17560.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

3 declarations, 3 contain sorry, 0 declared as axioms

Mathoverflow 17560

Reference: mathoverflow/17560 asked by user Alon-Amit

Imports (1)

FormalConjecturesUtil

theorem Mathoverflow17560.mathoverflow_17560 If $2^x$ and $3^x$ are integers, then $x$ must be an integer sorry line 35 source
Source documentation
If $2^x$ and $3^x$ are integers, then $x$ must be an integer.

depends on sorry what this means
Statement uses: Real, Exists, Nat, Eq, HPow.hPow, OfNat.ofNat
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
16 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem mathoverflow_17560 {x : ℝ} (hx : ∃ m : ℕ, (2 : ℝ) ^ x = m) (hx' : ∃ m : ℕ, (3 : ℝ) ^ x = m) :
    ∃ m : ℕ, x = m := by
…
```
theorem Mathoverflow17560.mathoverflow_17560.variants.all_nats If for each natural number $n$ the number $n^x$ is an integer then $x$ must also be an integer sorry line 43 source
Source documentation
If for each natural number $n$ the number $n^x$ is an integer then $x$ must also be an integer.

depends on sorry what this means
Statement uses: Real, Nat, Exists, Eq, HPow.hPow
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous, OfNat.ofNat
9 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem mathoverflow_17560.variants.all_nats {x : ℝ} (hx : ∀ n : ℕ, ∃ m : ℕ, (n : ℝ) ^ x = m) :
    ∃ m : ℕ, x = m := by
…
```
theorem Mathoverflow17560.mathoverflow_17560.variants.with_5 If $2^x$, $3^x$ and $5^x$ are integers, then $x$ must be an integer sorry line 51 source
Source documentation
If $2^x$, $3^x$ and $5^x$ are integers, then $x$ must be an integer.

depends on sorry what this means
Statement uses: Real, Exists, Nat, Eq, HPow.hPow, OfNat.ofNat
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
16 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem mathoverflow_17560.variants.with_5 {x : ℝ} (hx : ∃ m : ℕ, (2 : ℝ) ^ x = m)
    (hx' : ∃ m : ℕ, (3 : ℝ) ^ x = m) (hx'' : ∃ m : ℕ, (5 : ℝ) ^ x = m) : ∃ m : ℕ, x = m := by
…
```