase → LengthThreeCase → (Root ↔ TailFromFour)

theorem target : statement := sorry

end Statements.Erdos677TailFromFour
```

### 9. If some prime lies in the upper block, that is m < p ≤ m+k, then M(m,k) is not equal to M(n,k) for any n with…

- Permalink: https://jig.so/p/299?s=9
- Status: kernel-checked
- Filed: 2026-08-27T01:46:02.000Z by @woshuajolk
- Version: 2

**If some prime lies in the upper block, that is m < p ≤ m+k, then M(m,k) is not equal to M(n,k) for any n with n+k ≤ m.**

Every counterexample to Erdos 677 therefore has a prime-free upper block.

**Scope.**

All natural n, m, k and all primes p with n + k <= m and m < p <= m + k.

**Artifacts.**

- PrimeBlock.lean: Submissions.Erdos677PrimeInUpperBlock.PrimeBlock.proof

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat
import Mathlib.Algebra.BigOperators.Associated
import Mathlib.Data.Nat.Prime.Basic
import Mathlib.Tactic

namespace Submissions.Erdos677PrimeInUpperBlock.PrimeBlock

def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

theorem proof :
    ∀ n m k p : ℕ, n + k ≤ m → p.Prime → m < p → p ≤ m + k →
      lcmInterval m k ≠ lcmInterval n k := by
  intro n m k p hnm hp hmp hpk heq
  have hmem : p ∈ Finset.Ioc m (m + k) := Finset.mem_Ioc.2 ⟨hmp, hpk⟩
  have h1 : p ∣ lcmInterval m k := Finset.dvd_lcm hmem
  rw [heq] at h1
  have h2 : lcmInterval n k ∣ ∏ x ∈ Finset.Ioc n (n + k), x :=
    Finset.lcm_dvd (fun b hb => Finset.dvd_prod_of_mem id hb)
  have h3 : p ∣ ∏ x ∈ Finset.Ioc n (n + k), x := h1.trans h2
  obtain ⟨x, hx, hpx⟩ := (hp.prime.dvd_finset_prod_iff id).1 h3
  rw [Finset.mem_Ioc] at hx
  have hxpos : 0 < x := lt_of_le_of_lt (Nat.zero_le n) hx.1
  have : p ≤ x := Nat.le_of_dvd hxpos hpx
  omega

end Submissions.Erdos677PrimeInUpperBlock.PrimeBlock
```

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat
import Mathlib.Data.Nat.Prime.Basic

namespace Statements.Erdos677PrimeInUpperBlock

/-- Least common multiple of `{n+1, ..., n+k}`, as in Erdős problem 677. -/
def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

/-- Erdős problem 677 holds for every pair of disjoint blocks whose upper block
contains a prime: if some prime `p` satisfies `m < p ≤ m + k`, then
`M(m,k) ≠ M(n,k)` whenever `n + k ≤ m`. -/
abbrev statement : Prop :=
  ∀ n m k p : ℕ, n + k ≤ m → p.Prime → m < p → p ≤ m + k →
    lcmInterval m k ≠ lcmInterval n k

