686.lean`

Explanation legend Lean4 Domain Explanation Language (L4DEL)

L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.

L4DEL: Lean4 Domain Explanation Language

Annotation provenance. This overlay contains 10 curated AI-written annotations and 7190 mechanical source-derived annotations. They have not been human-reviewed as explanations; treat them as unverified and not mathematical authority. Retained generation history: The 71 original curated annotations were produced in a Claude Sonnet 5.5 session at medium effort; Claude Opus 5.5 is recorded in that session as the advisor model, not the direct author. The bulk source-derived expansion was generated mechanically from pinned source comments by tooling developed in a GPT-5.6 Sol session at ultra reasoning; GPT-5.6 Sol did not originate the quoted source prose. On 2026-10-17 the mechanically generated source-comment annotations of the Formal Conjectures, Carleson, FLT, PFR, CompFiles, Mathematics in Lean and Theorem Proving in Lean overlays were regenerated by the same generator after three corrections made in a Claude Sonnet 5.5 session: titles no longer cut a sentence inside mathematics or end on a dangling word, workflow markers are removed from the quoted prose, and declaration names follow the scanner's namespace resolution. The boilerplate step line was dropped, and comments that were only a row of dashes no longer yield annotations. No new prose was written and no human review is claimed.

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

8 declarations, 5 contain sorry, 0 declared as axioms

Erdős Problem 686

References:

erdosproblems.com/686
[Er79d] Erdős, P., Some unconventional problems in number theory. Acta Math. Acad. Sci. Hungar. (1979), 71-80.

Imports (1)

FormalConjecturesUtil

Imported by 1 file

FormalConjectures/Subsets/FC100SolvedSet1.lean

theorem Erdos686.erdos_686 Can every integer $N≥2$ be written as $$N=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$? sorry line 39 source
Source documentation
Can every integer $N≥2$ be written as $$N=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$?

depends on sorry what this means
Statement uses: Iff, True, Nat, GE.ge, OfNat.ofNat, Exists, And, HAdd.hAdd, Eq, Rat, HDiv.hDiv, Finset.prod, Finset.Icc
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
22 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_686 :
    answer(sorry) ↔ ∀ N ≥ (2 : ℕ), ∃ᵉ (k ≥ 2) (n : ℕ) (m ≥ n + k),
      (N : ℚ) = (∏ i ∈ Finset.Icc 1 k, (m + i)) / (∏ i ∈ Finset.Icc 1 k, (n + i)) := by
…
```
theorem Erdos686.erdos_686.variants.square Can every square $N≥2$ be written as $$N=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$? sorry line 50 source
Source documentation
Can every square $N≥2$ be written as $$N=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$?

depends on sorry what this means
Statement uses: Iff, True, Nat, GE.ge, OfNat.ofNat, IsSquare, Exists, And, HAdd.hAdd, Eq, Rat, HDiv.hDiv, Finset.prod, Finset.Icc
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
24 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_686.variants.square :
    answer(sorry) ↔ ∀ N ≥ (2 : ℕ),  (IsSquare N) → ∃ᵉ (k ≥ 2) (n : ℕ) (m ≥ n + k),
      (N : ℚ) = (∏ i ∈ Finset.Icc 1 k, (m + i)) / (∏ i ∈ Finset.Icc 1 k, (n + i)) := by
…
```
theorem Erdos686.erdos_686.variants.four Can $4$ be written as $$4=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$? sorry line 61 source
Source documentation
Can $4$ be written as $$4=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$?

depends on sorry what this means
Statement uses: Iff, True, Exists, Nat, And, GE.ge, OfNat.ofNat, HAdd.hAdd, Eq, Rat, HDiv.hDiv, Finset.prod, Finset.Icc
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
24 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_686.variants.four :
    answer(sorry) ↔ ∃ᵉ (k ≥ 2) (n : ℕ) (m ≥ n + k),
      (4 : ℚ) = (∏ i ∈ Finset.Icc 1 k, (m + i)) / (∏ i ∈ Finset.Icc 1 k, (n + i)) := by
…
```
theorem Erdos686.erdos_686.variants.four_two The number $4$ cannot be written as $$4=\frac{\prod_{1\leq i\leq 2}(m+i)}{\prod_{1\leq i\leq 2}(n+i)}$$ for $m≥n+2$! line 72 source
Source documentation
The number $4$ cannot be written as $$4=\frac{\prod_{1\leq i\leq 2}(m+i)}{\prod_{1\leq i\leq 2}(n+i)}$$ for $m≥n+2$!

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Not, Exists, Nat, And, GE.ge, HAdd.hAdd, OfNat.ofNat, Eq, Rat, HDiv.hDiv, Finset.prod, Finset.Icc
Proof uses: Eq.mpr, HMul.hMul, id, congrArg, funext, congr, Eq.trans, CommMonoid, Finset.prod_Icc_succ_top, of_decide_eq_true, LE.le, Bool, Decidable.decide, Bool.true, Eq.refl, congrFun', Singleton.singleton, Finset, Finset.prod_congr, Finset.Icc_self and 97 more
167 instances, coercions and auxiliary constants left out.

Tactic trace: 12 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_686.variants.four_two :
    ¬ ∃ᵉ (n : ℕ) (m ≥ n + 2),
      (4 : ℚ) = (∏ i ∈ Finset.Icc 1 2, (m + i)) / (∏ i ∈ Finset.Icc 1 2, (n + i)) := by
…
```
theorem Erdos686.erdos_686.variants.four_three The number $4$ cannot be written as $$4=\frac{\prod_{1\leq i\leq 3}(m+i)}{\prod_{1\leq i\leq 3}(n+i)}$$ for $m≥n+3$! sorry line 93 source
Source documentation
The number $4$ cannot be written as $$4=\frac{\prod_{1\leq i\leq 3}(m+i)}{\prod_{1\leq i\leq 3}(n+i)}$$ for $m≥n+3$!
See comment section on erdosproblems.com

