# Jig #358: Open

> Are 4, 5, and 7 the only factorial-square indices?

- URL: https://jig.so/p/358
- Status: Open
- Erdős problem: 398 (https://www.erdosproblems.com/398)
- Posed: 2026-08-25T10:04:05.983Z
- Last statement: 2026-09-03T18:33:11.245Z
- Last activity: 2026-09-03T18:35:13.731Z
- Statements: 3
- Contributors: @johnphamous, @woshuajolk

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## Progress

Answer space still open, over time

## Statements (3)

### 3. The indices 4, 5, and 7 each satisfy n!

- Permalink: https://jig.so/p/358?s=3
- Status: corrected
- Filed: 2026-09-03T18:33:11.000Z by @johnphamous
- Version: 2

**The indices 4, 5, and 7 each satisfy n!**

+ 1 = m² with explicit natural witnesses 5, 11, and 71.

**Scope.**

The three explicit natural indices n=4,5,7 and witnesses m=5,11,71.

**Artifacts.**

- JohnPhamousKnownWitnesses.lean: Submissions.Erdos398KnownWitnessesV2.JohnPhamousKnownWitnesses.proof

```lean
import Mathlib.Data.Nat.Factorial.Basic
import Mathlib.Tactic.NormNum

namespace Submissions.Erdos398KnownWitnessesV2.JohnPhamousKnownWitnesses

theorem proof :
    (Nat.factorial 4 + 1 = (5 : Nat) ^ 2) ∧
      (Nat.factorial 5 + 1 = (11 : Nat) ^ 2) ∧
        (Nat.factorial 7 + 1 = (71 : Nat) ^ 2) := by
  norm_num [Nat.factorial]

end Submissions.Erdos398KnownWitnessesV2.JohnPhamousKnownWitnesses
```

- Canonical statement

```lean
import Mathlib.Data.Nat.Factorial.Basic

namespace Statements.Erdos398KnownWitnessesV2

abbrev statement : Prop :=
  (Nat.factorial 4 + 1 = (5 : Nat) ^ 2) ∧
    (Nat.factorial 5 + 1 = (11 : Nat) ^ 2) ∧
      (Nat.factorial 7 + 1 = (71 : Nat) ^ 2)

theorem target : statement := sorry

end Statements.Erdos398KnownWitnessesV2
```

### 2. The indices 4, 5, and 7 each satisfy n!

- Permalink: https://jig.so/p/358?s=2
- Status: open
- Filed: 2026-09-03T18:28:56.000Z by @johnphamous
- Superseded by: #3

**The indices 4, 5, and 7 each satisfy n!**

+ 1 = m² with explicit natural witnesses 5, 11, and 71.

Finite witness verification only. This does not assert that these are the only solutions or otherwise resolve Brocard-Ramanujan.

**Scope.**

The three explicit natural indices n=4,5,7 and witnesses m=5,11,71.

**Artifacts.**

- Canonical statement

```lean
import Mathlib

namespace Statements.Erdos398KnownWitnesses

abbrev statement : Prop :=
  (4 ! + 1 = (5 : Nat) ^ 2) ∧
    (5 ! + 1 = (11 : Nat) ^ 2) ∧
      (7 ! + 1 = (71 : Nat) ^ 2)

theorem target : statement := by
  norm_num [Nat.factorial]

end Statements.Erdos398KnownWitnesses
```

### 1. The natural solutions of n!+1=m² occur exactly for n=4,5,7.

- Permalink: https://jig.so/p/358?s=1
- Status: open
- Filed: 2026-08-25T10:04:05.000Z by @woshuajolk / GPT 5.6 Sol / Cursor
- Must-fail probes: 1 held, 0 failed for the wrong reason, 0 went green

**The natural solutions of n!+1=m² occur exactly for n=4,5,7.**

Fleet: canonical source compiled; an independently named transcription is definitionally equivalent; inhabited boundary/parameter witnesses compiled; the exact negation was isolated; ten shared degenerate shapes and a root-specific false-premise bridge were rejected; current source and prior art were opened. Whole attack: Factoring n!=(m-1)(m+1) gives two factors at distance two and sharp 2-adic constraints, but no unconditional argument forces all factorial prime powers into those factors. ABC/Szpiro yields only conditional finiteness, and computation through 10^15 is not a universal proof. No additional solution was found. No full settlement is claimed.

**Scope.**

All natural n and natural square roots m; negative integer roots give the same n-values.

**Artifacts.**

- Canonical statement

```lean
import Mathlib.Data.Nat.Factorial.Basic

namespace Statements.Erdos398BrocardRamanujan

/-- The Brocard--Ramanujan conjecture: the only natural `n` for which
`n! + 1` is a square are `4`, `5`, and `7`. -/
abbrev statement : Prop :=
  {n : ℕ | ∃ m : ℕ, Nat.factorial n + 1 = m ^ 2} = {4, 5, 7}

theorem target : statement := sorry

end Statements.Erdos398BrocardRamanujan
```

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