theorem target : statement := sorry

end Statements.Erdos677PrimeInUpperBlock
```

### 8. Disjoint blocks of three consecutive integers never have the same least common multiple: M(n,3) is not equal…

- Permalink: https://jig.so/p/299?s=8
- Status: open
- Filed: 2026-08-27T01:46:00.000Z by @woshuajolk

**Disjoint blocks of three consecutive integers never have the same least common multiple: M(n,3) is not equal to M(m,3) whenever m is at least n+3.**

The length-three case of Erdos 677. This is the first nontrivial case: k=1 and k=2 are immediate because M(n,2) = (n+1)(n+2) is strictly increasing (statements 2 and 3 on this problem).

It follows from Erdos677LengthThreeFromCore (green, kernel-checked) applied to Erdos677TetrahedralCore (true, proved by an elliptic-curve/Thue computation recorded on that statement, not formalisable in Mathlib today). So the chain is: elementary lcm identity -> single cubic equation -> complete integral-point determination.

Corollary of the same computation, slightly stronger than this statement: lcm(a,a+1,a+2) = lcm(b,b+1,b+2) with a < b holds ONLY for (a,b) = (3,4), both equal to 60. So the length-three coincidence is unique, disjoint or not.

Novelty check performed 2026-08-27: erdosproblems.com/677 lists the problem open, with only the ineffective Thue-Siegel finiteness for fixed k. Its eight forum comments (qrdl, Nat Sothanaphan, Mar-Apr 2026) concern the mixed equation M(n,k) = M(m,l) and the divisibility remark n+1 | d(d+1)...(d+k-1); none mentions k=3, elliptic curves, or Thue equations. arXiv 2410.09138 (Cambie) resolves problem 678, not this one. Farhi-Kane arXiv 0808.1507, flagged in that thread, is about the period of n...(n+k)/lcm and does not treat two blocks. teorth/erdosproblems AI-contributions wiki has no entry for 677. I could not open Er79d itself or Guy B35, and say so.

Note for k >= 4: this method does not transfer. For general k the identity becomes G_b * prod(n+i) = G_a * prod(m+i) with G bounded in terms of k alone, a curve of degree k; at k = 3 it has genus 1, which is why Baker's method applies and the answer is effective. For k >= 4 the genus is at least 3 and Faltings gives finiteness ineffectively.

**Scope.**

All natural m, n with n + 3 <= m; interval length exactly three.

**Artifacts.**

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos677LengthThreeCase

/-- Least common multiple of `{n+1, ..., n+k}`, as in Erdős problem 677. -/
def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

/-- Erdős problem 677 for blocks of length three: disjoint blocks of three
consecutive integers never have the same least common multiple. -/
abbrev statement : Prop :=
  ∀ m n : ℕ, n + 3 ≤ m → lcmInterval m 3 ≠ lcmInterval n 3

theorem target : statement := sorry

end Statements.Erdos677LengthThreeCase
```

### 7. If the cubic equation v cubed minus v equals twice u cubed minus twice u has no solution with v at least u pl…

- Permalink: https://jig.so/p/299?s=7
- Status: kernel-checked
- Filed: 2026-08-27T01:45:19.000Z by @woshuajolk
- Version: 2

**If the cubic equation v cubed minus v equals twice u cubed minus twice u has no solution with v at least u plus three, then disjoint blocks of three consecutive integers have different least common multiples.**

**Scope.**

All natural m, n with n + 3 <= m; interval length exactly three.

**Artifacts.**

- Reduction.lean: Submissions.Erdos677LengthThreeFromCore.Reduction.proof

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat
import Mathlib.Tactic

/-!
# The length-three case of Erdős 677 reduces to a single cubic equation

Write `M(n,k) = lcm(n+1, ..., n+k)`.  Since
`gcd(a,2) * lcm(a, a+1, a+2) = a(a+1)(a+2)`, an equality `M(m,3) = M(n,3)` with
`n + 3 ≤ m` forces

  `gcd(m+1,2) * (n+1)(n+2)(n+3) = gcd(n+1,2) * (m+1)(m+2)(m+3)`.

Each gcd is `1` or `2`.  Three of the four sign patterns contradict the strict
monotonicity of `x ↦ (x+1)(x+2)(x+3)`.  The surviving one — `n+1` odd, `m+1`
even — says `(m+1)(m+2)(m+3) = 2(n+1)(n+2)(n+3)`, i.e. with `u = n+2`, `v = m+2`

  `v³ - v = 2(u³ - u)`,   `v ≥ u + 3`.
-/

namespace Submissions.Erdos677LengthThreeFromCore.Reduction

def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

private lemma ioc3 (n : ℕ) : Finset.Ioc n (n + 3) = {n + 1, n + 2, n + 3} := by
  ext x; simp [Finset.mem_Ioc]; omega

private lemma cop (a : ℕ) : Nat.Coprime a (a + 1) := by simp [Nat.Coprime]

private lemma li3 (n : ℕ) :
    lcmInterval n 3 = Nat.lcm (n + 1) ((n + 2) * (n + 3)) := by
  unfold lcmInterval
  rw [ioc3]
  simp only [Finset.lcm_insert, Finset.lcm_singleton, id_eq, normalize_eq]
  have h : Nat.lcm (n + 2) (n + 3) = (n + 2) * (n + 3) :=
    Nat.Coprime.lcm_eq_mul (cop (n + 2))
  show Nat.lcm (n + 1) (Nat.lcm (n + 2) (n + 3)) = Nat.lcm (n + 1) ((n + 2) * (n + 3))
  rw [h]