L4DEL explanation
The number $4$ cannot be written as $$4=\frac{\prod_{1\leq i\leq 3}(m+i)}{\prod_{1\leq i\leq 3}(n+i)}$$ for $m≥n+3$! See comment section on erdosproblems.com
Sources and attribution: Pinned Lean source (1646ca16afd6)

depends on sorry what this means
Statement uses: Not, Exists, Nat, And, GE.ge, HAdd.hAdd, OfNat.ofNat, Eq, Rat, HDiv.hDiv, Finset.prod, Finset.Icc
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
24 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_686.variants.four_three :
    ¬ ∃ᵉ (n : ℕ) (m ≥ n + 3),
      (4 : ℚ) = (∏ i ∈ Finset.Icc 1 3, (m + i)) / (∏ i ∈ Finset.Icc 1 3, (n + i)) := by
…
```
theorem Erdos686.erdos_686.variants.nine Can $9$ be written as $$9=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$? line 104 source
Source documentation
Can $9$ be written as $$9=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$?

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Iff, True, Exists, Nat, And, GE.ge, OfNat.ofNat, HAdd.hAdd, Eq, Rat, HDiv.hDiv, Finset.prod, Finset.Icc
Proof uses: Eq.mpr, id, true_iff, Exists.intro, And.intro, Mathlib.Meta.NormNum.isNat_le_true, Mathlib.Meta.NormNum.isNat_ofNat, Eq.refl, Bool, Bool.true, Mathlib.Meta.NormNum.isNat_add, of_eq_true, Eq.trans, congrArg, congr, HMul.hMul, Finset.prod_Icc_succ_top, Singleton.singleton, Finset, Finset.prod_congr and 16 more
62 instances, coercions and auxiliary constants left out.

Tactic trace: 5 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_686.variants.nine :
    answer(True) ↔ ∃ᵉ (k ≥ 2) (n : ℕ) (m ≥ n + k),
      (9 : ℚ) = (∏ i ∈ Finset.Icc 1 k, (m + i)) / (∏ i ∈ Finset.Icc 1 k, (n + i)) := by
…
```
theorem Erdos686.erdos_686.variants.twenty_five Can $25$ be written as $$25=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$? sorry line 118 source
Source documentation
Can $25$ be written as $$25=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$?

depends on sorry what this means
Statement uses: Iff, True, Exists, Nat, And, GE.ge, OfNat.ofNat, HAdd.hAdd, Eq, Rat, HDiv.hDiv, Finset.prod, Finset.Icc
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
24 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_686.variants.twenty_five :
    answer(sorry) ↔ ∃ᵉ (k ≥ 2) (n : ℕ) (m ≥ n + k),
      (25 : ℚ) = (∏ i ∈ Finset.Icc 1 k, (m + i)) / (∏ i ∈ Finset.Icc 1 k, (n + i)) := by
…
```
theorem Erdos686.erdos_686.variants.non_square Can every non-square $N≥2$ be written as $$N=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$? line 129 source
Source documentation
Can every non-square $N≥2$ be written as $$N=\frac{\prod_{1\leq i\leq k}(m+i)}{\prod_{1\leq i\leq k}(n+i)}$$ for some $k≥2$ and $m≥n+k$?

extra axioms: _private.0.Erdos686.erdos_686.variants.non_square._native.native_decide.ax_7, _private.0.Erdos686.erdos_686.variants.non_square._native.native_decide.ax_8 what this means
Statement uses: Iff, True, Nat, GE.ge, OfNat.ofNat, Not, IsSquare, Exists, And, HAdd.hAdd, Eq, Rat, HDiv.hDiv, Finset.prod, Finset.Icc
Proof uses: Iff.intro, HMul.hMul, False, Exists.intro, Eq.symm, And.intro, le_rfl, id, Eq.mpr, LE.le, congrArg, funext, congr, Eq.trans, Finset.prod_Icc_succ_top, of_eq_true, congrFun', Singleton.singleton, Finset, Finset.prod_congr and 208 more
255 instances, coercions and auxiliary constants left out.

Tactic trace: 49 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_686.variants.non_square :
    answer(True) ↔ ∀ N ≥ (2 : ℕ), (¬ IsSquare N) → ∃ᵉ (k ≥ 2) (n : ℕ) (m ≥ n + k),
      (N : ℚ) = (∏ i ∈ Finset.Icc 1 k, (m + i)) / (∏ i ∈ Finset.Icc 1 k, (n + i)) := by
…
```