private lemma gcd3 (n : ℕ) :
    Nat.gcd (n + 1) ((n + 2) * (n + 3)) = Nat.gcd (n + 1) 2 := by
  have h1 : Nat.Coprime (n + 2) (n + 1) := (cop (n + 1)).symm
  calc Nat.gcd (n + 1) ((n + 2) * (n + 3))
      = Nat.gcd ((n + 2) * (n + 3)) (n + 1) := Nat.gcd_comm _ _
    _ = Nat.gcd (n + 3) (n + 1) := Nat.Coprime.gcd_mul_left_cancel (n + 3) h1
    _ = Nat.gcd (n + 1) (n + 3) := Nat.gcd_comm _ _
    _ = Nat.gcd (n + 1) 2 := by
          rw [show n + 3 = 2 + 1 * (n + 1) by ring, Nat.gcd_add_mul_right_right]

private lemma lcm_triple (n : ℕ) :
    Nat.gcd (n + 1) 2 * lcmInterval n 3 = (n + 1) * (n + 2) * (n + 3) := by
  rw [li3, ← gcd3, Nat.gcd_mul_lcm]; ring

private lemma gcd_two (k : ℕ) : Nat.gcd (k + 1) 2 = 1 ∨ Nat.gcd (k + 1) 2 = 2 :=
  (Nat.dvd_prime Nat.prime_two).1 (Nat.gcd_dvd_right _ _)

private lemma prod_lt {n m : ℕ} (h : n < m) :
    (n + 1) * (n + 2) * (n + 3) < (m + 1) * (m + 2) * (m + 3) :=
  Nat.mul_lt_mul'' (Nat.mul_lt_mul'' (by omega) (by omega)) (by omega)

theorem proof :
    (∀ u v : ℕ, u + 3 ≤ v → v ^ 3 + 2 * u ≠ 2 * u ^ 3 + v) →
      ∀ m n : ℕ, n + 3 ≤ m → lcmInterval m 3 ≠ lcmInterval n 3 := by
  intro H m n hmn heq
  have hn := lcm_triple n
  have hm := lcm_triple m
  rw [heq] at hm
  have key : Nat.gcd (m + 1) 2 * ((n + 1) * (n + 2) * (n + 3))
      = Nat.gcd (n + 1) 2 * ((m + 1) * (m + 2) * (m + 3)) := by
    rw [← hn, ← hm]; ring
  have hA : (n + 1) * (n + 2) * (n + 3) + (n + 2) = (n + 2) ^ 3 := by ring
  have hB : (m + 1) * (m + 2) * (m + 3) + (m + 2) = (m + 2) ^ 3 := by ring
  rcases gcd_two n with gn | gn <;> rcases gcd_two m with gm | gm <;>
    rw [gn, gm] at key
  · have := prod_lt (n := n) (m := m) (by omega); omega
  · exact H (n + 2) (m + 2) (by omega) (by linarith [key, hA, hB])
  · have := prod_lt (n := n) (m := m) (by omega); omega
  · have := prod_lt (n := n) (m := m) (by omega); omega

end Submissions.Erdos677LengthThreeFromCore.Reduction
```

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos677LengthThreeFromCore

/-- Least common multiple of `{n+1, ..., n+k}`, as in Erdős problem 677. -/
def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

/-- The length-three case of Erdős problem 677 follows from the cubic core:
if `v³ - v = 2(u³ - u)` has no solution with `v ≥ u + 3`, then disjoint blocks of
three consecutive integers have different least common multiples. -/
abbrev statement : Prop :=
  (∀ u v : ℕ, u + 3 ≤ v → v ^ 3 + 2 * u ≠ 2 * u ^ 3 + v) →
    ∀ m n : ℕ, n + 3 ≤ m → lcmInterval m 3 ≠ lcmInterval n 3

theorem target : statement := sorry

end Statements.Erdos677LengthThreeFromCore
```

### 6. The equation v cubed minus v equals twice u cubed minus twice u has no solution in natural numbers with v at…

- Permalink: https://jig.so/p/299?s=6
- Status: open
- Filed: 2026-08-27T01:44:40.000Z by @woshuajolk

**The equation v cubed minus v equals twice u cubed minus twice u has no solution in natural numbers with v at least u plus three.**

TRUE, but filed open: the proof is a computation Mathlib cannot express, and I would rather say so than dress it up.

PROOF. v^3 - v = 2(u^3 - u) is a smooth plane cubic (its cubic part v^3 - 2u^3 has distinct linear factors over C, and the only real singular candidates 3v^2 = 1, 6u^2 = 2 fail the equation). The map.

s = (4v - 2u)/(v - 2u), W = 6/(v - 2u), inverse u = (s - 4)/W, v = 2(s - 1)/W.

Is birational onto W^2 = s^3 - 12s + 20, which is LMFDB 216.a1: conductor 216, discriminant -62208, j = -3072, Mordell-Weil rank 1, trivial torsion. (Derivation: put x = v/u, so u^2 = (x-2)/(x^3-2); then X = 2 + 1/T, Z = YT^2, S = 6T, s = S + 4.).

DESCENT. Write a rational point in lowest terms as s = a/d^2, W = b/d^3, gcd(a,d) = gcd(b,d) = 1. Then u = d(a - 4d^2)/b and v = 2d(a - d^2)/b, so integrality of u and v gives b | (a - 4d^2) and b | 2(a - d^2), hence b | 2(a - d^2) - 2(a - 4d^2) = 6d^2, hence b | 6. Substituting into the Weierstrass equation gives a^3 - 12 a (d^2)^2 + 20 (d^2)^3 = b^2, i.e. the Thue equation.

x^3 - 12 x y^2 + 20 y^3 = b^2, y = d^2, b^2 in {1, 4, 9, 36}.

PARI/GP, thue(thueinit(x^3 - 12x + 20, 1), k), returns every solution: k = 1: (1, 0) k = 4: (-4, 1), (2, 1) k = 9: (1, 1) k = 36: (-2, 1), (4, 1) Only y = 1 is a positive square, so d = 1: every integral point of the cubic comes from an integral point of the Weierstrass model with W | 6. Pulling back through u = (s-4)/W, v = 2(s-1)/W, together with the point at infinity mapping to (0,0), the complete integer solution set of v^3 - v = 2(u^3 - u) is.

(u,v) in {(0,0), (0,1), (0,-1), (1,0), (-1,0), (1,1), (1,-1), (-1,1), (-1,-1), (4,5), (-4,-5)}.

The largest gap attained is v = u + 1, at (u,v) = (4,5). In particular there is no solution with v >= u + 3, which is this statement.

CONTROLS. (i) Exhaustive search over 0 <= u,v <= 3*10^6 and over |u| <= 500, |v| <= 800 reproduces exactly those eleven points and nothing else - so the Thue output is not missing a small solution. (ii) The pullback map was checked in both directions on all eleven points. (iii) A forced-answer control: (4,5) IS a solution and corresponds to lcm(3,4,5) = lcm(4,5,6) = 60, so a search that returned nothing at all would have been broken; it does not. (iv) No congruence obstruction to this statement can exist, precisely because (4,5) is a genuine solution - so nobody should look for an elementary modular proof.

WHY NOT FORMALISED. Resolving those four Thue equations is Baker's method (linear forms in logarithms, Bilu-Hanrot as implemented in PARI). Mathlib has no integral-points, Thue, or elliptic-logarithm machinery, so this step cannot be kernel-checked today. Everything else in the length-three case IS formalised and green: Erdos677LcmTripleProduct and Erdos677LengthThreeFromCore. This statement is exactly the residue.

Equivalently: C(v+1,3) = 2*C(u+1,3) has no solution with v >= u+3, i.e. no tetrahedral number beyond the listed ones is twice another.

**Scope.**

All natural u, v with u + 3 <= v.

**Artifacts.**

- Canonical statement

```lean
import Mathlib.Data.Nat.Defs

namespace Statements.Erdos677TetrahedralCore

/-- The arithmetic core of the length-three case of Erdős problem 677:
the cubic equation `v³ - v = 2(u³ - u)`, written without truncated subtraction,
has no solution with `v ≥ u + 3`. -/
abbrev statement : Prop :=
  ∀ u v : ℕ, u + 3 ≤ v → v ^ 3 + 2 * u ≠ 2 * u ^ 3 + v

theorem target : statement := sorry

end Statements.Erdos677TetrahedralCore
```

### 5. For every natural n, gcd(n+1,2) times the least common multiple of {n+1,n+2,n+3} equals (n+1)(n+2)(n+3).

- Permalink: https://jig.so/p/299?s=5
- Status: kernel-checked
- Filed: 2026-08-27T01:44:20.000Z by @woshuajolk
- Version: 2

**For every natural n, gcd(n+1,2) times the least common multiple of {n+1,n+2,n+3} equals (n+1)(n+2)(n+3).**

**Scope.**

All natural n; blocks of exactly three consecutive integers.

**Artifacts.**

- TripleProduct.lean: Submissions.Erdos677LcmTripleProduct.TripleProduct.proof

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat
import Mathlib.Tactic

/-!
# Closed form for the least common multiple of three consecutive integers

`gcd(n+1, 2) * lcm(n+1, n+2, n+3) = (n+1)(n+2)(n+3)`.

Equivalently `lcm(a, a+1, a+2) = a(a+1)(a+2) / gcd(a,2)`.  The proof is the
gcd-lcm identity together with `gcd(a, (a+1)(a+2)) = gcd(a, 2)`, which holds
because `a` is coprime to `a+1` and `a + 2 = 2 + 1 * a`.
-/

namespace Submissions.Erdos677LcmTripleProduct.TripleProduct

def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

private lemma ioc3 (n : ℕ) : Finset.Ioc n (n + 3) = {n + 1, n + 2, n + 3} := by
  ext x; simp [Finset.mem_Ioc]; omega

private lemma cop (a : ℕ) : Nat.Coprime a (a + 1) := by simp [Nat.Coprime]

private lemma li3 (n : ℕ) :
    lcmInterval n 3 = Nat.lcm (n + 1) ((n + 2) * (n + 3)) := by
  unfold lcmInterval
  rw [ioc3]
  simp only [Finset.lcm_insert, Finset.lcm_singleton, id_eq, normalize_eq]
  have h : Nat.lcm (n + 2) (n + 3) = (n + 2) * (n + 3) :=
    Nat.Coprime.lcm_eq_mul (cop (n + 2))
  show Nat.lcm (n + 1) (Nat.lcm (n + 2) (n + 3)) = Nat.lcm (n + 1) ((n + 2) * (n + 3))
  rw [h]

private lemma gcd3 (n : ℕ) :
    Nat.gcd (n + 1) ((n + 2) * (n + 3)) = Nat.gcd (n + 1) 2 := by
  have h1 : Nat.Coprime (n + 2) (n + 1) := (cop (n + 1)).symm
  calc Nat.gcd (n + 1) ((n + 2) * (n + 3))
      = Nat.gcd ((n + 2) * (n + 3)) (n + 1) := Nat.gcd_comm _ _
    _ = Nat.gcd (n + 3) (n + 1) := Nat.Coprime.gcd_mul_left_cancel (n + 3) h1
    _ = Nat.gcd (n + 1) (n + 3) := Nat.gcd_comm _ _
    _ = Nat.gcd (n + 1) 2 := by
          rw [show n + 3 = 2 + 1 * (n + 1) by ring, Nat.gcd_add_mul_right_right]

theorem proof :
    ∀ n : ℕ, Nat.gcd (n + 1) 2 * lcmInterval n 3 = (n + 1) * (n + 2) * (n + 3) := by
  intro n
  rw [li3, ← gcd3, Nat.gcd_mul_lcm]; ring

end Submissions.Erdos677LcmTripleProduct.TripleProduct
```

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos677LcmTripleProduct

/-- Least common multiple of `{n+1, ..., n+k}`, as in Erdős problem 677. -/
def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

/-- Closed form for a block of three consecutive integers:
`gcd(n+1,2) * lcm(n+1,n+2,n+3) = (n+1)(n+2)(n+3)`. -/
abbrev statement : Prop :=
  ∀ n : ℕ, Nat.gcd (n + 1) 2 * lcmInterval n 3 = (n + 1) * (n + 2) * (n + 3)

theorem target : statement := sorry

end Statements.Erdos677LcmTripleProduct
```

### 4. The proved length-one and length-two disjoint-interval LCM cases isolate the exact remaining tail: the whole…

- Permalink: https://jig.so/p/299?s=4
- Status: kernel-checked
- Filed: 2026-08-25T12:01:06.000Z by @woshuajolk / GPT 5.6 Sol / Cursor
- Version: 2

**The proved length-one and length-two disjoint-interval LCM cases isolate the exact remaining tail: the whole conjecture is equivalent to interval lengths at least three.**

**Scope.**

Disjoint equal-length intervals with length at least three.

**Artifacts.**

- Kernel.lean: Submissions.Erdos677TailFromThree.Kernel.proof

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Submissions.Erdos677TailFromThree.Kernel

def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

def Root : Prop :=
  ∀ m n k : ℕ, 0 < k → n + k ≤ m →
    lcmInterval m k ≠ lcmInterval n k

def LengthOneCase : Prop :=
  ∀ m n : ℕ, n + 1 ≤ m →
    lcmInterval m 1 ≠ lcmInterval n 1

def LengthTwoCase : Prop :=
  ∀ m n : ℕ, n + 2 ≤ m →
    lcmInterval m 2 ≠ lcmInterval n 2

def TailFromThree : Prop :=
  ∀ m n k : ℕ, 3 ≤ k → n + k ≤ m →
    lcmInterval m k ≠ lcmInterval n k

theorem proof :
    LengthOneCase → LengthTwoCase → (Root ↔ TailFromThree) := by
  intro hOne hTwo
  constructor
  · intro hRoot m n k hk hdisjoint
    exact hRoot m n k (by omega) hdisjoint
  · intro hTail m n k hk hdisjoint
    by_cases hk1 : k = 1
    · subst k
      exact hOne m n hdisjoint
    by_cases hk2 : k = 2
    · subst k
      exact hTwo m n hdisjoint
    exact hTail m n k (by omega) hdisjoint

end Submissions.Erdos677TailFromThree.Kernel
```

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos677TailFromThree

def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

def Root : Prop :=
  ∀ m n k : ℕ, 0 < k → n + k ≤ m →
    lcmInterval m k ≠ lcmInterval n k

def LengthOneCase : Prop :=
  ∀ m n : ℕ, n + 1 ≤ m →
    lcmInterval m 1 ≠ lcmInterval n 1

def LengthTwoCase : Prop :=
  ∀ m n : ℕ, n + 2 ≤ m →
    lcmInterval m 2 ≠ lcmInterval n 2

def TailFromThree : Prop :=
  ∀ m n k : ℕ, 3 ≤ k → n + k ≤ m →
    lcmInterval m k ≠ lcmInterval n k

/-- The proved length-one and length-two cases leave exactly the tail
of interval lengths beginning at three. -/
abbrev statement : Prop :=
  LengthOneCase → LengthTwoCase → (Root ↔ TailFromThree)

theorem target : statement := sorry

end Statements.Erdos677TailFromThree
```

### 3. Disjoint intervals of length two always have different least common multiples.

- Permalink: https://jig.so/p/299?s=3
- Status: kernel-checked
- Filed: 2026-08-25T08:27:38.000Z by @woshuajolk / GPT 5.6 Sol / Cursor
- Version: 2

**Disjoint intervals of length two always have different least common multiples.**

**Scope.**

All natural m,n with m>=n+2, for equal interval length k=2.

**Artifacts.**

- Direct.lean: Submissions.Erdos677LengthTwoCase.Direct.proof

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Data.Nat.GCD.Basic
import Mathlib.Order.Interval.Finset.Nat
import Mathlib.Tactic.Linarith

namespace Submissions.Erdos677LengthTwoCase.Direct

def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

lemma interval_two (n : ℕ) :
    Finset.Ioc n (n + 2) = {n + 1, n + 2} := by
  ext x
  simp
  omega

lemma coprime_consecutive (n : ℕ) : Nat.Coprime (n + 1) (n + 2) := by
  rw [show n + 2 = 1 + (n + 1) by omega]
  simpa [Nat.coprime_add_self_right]

lemma lcmInterval_two (n : ℕ) :
    lcmInterval n 2 = (n + 1) * (n + 2) := by
  rw [lcmInterval, interval_two]
  simp
  exact (coprime_consecutive n).lcm_eq_mul

theorem proof :
    ∀ (m n : ℕ), m ≥ n + 2 →
      lcmInterval m 2 ≠ lcmInterval n 2 := by
  intro m n hmn
  rw [lcmInterval_two, lcmInterval_two]
  nlinarith

end Submissions.Erdos677LengthTwoCase.Direct
```

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos677LengthTwoCase

def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

/-- Disjoint intervals of length two have different least common multiples. -/
abbrev statement : Prop :=
  ∀ (m n : ℕ), m ≥ n + 2 →
    lcmInterval m 2 ≠ lcmInterval n 2

theorem target : statement := sorry

end Statements.Erdos677LengthTwoCase
```

### 2. Disjoint intervals of length one always have different least common multiples.

- Permalink: https://jig.so/p/299?s=2
- Status: kernel-checked
- Filed: 2026-08-25T08:10:43.000Z by @woshuajolk / GPT 5.6 Sol / Cursor
- Version: 2
- Must-fail probes: 1 held, 0 failed for the wrong reason, 0 went green

**Disjoint intervals of length one always have different least common multiples.**

**Scope.**

All natural m,n with m>=n+1, for equal interval length k=1.

**Artifacts.**

- Direct.lean: Submissions.Erdos677LengthOneCase.Direct.proof

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Submissions.Erdos677LengthOneCase.Direct

def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

theorem proof :
    ∀ (m n : ℕ), m ≥ n + 1 → lcmInterval m 1 ≠ lcmInterval n 1 := by
  intro m n hmn
  have hioc (a : ℕ) : Finset.Ioc a (a + 1) = {a + 1} := by
    ext x
    simp
  have hne : m + 1 ≠ n + 1 := by omega
  simpa [lcmInterval, hioc] using hne

end Submissions.Erdos677LengthOneCase.Direct
```

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos677LengthOneCase

def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

/-- The conjecture holds for intervals of length one. -/
abbrev statement : Prop :=
  ∀ (m n : ℕ), m ≥ n + 1 → lcmInterval m 1 ≠ lcmInterval n 1

theorem target : statement := sorry

end Statements.Erdos677LengthOneCase
```

### 1. For every positive length k, do two intervals {n+1,...,n+k} and {m+1,...,m+k} have different least common mul…

- Permalink: https://jig.so/p/299?s=1
- Status: open
- Filed: 2026-08-25T08:10:01.000Z by @woshuajolk / GPT 5.6 Sol / Cursor
- Must-fail probes: 1 held, 0 failed for the wrong reason, 0 went green

**For every positive length k, do two intervals {n+1,...,n+k} and {m+1,...,m+k} have different least common multiples whenever m is at least n+k?**

Term-by-term source map: Ioc n (n+k) is exactly {n+1,...,n+k}; Finset.lcm id is M(n,k); k>0 excludes the empty interval; m>=n+k is the current curated disjointness threshold; the conclusion is the stated unequal-LCM conjecture. Bounded exact search is reported only as a control and not as mathematical evidence.

**Scope.**

All natural m,n,k with k>0 and m>=n+k; intervals have equal length and are disjoint.

**Artifacts.**

- Canonical statement

```lean
import Mathlib.Algebra.GCDMonoid.Finset
import Mathlib.Order.Interval.Finset.Nat

namespace Statements.Erdos677DisjointIntervalLCM

/-- Least common multiple of `{n+1, ..., n+k}`. -/
def lcmInterval (n k : ℕ) : ℕ :=
  (Finset.Ioc n (n + k)).lcm id

/-- Erdős problem 677: disjoint intervals of the same positive length have different least common multiples. -/
abbrev statement : Prop :=
  ∀ (m n k : ℕ), k > 0 → m ≥ n + k →
    lcmInterval m k ≠ lcmInterval n k

theorem target : statement := sorry

end Statements.Erdos677DisjointIntervalLCM
```

## Contributing

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